536 PUZZLES CURIOUS PROBLEMS

536 PUZZLES &CURIOUS PROBLEMS

Arithmetic & Algebraic Problems
Arithmetic & Algebraic Problems

1. CONCERNING A CHECKA man went into a bank to cash a check. In handing over the money thecashier, by mistake, gave him dollars for cents and cents for dollars. Hepocketed the money without examining it, and spent a nickel on his way home.He then found that he possessed exactly twice the amount of the check. Hehad no money in his pocket before going to the bank. What was the exactamount of that check?

2. DOLLARS AND CENTSA man entered a store and spent one-half ofthe money that was in his pocket.When he came out he found that he had just as many cents as he had dollarswhen he went in and half as many dollars as he had cents when he went in.How much money did he have on him when he entered?

3. LOOSE CASHWhat is the largest sum of money-all in current coins and no silverdollars-that I could have in my pocket without being able to give changefor a dollar, half dollar, quarter, dime, or nickel?

4. GENEROUS GIFTSA generous man set aside a certain sum of money for equal distributionweekly to the needy of his acquaintance. One day he remarked, "If there arefive fewer applicants next week, you will each receive two dollars more." Unfortunately, instead of there being fewer there were actually four more persons applying for the gift."This means," he pointed out, "that you will each receive one dollar less."How much did each person receive at that last distribution?

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5. BUYING BUNSBuns were being sold at three prices: one for a penny, two for a penny,and three for a penny. Some children (there were as many boys as girls) weregiven seven pennies to spend on these buns, each child to receive exactly thesame value in buns. Assuming that all buns remained whole, how manybuns, and of what types, did each child receive?

6. UNREWARDED LABORA man persuaded Weary Willie, with some difficulty, to try to work on ajob for thirty days at eight dollars a day, on the condition that he would forfeit ten dollars a day for every day that he idled. At the end of the monthneither owed the other anything, which entirely convinced Willie of the follyof labor. Can you tell just how many days' work he put in and on how manydays he idled?

7. THE PERPLEXED BANKERA man went into a bank with a thousand dollars, all in dollar bills, and tenbags. He said, "Place this money, please, in the bags in such a way that if Icall and ask for a certain number of dollars you can hand me over oneor more bags, giving me the exact amount called for without opening any ofthe bags."How was it to be done? We are, of course, only concerned with a singleapplication, but he may ask for any exact number of dollars from one to onethousand.

8. A WEIRD GAMESeven men engaged in play. Whenever a player won a game he doubledthe money of each of the other players. That is, he gave each player just asmuch money as each had in his pocket. They played seven games and,strange to say, each won a game in turn in the order of their names, whichbegan with the letters A, B, C, D, E, F, and G.Money Puzzles 5When they had finished it was found that each man had exactly $1.28 in hispocket. How much had each man in his pocket before play?

9. DIGGING A DITCHHere is a curious question that is more perplexing than it looks at first sight.Abraham, an infirm old man, undertook to dig a ditch for two dollars. Heengaged Benjamin, an able-bodied fellow, to assist him and share the moneyfairly according to their capacities. Abraham could dig as fast as Benjamincould shovel out the dirt, and Benjamin could dig four times as fast as Abraham could do the shoveling.How should they divide the money? Of course, we must assume their relative abilities for work to be the same in digging or shoveling.

10. NAME THEIR WIVESA man left a legacy of $1 ,000.00 to three relatives and their wives. The wivesreceived together $396.00. Jane received $10.00 more than Catherine, andMary received $10.00 more than Jane. John Smith was given just as much ashis wife, Henry Snooks got half as much again as his wife, and Tom Crowereceived twice as much as his wife. What was the Christian name of each man'swife?II. MARKET TRANSACTIONSA farmer goes to market and buys a hundred animals at a total costof $1,000.00. The price of cows being $50.00 each, sheep $10.00 each,and rabbits 50¢ each, how many of each kind does he buy? Most people willsolve this, if they succeed at all, by more or less laborious trial, but there areseveral direct ways of getting the solution.

12. THE SEVEN APPLEWOMENHere is an old puzzle that people are frequently writing to me about.Seven applewomen, possessing respectively 20, 40, 60, 80, 100, 120, and 140

6 Arithmetic & Algebraic Problemsapples, went to market and sold all theu apples at the same price, and eachreceived the same sum of money. What was the price?

13. A LEGACY PUZZLEA man left legacies to his three sons and to a hospital, amounting in all to$1,320.00. If he had left the hospital legacy also to his first son, that sonwould have received as much as the other two sons together. If he had left itto his second son, he would have received twice as much as the other two sonstogether. If he had left the hospital legacy to his third son, he would have received then thrice as much as the first son and second son together. Find theamount of each legacy.

14. PUZZLING LEGACIESA man bequeathed a sum of money, a little less than $1,500.00, to bedivided as follows: The five children and the lawyer received such sums thatthe square root of the eldest son's share, the second son's share dividedby two, the third son's share minus $2.00, the fourth son's share plus $2.00,the daughter'S share multiplied by two, and the square of the lawyer's fee allworked out at exactly the same sum of money. No dollars were divided, andno money was left over after the division. What was the total amountbequeathed?

15. DIVIDING THE LEGACYA man left $100.00 to be divided between his two sons Alfred andBenjamin. If one-third of Alfred's legacy be taken from one-fourth of Benjamin's, the remainder would be $11.00. What was the amount of eachlegacy?

16. A NEW PARTNERTwo partners named Smugg and Williamson have decided to take a Mr.Rogers into partnership. Smugg has 116 times as much capital invested in thebusiness as Williamson, and Rogers has to pay down $2,500.00, which sumshall be divided between Smugg and Williamson, so that the three partnersshall have an equal interest in the business. How shall the sum be divided?Money Puzzles 7

17. POCKET MONEY"When I got to the station this morning," said Harold Tompkins, athis club, "I found I was short of cash. I spent just one-half of what I had onmy railway ticket, and then bought a nickel's worth of candy. When I got tothe terminus I spent half of what I had left and ten cents for a newspaper. Then I spent half of the remainder on a bus and gave fifteen cents tothat old beggar outside the club. Consequently I arrive here with this singlenickel. How much did I start out with?"

18. DISTRIBUTIONNine persons in a party, A, B, C, D, E, F, G, H, K, did as follows: First Agave each of the others as much money as he (the receiver) alreadyheld; then B did the same; then C; and so on to the last, K giving to each ofthe other eight persons the amount the receiver then held. Then it was foundthat each of the nine persons held the same amount.Can you find the smallest amount in cents that each person couldhave originally held?

19. REDUCTIONS IN PRICE"I have often been mystified," said Colonel Crackham, "at the startling reductions some people make in their prices, and wondered on what principlethey went to work. For example, a man offered me a motorcycle two yearsago for $1,024.00; a year later his price was $640.00; a little while after heasked a level $400.00; and last week he was willing to sell for $250.00. Thenext time he reduces I shall buy. At what price shall I purchase if he makesa consistent reduction?"

20. HORSES AND BULLOCKSA dealer bought a number of horses at $344.00 each, and a numberof bullocks at $265.00 each. He then discovered that the horses had cost himin all $33.00 more than the bullocks. Now, what is the smallest numberof each that he must have bought?

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21. BUYING TURKEYSA man bought a number of turkeys at a cost of $60.00, and after reservingfifteen of the birds he sold the remainder for $54.00, thus gaining 1O¢ a headby these. How many turkeys did he buy?

22. THE THRIFTY GROCERA grocer in a small business had managed to put aside (apart fromhis legitimate profits) a little sum in dollar bills, half dollars, and quarters,which he kept in eight bags, there being the same number of dollar bi1ls andof each kind of coin in every bag. One night he decided to put the money intoonly seven bags, again with the same number of each kind of currency inevery bag. And the following night he further reduced the number of bags tosix, again putting the same number of each kind of currency in every bag.The next night the poor demented miser tried to do the same withfive bags, but after hours of trial he utterly failed, had a fit, and died, greatlyrespected by his neighbors. What is the smallest possible amount of moneyhe had put aside?

23. THE MISSING PENNYHere is an ancient puzzle that has always perplexed some people. Twomarket women were selling their apples, one at three for a penny andthe other at two for a penny. One day they were both called away when eachhad thirty apples unsold: these they handed to a friend to sell at five for 2¢.It will be seen that if they had sold their apples separately they would havefetched 25¢, but when they were sold together they fetched only 24¢."Now," people ask, "what in the world has become of that missingpenny?" because, it is said, three for l¢ and two for l¢ is surely exactly thesame as five for 2¢.Can you explain the little mystery?

24. THE RED DEATH LEAGUEThe police, when making a raid on the headquarters of a secret society,secured a scrap of paper similar to the one pictured."That piece of paper," said thedetective, throwing it on the table,"has worried me for two or three days.You see it gives the total of the subscriptions for the present year as$3,007.37, but the number of members(I know it is under 500) and theamount of the subscription have beenobliterated. How many members werethere in the Red Death League, andMoney Puzzles 9THE "ED~ DEATH LEAGUEwhat was the uniform subscription?"Of course, no fraction of a cent ispermitted.

25. A POULTRY POSERThree chickens and one duck sold for as much as two geese; one chicken,two ducks, and three geese were sold together for $25.00. What was the priceof each bird in an exact number of dollars?

26. BOYS AND GIRLSNine boys and three girls agreed to share equally their pocket money.Every boy gave an equal sum to every girl, and every girl gave another equalsum to every boy. Every child then possessed exactly the same amount. Whatwas the smallest possible amount that each then possessed?

27. THE COST OF A SUIT"Hello, old chap," cried Russell as Henry Melville came into the clubarrayed in a startling new tweed suit, "have you been successful in the cardroom lately? No? Then why these fine feathers?""Oh, I just dropped into my tailor's the other day," he explained, "and thiscloth took my fancy. Here is a little puzzle for you. The coat cost as much asthe trousers and vest. The coat and two pairs of trousers would cost $175.00.The trousers and two vests would cost $100.00. Can you tell me the cost ofthe suit?"

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28. A QUEER SETILING UPProfessor Rackbrane told his family at the breakfast table that he hadheard the following conversation in a railway carriage the night before.One passenger said to another, "Here is my purse: give me just as muchmoney, Richard, as you find in it."Richard counted the money, added an equal value from his own pocket,and replied, "Now, John, if you give me as much as I have left of my own weshall be square."John did so, and then stated that his own purse contained $3.50, whileRichard said that he now had $3.00. How much did each man possessat first?

29. APPLE TRANSACTIONSA man was asked what price per 100 he paid for some apples, and his reply was as follows: "If they had been 4¢ more per 100 I should have got fiveless for $1.20." Can you say what was the price per lOO?

30. PROSPEROUS BUSINESSA man started business with a capital of $2,000.00, and increased hiswealth by 50 per cent every three years. How much did he possess at the expiration of eighteen years?

31. THE BANKER AND THE COUNTERFEIT BILLA banker in a country town was walking down the street when he saw afive-dollar bill on the curb. He picked it up, noted the number, and went tohis home for luncheon. His wife said that the butcher had sent in his bill forfive dollars, and, as the only money he had was the bill he had found,he gave it to her, and she paid the butcher. The butcher paid it to a farmerin buying a calf, the farmer paid it to a merchant who in turn paid itto a laundry woman, and she, remembering that she owed the bank five dollars, went there and paid the debt.The banker recognized the bill as the one he had found, and by that time ithad paid twenty-five dollars worth of debts. On careful examination he dis-Age Puzzles 11covered that the bill was counterfeit. What was lost in the whole transaction,and by whom?

32. THEIR AGESIf you add the square of Tom's age to the age of Mary, the sum is 62; butif you add the square of Mary's age to the age of Tom, the result is 176. Canyou say what are the ages of Tom and Mary?

33. MRS. WILSON'S FAMILYMrs. Wilson had three children: Edgar, James, and John. Their combinedages were half of hers. Five years later, during which time Ethel was born,Mrs. Wilson's age equalled the total of all her children's ages. Ten years morehave now passed, Daisy appearing during that interval. At the latter eventEdgar was as old as John and Ethel together. The combined ages of all thechildren are now double Mrs. Wilson's age, which is, in fact, only equal tothat of Edgar and James together. Edgar's age also equals that of thetwo daughters.Can you find all their ages?

34. DE MORGAN AND ANOTHERAugustus De Morgan, the mathematician, who died in 1871, used to boastthat he was x years old in the year x 2• Jasper Jenkins, wishing to improve onthis, told me in 1925 that he was a2 + b2 in a4 + b4 ; that he was 2m in theyear 2m2; and that he was 3n years old in the year 3n4 • Can you givethe years in which De Morgan and Jenkins were respectively born?

35. "SIMPLE" ARITHMETICWhen visiting an insane asylum, I asked two inmates to give me their ages.They did so, and then, to test their arithmetical powers, I asked them to addthe two ages together. One gave me 44 as the answer, and the othergave 1,280. I immediately saw that the first had subtracted one age from theother, while the second person had multiplied them together. What were theirages?

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36. ANCIENT PROBLEMHere is an example of the sort of "Breakfast Problem" propounded byMetrodorus in 310 A.D.Demochares has lived one-fourth of his life as a boy, one-fifth as a youth,one-third as a man, and has spent thirteen years in his dotage. How old isthe gentleman?

37. FAMILY AGESA man and his wife had three children, John, Ben, and Mary, and the difference between their parents' ages was the same as between John and Benand between Ben and Mary. The ages of John and Ben, multiplied together,equalled the age of the father, and the ages of Ben and Mary multiplied together equalled the age of the mother. The combined ages of the familyamounted to ninety years. What was the age of each person?

38. MIKE'S AGE"Pat O'Connor," said Colonel Crackham, "is now just one and one-thirdtimes as old as he was when he built the pig sty under his drawing-roomwindow. Little Mike, who was forty months old when Pat built the sty,is now two years more than half as old as Pat's wife, Biddy, was when Patbuilt the sty, so that when little Mike is as old as Pat was when he built thesty, their three ages combined will amount to just one hundred years. Howold is little Mike?"

39. THEIR AGESRackbrane said the other morning that a man on being asked the ages ofhis two sons stated that eighteen more than the sum of their ages is doublethe age of the elder, and six less than the difference of their ages is the age ofthe younger. What are their ages?

40. BROTHER AND SISTERA boy on being asked the age of himself and of his sister replied:"Three years ago I was seven times as old as my sister; two years ago I wasAge Puzzles 13four times as old; last year I was three times as old; and this year I am twoand one-half times as old."What are their ages?

41. A SQUARE FAMILYA man had nine children, all born at regular intervals, and the sum of thesquares of their ages was equal to the square of his own. What was the age ofeach? Every age was an exact number of years.

42. IN THE YEAR 1900A correspondent, in 1930, proposed the following question. The readermay think, at first sight, that there is insufficient data for an answer, but hewill be wrong:A man's age at death was one twenty-ninth of the year of his birth. Howold was he in the year 1900?

43. FINDING A BIRTHDAYA correspondent informs us that on Armistice Day (November 11, 1928)he had lived as long in the twentieth century as he had lived in thenineteenth. This tempted us to work out the day of his birth. Perhapsthe reader may like to do the same. We will assume he was born at midday.

44. THE BIRTH OF BOADICEAA correspondent (R. D.) proposes the following little puzzle:Boadicea died one hundred and twenty-nine years after Cleopatra wasborn. Their united ages (that is, the combined years of their complete lives)were one hundred years. Cleopatra died 30 B.C. When was Boadicea born?

45. ROBINSON'S AGE"How old are you, Robinson?" asked Colonel Crackham one morning."Well, I forget exactly," was the reply; "but my brother is two years olderthan I; my sister is four years older than he; my mother was twenty when I

14 Arithmetic & Algebraic Problemswas born; and I was told yesterday that the average age of the four of us isthirty-nine years."What was Robinson's age?

46. A DREAMLAND CLOCKIn a dream, I was travelling in a country where they had strange ways ofdoing things. One little incident was fresh in my memory when I awakened. Isaw a clock and announced the time as it appeared to be indicated, butmy guide corrected me.He said, "You are apparently not aware that the minute hand always movesin the opposite direction to the hour hand. Except for this improvement, ourclocks are precisely the same as. those you have been accustomed to."Since the hands were exactly together between the hours of four and fiveoclock, and they started together at noon, what was the real time?

47. WHAT IS THE TIME?At what time are the two hands of a clock so situated that, reckoningas minute points past XII, one is exactly the square of the distance of the other?

48. THE AMBIGUOUS CLOCKA man had a clock with an hour hand and minute hand of the same lengthand indistinguishable. If it was set going at noon, what would be the first timethat it would be impossible, by reason of the similarity of the hands, to besure of the correct time?Readers will remember that with these clock puzzles there is the conventionthat we may assume it possible to indicate fractions of seconds. On thisassumption an exact answer can be given.

49. THE BROKEN CLOCK FACEColonel Crackham asked his family at the breakfast table if, without having a dial before them, they could correctly draw in Roman numerals thehours round a clock face. George fell into the trap that catches so manypeople, of writing the fourth hour as IV, instead of I1II.Colonel Crackham then asked themto show how a dial may be broken intofour parts so that the numerals oneach part shall in every case sum to 20.As an example he gave our illustration, where it will be found that theseparated numerals on two parts sumto 20, but on the other parts they addup to 19 and 21 respectively, so it fails.Clock Puzzles 15so. WHEN DID THE DANCING BEGIN?"The guests at that ball the other night," said Dora at the breakfast table,"thought that the clock had stopped, because the hands appeared in exactlythe same position as when the dancing began. But it was found that they hadreally only changed places. As you know, the dancing commenced betweenten and eleven oclock. What was the exact time of the start?"

51. MISTAKING THE HANDS"Between two and three oclock yesterday," said Colonel Crackham, "Ilooked at the clock and mistook the minute hand for the hour hand, andconsequently the time appeared to be fifty-five minutes earlier than it actuallywas. What was the correct time?"

52. EQUAL DISTANCESA few mornings ago the following clock puzzle was sprung on his pupils byProfessor Rackbrane. At what time between three and four oclock is theminute hand the same distance from VIII as ·the hour hand is from XII?

53. RIGHT AND LEFTAt what time between three and four oclock will the minute hand be as farfrom twelve on the left side of the dial plate as the hour hand is from twelveon the right side of the dial plate?

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54. AT RIGHT ANGLESRackbrane asked his young friends at the breakfast table one morning thislittle question:"How soon between the hours of five and six will the hour and minutehands of a clock be exactly at right angles?"

55. WESTMINSTER CLOCKA man crossed over Westminster Bridge one morning between eight andnine oclock by the tower clock (often mistakenly called Big Ben, which is thename of the large bell only, but this by the way). On his return between fourand five oclock he noticed that the hands were exactly reversed. What werethe exact times that he made the two crossings?

56. HILL CLIMBINGWeary Willie went up a certain hill at the rate of one and a half miles perhour and came down at the rate of four and a half miles per hour, so that ittook him just six hours to make the double journey. How far was it to the topof the hill?

57. TIMING THE CAR"I was walking along the road at three and a half miles an hour," said Mr.Pipkins, "when the car dashed past me and only missed me by a few inches.""Do you know at what speed it was going?" asked his friend."Well, from the moment it passed me to its disappearance round a cornertook twenty-seven steps and walking on reached that corner with onehundred and thirty-five steps more.""Then, assuming that you walked, and the car ran, each at a uniform rate,we can easily work out the speed."

58. THE STAIRCASE RACEThis is a rough sketch of the finish of a race up a staircase in which threemen took part. Ackworth, who is leading, went up three steps at a time, asarranged; Barnden, the second man, went four steps at a time, and Croft, whoSpeed & Distance Puzzles 17is last, went five at a time. Undoubtedly Ackworth wins. But the point is, howmany steps are there in the stairs, counting the top landing as a step?I have only shown the top of the stairs. There may be scores, or hundreds,of steps below the line. It was not necessary to draw them, as I only wantedto show the finish. But it is possible to tell from the evidence the fewest possible steps in that staircase. Can you do it?

59. A WALKING PUZZLEA man set out at noon to walk from Appleminster to Boneyham, anda friend of his started at two P.M. on the same day to walk from Boneyhamto Appleminster. They met on the road at five minutes past four oclock, andeach man reached his destination at exactly the same time. Can you sayat what time they both arrived?

60. RIDING IN THE WINDA man on a bicycle rode a mile in three minutes with the wind at his back,but it took him four minutes to return against the wind. How long would ittake him to ride a mile if there was no wind? Some will say that the average

18 Arithmetic & Algebraic Problemsof three and four is three and one-half, and it would take him three and onehalf minutes. That answer is entirely wrong.

61. A ROWING PUZZLEA crew can row a certain course upstream in eight and four-seventhsminutes, and, if there were no stream, they could row it in seven minutes lessthan it takes them to drift down the stream. How long would it take torow down with the stream?

62. THE ESCALATOROn one of the escalators on the London subway I find that if I walk downtwenty-six steps I require thirty seconds to get to the bottom, but if I makethirty-four steps I require only eighteen seconds to reach the bottom. What isthe height of the stairway in steps? The time is measured from the momentthe top step begins to descend to the time I step off the last step at the bottomonto the level platform.

63. SHARING A BICYCLETwo brothers had to go on a journey and arrive at the same time. They hadonly a single bicycle, which they rode in turns, each rider leaving it in thehedge when he dismounted for the one walking behind to pick up, and walking ahead himself, to be again overtaken. What was their best way of arranging their distances? As their walking and riding speeds were the same, itis extremely easy. Simply divide the route into any even number of equal stagesand drop the bicycle at every stage, using the cyclometer. Each man wouldthen walk half way and ride half way.But here is a case that will require a little more thought. Anderson and Brownhave to go twenty miles and arrive at exactly the same time. They have onlyone bicycle. Anderson can only walk four miles an hour, while Brown canwalk five miles an hour, but Anderson can ride ten miles an hour to Brown'seight miles an hour.How are they to arrange the journey? Each man always either walks orrides at the speeds mentioned, without any rests.Speed & Distance Puzzles 19

64. MORE BICYCLINGReferring to the last puzzle, let us now consider the case where a third riderhas to share the same bicycle. As a matter of fact, I understand that Andersonand Brown have taken a man named Carter into partnership, and the positiontoday is this: Anderson, Brown, and Carter walk respectively four, five, andthree miles per hour, and ride respectively ten, eight, and twelve miles perhour. How are they to use that single bicycle so that all shall completethe twenty miles' journey at the same time?

65. A SIDECAR PROBLEMAtkins, Baldwin, and Clarke had to go on a journey of fifty-two milesacross country. Atkins had a motorcycle with a sidecar for one passenger.How was he to take one of his companions a certain distance, drop him on theroad to walk the remainder of the way, and return to pick up the second friend,who, starting at the same time, was already walking on the road, so that theyshould all arrive at their destination at exactly the same time?The motorcycle could do twenty miles an hour, Baldwin could walk fivemiles an hour, and Clarke could walk four miles an hour. Of course, each wentat his proper speed throughout and there was no waiting.I might have complicated the problem by giving more passengers, butI have purposely made it easy, and all the distances are an exact number ofmiles-without fractions.

66. THE DISPATCH RIDERIf an army forty miles long advances forty miles while a dispatch ridergallops from the rear to the front, delivers a dispatch to the commandinggeneral, and returns to the rear, how far has he to travel?

67. THE TWO TRAINSTwo railway trains, one four hundred feet long and the other two hundredfeet long, ran on parallel rails. It was found that when they went in oppositedirections they passed each other in five seconds, but when they ran in the samedirection the faster train would pass the other in fifteen seconds. A curious

20 Arithmetic & Algebraic Problemspassenger worked out from these facts the rate per hour at which each trainran.Can the reader discover the correct answer? Of course, each train ran witha uniform velocity.

68. PICKLEMINSTER TO QUICKVILLETwo trains, A and B, leave Pickleminster for Quickville at the same time astwo trains, C and D, leave Quickville for Pickleminster. A passes C 120 milesfrom Pickleminster and D 140 miles from Pickleminster. B passes C 126 milesfrom Quickville and D half way between Pickleminster and Quickville. Now,what is the distance from Pickleminster to Quickville? Every train runs uniformly at an ordinary rate.

69. THE DAMAGED ENGINEWe were going by train from Anglechester to Clinkerton, and an hourafter starting an accident happened to the engine.We had to continue the journey at three-fifths of the former speed. It madeus two hours late at Clinkerton, and the driver said that if only the accidenthad happened fifty miles farther on the train would have arrived forty minutessooner. Can you tell from that statement just how far it is from Anglechesterto Clinkerton?

70. THE PUZZLE OF THE RUNNERSTwo men ran a race round a circular course, going in opposite directions.Brown was the best runner and gave Tompkins a start of one-eighth of thedistance. But Brown, with a contempt for his opponent, took things too easilyat the beginning, and when he had run one-sixth of his distance he metTompkins, and saw that his chance of winning the race was very small.How much faster than he went before must Brown now run in order to tiewith his competitor? The puzzle is quite easy when once you have grasped itssimple conditions.

71. THE TWO SHIPSA correspondent asks the following question. Two ships sail from one portto another-two hundred nautical miles-and return. The Mary Jane travelsSpeed & Distance Puzzles 21outwards at twelve miles an hour and returns at eight miles an hour, thustaking forty-one and two-third hours for the double journey. The ElizabethAnn travels both ways at ten miles an hour, taking forty hours on the doublejourney.Seeing that both ships travel at the average speed often miles per hour, whydoes the Mary Jane take longer than the Elizabeth Ann? Perhaps the readercould explain this little paradox.

72. FIND THE DISTANCEA man named Jones set out to walk from A to B , andon the road he met his friend Kenward, ten miles from A , who hadleft B at exactly the same time. Jones executed his commissionat B and, without delay, set out on his return journey, while Kenwardas promptly returned from A to B . They met twelve milesfrom B . Of course, each walked at a uniform rate throughout. Howfar is A from B ?I will show the reader a simple rule by which the distance may be foundby anyone in a few seconds without the use of a pencil. In fact, it is quiteabsurdly easy-when you know how to do it.

73. THE MAN AND THE DOG"Yes, when I take my dog for a walk," said a mathematical friend, "he frequently supplies me with some interesting puzzle to solve. One day, forexample, he waited, as I left the door, to see which way I should go, and whenI started he raced along to the end of the road, immediately returning to me;again racing to the end of the road and again returning. He did this four timesin all, at a uniform speed, and then ran at my side the remaining distance,which according to my paces measured 27 yards. I afterwards measured thedistance from my door to the end of the road and found it to be 625 feet. Now,ifl walk 4 miles per hour, what is the speed of my dog when racing to and fro?"

74. BAXTER'S DOGThis is an interesting companion to the "Man and Dog" puzzle. Andersonset off from an hotel at San Remo at nine oc1ock and had been walking an

22 Arithmetic & Algebraic Problemshour when Baxter went after him along the same road. Baxter's dog startedat the same time as his master and ran uniformly forwards and backwardsbetween him and Anderson until the two men were together. Anderson'sspeed is two, Baxter's four, and the dog's ten miles an hour. How far had thedog run when Baxter overtook Anderson?My correspondent in Italy who sends me this is an exact man, and he says,"Neglect length of dog and time spent in turning." I will merely add, neglectalso the dog's name and the day of the month.

75. THE RUNNER'S REFRESHMENTA man runs n times round a circular track whose radius is t miles. He drinkss quarts of beer for every mile that he runs. Prove that he will only need onequart!

76. EXPLORING THE DESERTNine travellers, each possessing a car, meet on the eastern edge of a desert.They wish to explore the interior, always going due west. Each car can travelforty miles on the contents of the engine tank, which holds a gallon of fuel,and each can carry nine extra gallon cans of fuel and no more. Unopenedcans can alone be transferred from car to car. What is the greatest distanceat which they can enter the desert without making any depots of fuel for thereturn journey?

77. EXPLORING MOUNT NEVERESTProfessor Walkingholme, one of the exploring party, was allotted the special task of making a complete circuit of the base of the mountain at a certainlevel. The circuit was exactly a hundred miles in length and he had to do itall alone on foot. He could walk twenty miles a day, but he could only carryrations for two days at a time, the rations for each day being packed in sealedboxes for convenience in dumping. He walked his full twenty miles every dayand consumed one day's ration as he walked. What is the shortest timein which he could complete the circuit?This simple question will be found to form one of the most fascinatingSpeed & Distance Puzzles 23puzzles that we have considered for some time. It made a considerable demandon Professor Walkingholme's well-known ingenuity. The idea was suggestedto me by Mr. H. F. Heath.

78. THE BATH CHAIRA correspondent informs us that a friend's house at A, where he was invitedto lunch at 1 P.M., is a mile from his own house at B. He is an invalid, and at

12 noon started in his Bath chair from B towards C. His friend, who hadarranged to join him and help push back, left A at 12: 15 P.M., walking at fivemiles per hour towards C. He joined him, and with his help they went backat four miles per hour, and arrived at A at exactly 1 P.M. How far did ourcorrespondent go towards C?-----c

79. THE PEDESTRIAN PASSENGERA train is travelling at the rate of sixty miles per hour. A passenger at theback of the train wishes to walk to the front along the corridor and in doingso walks at the rate of three miles per hour. At what rate is the man travellingover the permanent way? We will not involve ourselves here in quibbles anddifficulties similar to Zeno's paradox of the arrow and Einstein's theory ofrelativity, but deal with the matter in the simple sense of motion in referenceto the permanent way.SO. MEETING TRAINSAt Wurzletown Junction an old lady put her head out of the window andshouted:"Guard! how long will the journey be from here to Mudville?"

24 Arithmetic & Algebraic Problems"All the trains take five hours, ma'am, either way," replied the official."And how many trains shall I meet on the way?"This absurd question tickled the guard, but he was ready with his reply:"A train leaves Wurzletown for Mudville, and also one from Mudvilleto Wurzletown, at five minutes past every hour. Right away!"The old lady induced one of her fellow passengers to work out the answerfor her. What is the correct number of trains?

81. CARRYING BAGSA gentleman had to walk to his railway station, four miles from his house,and was encumbered by two bags of equal weight, but too heavy for him tocarry alone. His gardener and a boy both insisted on carrying the luggage; butthe gardener is an old man and the boy not sufficiently strong, while thegentleman believes in a fair division oflabor and wished to take his own share.They started off with the gardener carrying one bag and the boy the other,while the gentleman worked out the best way of arranging that the three shouldshare the burden equally among them. How would you have managed it?

82. THE ESCALATOR"I counted fifty steps that I made in going down the escalator," said Walker."I counted seventy-five steps," said Trotman; "but I was walking down threetimes as quickly as you."If the staircase were stopped, how many steps would be visible? It isassumed that each man travelled at a uniform rate, and the speed of thestaircase was also constant.

83. THE FOUR CYCLISTSThe four circles represent cinder paths. The four cyclists started at noon.Each person rode round a different circle, one at the rate of six miles an hour,another at the rate of nine miles an hour, another at the rate of twelve milesan hour, and the fourth at the rate of fifteen miles an hour. They agreed to rideSpeed & Distance Puzzles 25until all met at the center, from which they started, for the fourth time.The distance round each circle was exactly one-third of a mile. When did theyfinish their ride?

84. THE DONKEY CART"Three men," said Crackham, "Atkins, Brown, and Cranby, had to go ajourney of forty miles. Atkins could walk one mile an hour, Brown could walktwo miles an hour, and Cranby could go in his donkey cart at eight miles anhour. Cranby drove Atkins a certain distance, and, dropping him to walk theremainder, drove back to meet Brown on the way and carried him to theirdestination, where they all arrived at the same time."How long did the journey take? Of course each went at a uniform ratethroughout."

26 Arithmetic & Algebraic Problems

85. THE THREE CARSThree cars travelling along a road in the same direction are, at a certainmoment, in the following positions in relation to one another. Andrews is acertain distance behind Brooks, and Carter is twice that distance in front ofBrooks. Each car travels at its own uniform rate of speed, with the result thatAndrews passes Brooks in seven minutes, and passes Carter five minutes later.In how many minutes after Andrews would Brooks pass Carter?

86. THE FLY AND THE CARSA road is 300 miles long. A car, A, starts at noon from one end and goesthroughout at 50 miles an hour, and at the same time another car, B, goinguniformly at 100 miles an hour, starts from the other end together with a flytravelling 150 miles an hour. When the fly meets car A, it immediately turnsand flies towards B.(1) When does the fly meet B?The fly then turns towards A and continues flying backwards and forwardsbetween A and B.(2) When will the fly be crushed between the two cars if they collide and itdoes not get out of the way?

87. THE SUBWAY STAIRSWe ran up against Percy Longman, a young athlete, the other day whenleaving Curley Street subway station. He stopped at the elevator, saying,"I always go up by the stairs. A bit of exercise, you know. But this is the longest stairway on the line-nearly a thousand steps. I will tell you a queer thingabout it that only applies to one other smaller stairway on the line. If I go uptwo steps at a time, there is one step left for the last bound; if I go up threeat a time, there are two steps left; ifl go up four at a time, there are three stepsleft; five at a time, four are left; six at a time, five are left; and if I went upseven at a time there would be six steps left over for the last bound. Now,why is that?"As he went flying up the stairs, three steps at a time, we laughed and said,"He little suspects that if he went up twenty steps at a time there would beSpeed & Distance Puzzles 27nineteen steps for his last bound!" How many steps are there in the CurleyStreet subway stairway? The bottom floor does not count as a step, and thetop landing does.

88. THE BUS RIDEGeorge treated his best girl to a bus ride, but on account of his limited resources it was necessary that they should walk back. Now, if the bus goes atthe rate of nine miles an hour and they walk at the rate of three miles anhour, how far can they ride so that they may be back in eight hours?

89. A QUESTION OF TRANSPORTTwelve soldiers had to get to a place twenty miles distant with the quickestpossible dispatch, and all had to arrive at exactly the same time. They requisitioned the services of a man with a small car."I can do twenty miles an hour," he said, "but I cannot carry more thanfour men at a time. At what rate can you walk?""All of us can do a steady four miles an hour," they replied."Very well," exclaimed the driver, "then I will go ahead with four men,drop them somewhere on the road to walk, then return and pick up four more(who will be somewhere on the road), drop them also, and return for the lastfour. So all you have to do is to keep walking while you are on your feet, andI will do the res t."They started at noon. What was the exact time that they all arrived together?

90. HOW FAR WAS IT?"The steamer," remarked one of our officers home from the East, "was ableto go twenty miles an hour downstream, but could only do fifteen miles anhour upstream. So, of course, she took five hours longer in coming up thanin going down."One could not resist working out mentally the distance from point to point.What was it?

28 Arithmetic & Algebraic Problems

91. OUT AND HOMEColonel Crackham says that his friend, Mr. Wilkinson, walks from hiscountry house into the neighboring town at the rate of five miles per hour,and, because he is a little tired, he makes the return journey at the rate of threemiles per hour. The double journey takes him exactly seven hours. Can youtell the distance from his house to the town?

92. THE MEETING CARSThe Crackhams made their first stop at Bugleminster, where they were tospend the night at a friend's house. This friend was to leave home at thesame time and ride to London to put up at the Crackhams' house. They tookthe same route, and each car went at its own uniform speed. They kept alook-out for one another, and met forty miles from Bugleminster. Georgethat evening worked out the following little puzzle:"I find that if, on our respective arrivals, we had each at once proceededon the return, journey at the same speeds we should meet at forty-eight milesfrom London."If this were so, what is the distance from London to Bugleminster?

93. A BICYCLE RACETwo cyclists race on a circular track. Brown can ride once round the trackin six minutes, and Robinson in four minutes. In how many minutes willRobinson overtake Brown?

94. A LITTLE TRAIN PUZZLEA nonstop express going sixty miles an hour starts from Bustletown forIronchester, and another nonstop express going forty miles an hour starts atthe same time from Ironchester for Bustletown. How far apart are theyexactly an hour before they meet?Because I have failed to find these cities on any map or in any gazetteer, Icannot state the distance between them, so we will just assume that it is morethan 250 miles.Speed & Distance Puzzles 29If this little puzzle gives the reader much trouble he will certainly smilewhen he sees the answer.

95. AN IRISH JAUNT"It was necessary," said Colonel Crackham, "for me to go one day fromBOghooley to Ballyfoyne, where I had to meet a friend. But the only conveyance obtainable was old Pat Doyle's rickety little cart, propelled by a marewhose working days, like her legs, were a bit over."It was soon evident that our rate of progress was both safe and steady,though unquestionably slow."'I say, Pat,' I inquired after a few minutes' ride, 'has your engine gotanother speed?'"'Yes, begorra,' the driver replied, 'but it's not so fast.'"'Then we'll keep her on this gear,' said I."Pat assured me that she would keep going at one pace until she got to herjourney's end. She wouldn't slow down and she wouldn't put on any spurts." 'We have been on the road twenty minutes,' I remarked, looking at mywatch. 'How far have we come from Boghooley?'" 'Just half as far as it is from here to Pigtown,' said Pat."After a rapid refreshment at Pig town we went on another five miles, andthen I asked Pat how far it was to Ballyfoyne. I got exactly the same reply.It was clear he could only think in terms of Pigtown." 'Just half as far as it is from here to Pigtown.'"Another hour's ride and we were at the end of our journey."What is the distance from Boghooley to Ballyfoyne?

96. A WALKING PROBLEMA man taking a walk in the country on turning round saw a friend of hiswalking 400 yards behind in his direction. They each walked 200 yards in adirect line, with their faces towards each other, and you would suppose thatthey must have met. Yet they found after their 200 yards' walk that they werestill 400 yards apart. Can you explain?

30 Arithmetic & Algebraic Problems

97. THE FALSE SCALESA pudding, when put into one ofthe pans of these scales, appeared toweigh four ounces more than nineelevenths of its true weight, but whenput into the other pan it appeared toweigh three pounds more than in thefirst pan. What was its true weight?

98. WEIGHING THE GOODSA tradesman whose morals had become corrupted during the war bya course of profiteering went to the length of introducing a pair of falsescales. It will be seen from the illustration below that one arm is longer than theother, though they are purposely so drawn as to give no clue to the answer.As a consequence, it happened that in one of the cases exhibited eight of thelittle packets (it does not matter what they contain) exactly balanced three ofthe canisters, while in the other case one packet appeared to be of the sameweight as six canisters.The true weight of one canister was known to be exactly one ounce. Whatwas the true weight of the eight packets?

1~

99. WEIGHING THE BABY"I saw a funny incident at the railway station last summer," said a friend."There was a little family group in front of the automatic weighing machine,that registered up to 200 lb., and they were engaged in the apparently difficulttask of weighing the baby. Wheneverthey attempted to put the baby aloneon the machine she always yelled androlled off, while the father was holdingoff the dog, who always insisted onbeing included in the operations. Atlast the man with the baby and Fidowere on the machine together, and Itook this snapshot of them with mycamera."He produced a photograph, fromwhich I have simply copied the dial,as that is all we need."Then the man turned to his wifeand said, 'It seems to me, my dear,that baby and I together weigh 162 lb.more than the dog, while the dogweighs 70 per cent less than the baby.Weight Puzzles 31We must try to work it out at home.'I also amused myself by working itout from those figures. What do yousuppose was the actual weight of thatdear infant?"

100. FRESH FRUITSSome fresh fruit was being weighed for some domestic purpose, when it wasfound that the apples, pears, and plums exactly balanced one another, as shownin the sketch. Can you say how many plums were equal in weight to one pear?The relative sizes of the fruits in the drawing must not be taken to be correct(they are purposely not so), but we must assume that every fruit is exactlyequal in weight to every other of its own kind.It is clear that three apples and one pear are equal in weight to ten plums,and that one apple and six plums weigh the same as a single pear, but howmany plums alone would balance that pear?

32 Arithmetic & Algebraic ProblemsThis appears to be an excellent method of introducing the elements ofalgebra to the untutored mind. When the novice starts working it out he willinevitably be adopting algebraical methods, without, perhaps, being consciousof the fact. The two weighings show nothing more than two simultaneousequations, with three unknowns.

101. WEIGHING THE TEAA grocer proposed to put up 20 Ibs. of China tea into 2-lb. packets, but hisweights had been misplaced by somebody, and he could only find the 5-lb.and the 9-lb. weights. What is the quickest way for him to do the business?We will say at once that only nine weighings are really necessary.

102. AN EXCEPTIONAL NUMBERA number is formed of five successive digits (not necessarily in regularorder) so that the number formed by the first two multiplied by the centraldigit will produce the number expressed by the last two. Thus, if it were

1 2 8 9 6, then 12 multiplied by 8 produces 96. But, unfortunately, I, 2, 6, 8, 9are not successive numbers, so it will not do.

103. THE FIVE CARDSI have five cards bearing the figuresI, 3, 5, 7, and 9. How can I arrangethem in a row so that the numberformed by the first pair multiplied bythe number formed by the last pair,with the central number subtracted,will produce a number composed ofrepetitions of one figure? Thus, in theexample I have shown, 31 multipliedby 79 and 5 subtracted will produce

2444, which would have been allright if that 2 had happened to beanother 4. Of course, there must betwo solutions, for the pairs are clearlyinterchangeable.

104. SQUARES AND DIGITSWhat is the smallest square number that terminates with the greatest possible number of similar digits? Thus the greatest possible number might beDigital Puzzles 33five and the smallest square number with five similar digits at the end mightbe 24677777. But this is certainly not a square number. Of course, 0 is not tobe regarded as a digit.

105. THE TWO ADDITIONSCan you arrange the following figures in two groups of four figures each sothat each group shall add to the same sum?

12345 789If you were allowed to reverse the 9 so as to change it into the missing 6 itwould be very easy. For example, 1,2,7,8 and 3, 4,5,6 add up to 18 in bothcases. But you are not allowed to make any such reversal.

106. THE REPEATED QUARTETTEIf we multiply 64253 by 365 we get the product 23452345, where the firstfour figures are repeated. What is the largest number that we can multiply by

365 in order to produce a similar product of eight figures with the first fourfigures repeated in the same order? There is no objection to a repetition offigures-that is, the four that are repeated need not be all different, as in thecase shown.

107. EASY DIVISIONTo divide the number 8,1 0 1,265,822,784 by 8, all we need do is totransfer the 8 from the beginning to the end! Can you find a number beginning with 7 that can be divided by 7 in the same simple manner?

108. A MISUNDERSTANDINGAn American correspondent asks me to find a number composed of anynumber of digits that may be correctly divided by 2 by simply transferring thelast figure to the beginning. He has apparently come across our last puzzle withthe conditions wrongly stated. If you are to transfer the first figure to the endit is solved by 3 I 5 7 8 9 4 7 3 6 84 2 I 0 5 2 6, and a solution may easily befound from this with any given figure at the beginning. But if the figure is tobe moved from the end to the beginning, there is no possible solution for thedivisor 2. But there is a solution for the divisor 3. Can you find it?

34 Arithmetic & Algebraic Problems

109. THE TWO FOURSI am perpetually receiving inquiries about the old "Four Fours" puzzle. Ipublished it in IS99, but have since found that it first appeared in the firstvolume of Knowledge (lSSI). It has since been dealt with at some length byvarious writers. The point is to express all possible whole numbers with fourfours (no more and no fewer), using the various arithmetical signs. Thus

4 X 4 + % equals 17, and 44 + 4 + v'4 equals 50. All numbers up to 112inclusive may be solved, using only the signs for addition, subtraction, multiplication, division, square root, decimal points, and the factorial sign 4! whichmeans I X 2 X 3 X 4, or 24, but 113 is impossible.It is necessary to discover which numbers can be formed with one four,with two fours, and with three fours, and to record these for combination asrequired. It is the failure to find some of these that leads to so much difficulty.For example, I think very few discover that 64 can be expressed with onlytwo fours. Can the reader do it?

110. THE TWO DIGITSWrite down any two-figure number (different figures and no 0) and thenexpress that number by writing the same figures in reverse order, with orwithout arithmetical signs. For example, 45 = 5 X 9 would be correct if onlythe 9 had happened to be a 4. Or SI = (I + S)2 would do, except for the factthat it introduces a third figure-the 2.

111. DIGITAL COINCIDENCESIf! multiply, and also add, 9 and 9, I get SI and IS, which contain the samefigures. If I multiply and add 2 and 47, I get 94 and 49-the same figures. IfI multiply and add 3 and 24, I get the same figures-72 and 27. Can you findtwo numbers that when multiplied and added will, in this simple manner,produce the same three figures? There are two cases.

112. PALINDROMIC SQUARE NUMBERSThis is a curious subject for investigation-the search for square numbersthe figures of which read backwards and forwards alike. Some of them areDigital Puzzles 35very easily found. For example, the squares of 1, 11, Ill, and 1111 are respectively 1, 121, 12321, and 1234321, all palindromes, and the rule applies for anynumber of l's provided the number does not contain more than nine. But thereare other cases that we may call irregular, such as the square of 264 = 69696and the square of 2285 = 5221225.Now, all the examples I have given contain an odd number of digits. Canthe reader find a case where the square palindrome contains an even numberof figures?

113. FACTORIZINGWhat are the factors (the numbers that will divide it without any remainder)of this number-l 0 0 0 0 0 0 0 0 0 0 0 I? This is easily done if you happen toknow something about numbers of this peculiar form. In fact, it is just as easyfor me to give two factors if you insert, say, one hundred and one noughts,instead of eleven, between the two ones.There is a curious, easy, and beautiful rule for these cases. Can you find it?

114. FIND THE FACTORSFind two whole numbers with the smallest possible difference betweenthem which, when multiplied together, will produce 1234567890.

115. DIVIDING BY ELEVENIf the nine digits are written at haphazard in any order, for example,

4 1 2 5 3 97 6 8, what are the chances that the number that happens to beproduced will be divisible by 11 without remainder? The number I havewritten at random is not, I see, so divisible, but if I had happened to make the

1 and the 8 change places it would be.

116. DIVIDING BY 37I want to know whether the number 49,129,308,213 is exactly divisibleby 37, or, if not, what is the remainder when so divided. How may I do thisquite easily without any process of actual division whatever? It can be doneby inspection in a few seconds-if you know how.

36 Arithmetic & Algebraic Problems

117. ANOTHER 37 DIVISIONHere is an interesting extension of the last puzzle. If the nine digits arewritten at haphazard in any order, for example, 4 1 2539768, what are thechances that the number that happens to be produced will be divisible by 37without remainder?

118. A DIGITAL DIFFICULTYArrange the ten digits, 1 2 3 4 5 678 9 0, in such order that they shall forma number that may be divided by every number from 2 to 18 without in anycase a remainder. As an example, if I arrange them thus, 1 2 7 4 9 5 3 6 8 0,this number can be divided by 2, 3,4, 5, and so on up to 16, without any remainder, but it breaks down at 17.

119. THREES AND SEVENSWhat is the smallest number composed only of the digits 3 and 7 that maybe divided by 3 and by 7, and also the sum of its digits by 3 and by 7, without any remainder? Thus, for example, 7733733 is divisible by 3 and by 7without remainder, but the sum of its digits (33), while divisible by 3, is notdivisible by 7 without remainder. Therefore it is not a solution.

120. ROOT EXTRACTIONIn a conversation I had with Professor Simon Greathead, the eminentmathematician, now living in retirement at Colney Hatch, * I had occasion torefer to the extraction of the cube root."Ah," said the professor, "it is astounding what ignorance prevails on thatelementary matter! The world seems to have made little advance in theprocess of the extraction of roots since the primitive method of employingspades, forks, and trowels for the purpose. For example, nobody but myselfhas ever discovered the simple fact that, to extract the cube root of a number,all you have to do is to add together the digits. Thus, ignoring the obvious caseof the number 1, if we want the cube root of 512, add the digits-8, and thereyou are!"I suggested that that was a special case.°A large mental hospital in Middlesex, near London.-M. G.Digital Puzzles 37"Not at all," he replied. "Take another number at random-4913-and thedigits add to 17, the cube of which is 4913."I did not presume to argue the point with the learned man, but I will justask the reader to discover all the other numbers whose cube root is the sameas the sum of their digits. They are so few that they can be counted onthe fingers of one hand.

121. QUEER DIVISIONThe following is a rather curious puzzle. Find the smallest number that,when divided successively by 45, 454, 4545 and 45454, leaves the remainders

4, 45, 454, and 4,545 respectively. This is perhaps not easy but it affordsa good arithmetical exercise.

122. THREE DIFFERENT DIGITSThe professor, a few mornings ago, proposed that they should find all thosenumbers composed of three different digits such that each is divisible withoutremainder by the square of the sum of those digits. Thus, in the case of 112,the digits sum to 4, the square of which is 16, and 112 can be divided by 16without remainder, but unfortunately 112 does not contain three differentdigits.Can the reader find all the possible answers?

123. DIGITS AND CUBESProfessor Rackbrane recently asked his young friends to find all those fivefigure squares such that the number formed by the first two figures added tothat formed by the last two figures should equal a cube. Thus with the squareof 141, which is 19,881, if we add 19 and 81 together we get 100, whichis a square but unfortunately not a cube.How many solutions are there altogether?

124. REVERSING THE DIGITSWhat number composed of nine figures, if multiplied by 1,2, 3,4, 5, 6, 7,

8, 9, will give a product with 9, 8, 7, 6, 5, 4, 3, 2, 1 (in that order), in the lastnine places to the right?

38 Arithmetic & Algebraic Problems

125. DIGITAL PROGRESSION"If you arrange the nine digits," said Professor Rackbrane, "in three numbers thus, 147, 258, 369, they have a common difference of III, and are,therefore, in arithmetical progression."Can you find four ways of rearranging the nine digits so that in each casethe three numbers shall have a common difference and the middle number bein every case the same?

126. FORMING WHOLE NUMBERSCan the reader give the sum of all the whole numbers that can be formedwith the four figures I, 2, 3, 4? That is, the addition of all such numbers as

1,234,1,423,4,312, etc. You can, of course, write them all out and make theaddition, but the interest lies in finding a very simple rule for the sum of allthe numbers that can be made with four different digits selected in everypossible way, but zero excluded.

127. SUMMING THE DIGITSProfessor Rackbrane wants to know what is the sum of all the numbers thatcan be formed with the complete nine digits (0 excluded), using each digitonce, and once only, in every number?

128. SQUARING THE DIGITSTake nine counters numbered I to 9, and place them in a row as shown. Itis required in as few exchanges of pairs as possible to convert this into a squarenumber. As an example in six pairs we give the following: 78 (exchanging 7and 8), 8 4, 4 6, 6 9, 9 3, 3 2, which gives us the number 139,854,276, which isthe square of 11,826. But it can be done in much fewer moves.Digital Puzzles 39

129. DIGITS AND SQUARESOne of Rackbrane's little Christmas puzzles was this: (1) What is thesmallest square number, and (2) what is the largest square number that contains all the ten digits (I to 9 and 0) once, and once only?

130. DIGITAL SQUARESIt will be found a very good puzzle to try to discover a number which, together with its square, shall contain all the nine digits once, and once only, thezero disallowed. Thus, if the square of 378 happened to be 152,694, it wouldbe a perfect solution. But unfortunately the actual square is 142,884, whichgives us those two repeated 4's and 8's, and omits the 6, 5, and 9.There are only two possible cases, and these may be discovered in about aquarter of an hour if you proceed in the right way.

131. FINDING A SQUAREHere are six numbers: 4,784,887,2,494,651,8,595,087, 1,385,287,9,042,451,

9,406,087. It is known that three of these numbers added together will form asquare. Which are they?The reader will probably see no other course but rather laborious trial, andyet the answer may be found directly by very simple arithmetic and withoutany experimental extraction of a square root.

132. JUGGLING WITH DIGITSArrange the ten digits in three arithmetical sums, employing three of thefour operations of addition, subtraction, multiplication, and division, andusing no signs except the ordinary ones implying those operations. Here is anexample to make it quite clear:

3 + 4 = 7; 9 - 8 = 1; 30 + 6 = 5.But this is not correct, because 2 is omitted, and 3 is repeated.

133. EQUAL FRACTIONSCan you construct three ordinary vulgar fractions (say, '12, Ih, or 114, or anything up to ¥l inclusive) all of the same value, using in every group all the

40 Arithmetic & Algebraic Problemsnine digits once, and once only? The fractions may be formed in one of thefollowing ways:We have only found five cases, but the fifth contains a simple little trickthat may escape the reader.

134. DIGITS AND PRIMESUsing the nine digits once, and once only, can you find prime numbers(numbers that cannot be divided, without remainder, by any number except

1 and itself) that will add up to the smallest total possible? Here is an example.The four prime numbers contain all the nine digits once, and once only, andadd up to 450, but this total can be considerably reduced. It is quite an easypuzzle.

61

283

47

59

450

135. A SQUARE OF DIGITSThe nine digits may be arranged in a square in many ways, so that thenumbers formed in the first row and second row will sum to the third row.We give three examples, and it will be found that the difference between theZ '1 3 3 Z r

5 4 6 6 5 4

8 1 ~ i~ B 1Digital Puzzles 41first total, 657, and the second, 819, is the same as the difference betweenthe second, 819, and the third, 981-that is, 162.Can you form eight such squares, everyone containing the nine digits, sothat the common difference between the eight totals is throughout the same?Of course it will not be 162.

136. THE NINE DIGITSIt will be found that 32,547,891 multiplied by 6 (thus using all the nine digitsonce, and once only) gives the product 195,287,346 (also containing all thenine digits once, and once only). Can you find another number to be multiplied by 6 under the same conditions? Remember that the nine digits mustappear once, and once only, in the numbers multiplied and in the product.

137. EXPRESSING TWENTY-FOURIn a book published in America was the following: "Write 24 with threeequal digits, none of which is 8. (There are two solutions to this problem.)"The answers given are 22 + 2 = 24, and 33 - 3 = 24. Readers familiarwith the old "Four Fours" puzzle, and others of the same class, will ask whythere are supposed to be only these solutions. With which of the remainingdigits is a solution equally possible?

138. THE NINE BARRELSIn how many different ways maythese nine barrels be arranged in threetiers of three so that no barrel shallhave a smaller number than its ownbelow it or to the right of it? The firstcorrect arrangement that will occurto you is I 23 at the top, then 4 5 6 inthe second row, and 789 at the bottom, and my sketch gives a secondarrangement. How many are therealtogether?

42 Arithmetic & Algebraic Problems

139. THE EIGHT CARDSColonel Crackham placed eightnumbered cards on the breakfasttable, as here shown, and asked hisyoung friends to rearrange them,moving as few as possible, so thatthe two columns should add up alike.Can it be done?

140. FIND THE NUMBERSCan you find two numbers composed only of ones which give the sameresult by addition and multiplication? Of course 1 and 11 are very near, butthey will not quite do, because added they make 12, and multiplied they makeonlyll.

141. MULTIPLYING THE NINE DIGITSThey were discussing mental problems at the Crackhams' breakfast tablewhen George suddenly asked his sister Dora to multiply as quickly as possibleIX2X3X4X5X6X7X8X9XQHow long would it have taken the reader?

142. CURIOUS MULTIPLICANDReaders who remember the Ribbon Problem No. 83 in The CanterburyPuzzles, may be glad to have this slightly easier variation of it:Skeleton Puzzles 43What number is it that can be multiplied by I, 2, 3, 4, 5, or 6 and no newfigures appear in the result?

143. ADDING THEIR CUBESThe numbers 407 and 370 have this peculiarity, that they exactly equal thesum of the cubes of their digits. Thus the cube of 4 is 64, the cube of 0 is 0,and the cube of7 is 343. Add together 64, 0, and 343, and you get 407. Again,the cube of 3 (27), added to the cube of 7 (343), is 370. Can you find anumber not containing a zero that will work in the same way? Of course, webar the absurd case of I.

144. THE SOLITARY SEVEN***)********(*7**** * * ** * ** * ** * * ** * ** * * ** * * ofcHere is a puzzle, sent me by the Rev. E. F. O. It is the first example I haveseen of one of these missing-figure puzzles in which only one figure is given,and there appears to be only one possible solution. And, curiously enough, itis not difficult to reconstruct the simple division sum. For example, as the divisor when multiplied by 7 produces only three figures we know the first figurein the divisor must be I. We can then prove that the first figure in the dividendmust be I; that, in consequence of bringing down together the last two figuresof the dividend, the last but one figure in the quotient must be 0, that thefirst and last figures in the quotient must be greater than 7, because they eachproduce four figures in the sum, and so on.

44 Arithmetic & Algebraic Problems

145. A COMPLETE SKELETONThis is an arrangement without any figure at all, constructed by Mr. A.Corrigan. Note the decimal dot in the quotient. The extension to four placesof decimals makes it curiously easy to solve.***)******(****-***** * ** * ** * ** * ** * ** * ** * ** * * ** * * *

146. SIMPLE MULTIPLICATIONGeorge Crackham produced this puzzle at the breakfast table one morning:**********

2**********He asked them to substitute for the stars all the ten digits in each row, soarranged as to form a correct little sum in multiplication. He said that the 0was not to appear at the beginning or end of either number. Can the readerfind an answer?Skeleton Puzzles 45

147. AN ABSOLUTE SKELETONHere is a good skeleton puzzle.The only conditions are:(l) No digit appears twice inany row of figures except in thedividend.(2) If 2 be added to the lastfigure in the quotient it equals thelast but one, and if 2 be added tothe third figure from the end itgives the last figure but three inthe quotient. That is to say, thequotient might end in, say, 9,742,or in 3,186.We have only succeeded infinding a single solution.***)***********(********* * ** * ** * ** * * ** * * ** * ** * ** * * ** * * ** * * ** * * ** * * ** * * ** * * ** * * *

148. ODDS AND EVENS***)*********(*******OE** * * *00*** * *EE** * *EO*****EE*** * *00*The Rev. E. F. O. devised thisskeleton problem in simple division.Every asterisk and letter represents afigure, and "0" stands for an oddfigure (1, 3, 5, 7, or 9), while "E"represents an even figure (2, 4, 6, 8,or 0). Can you construct an arrangement complying with these conditions? There are six solutions. Canyou find one, or all of them?

46 Arithmetic & Algebraic Problems

149. SIMPLE DIVISIONCan you restore this divisionproblem by substituting a figurefor every asterisk without alteringor removing the sevens? If youstart out with the assumptionthat all the sevens are given andthat you must not use another,you will attempt an impossibility,though the proof is difficult; butwhen you are told that though noadditional sevens may be used indivisor, dividend, or quotient, anynumber of extra sevens may beused in the working, it is comparatively easy.****7*)**7*******(**7*************7*********7*****7**************7**************

150. A COMPLETE SKELETON***)*********(******* * ** * * ** * ** * ** * ** * * ** * * ***)******(****** ** * ** ** * ** * ** * ** * *It will be remembered that askeleton puzzle, where the figures arerepresented by stars, has not beenconstructed without at least one figure,or some added condition, being used.Perhaps the following (received fromW. J. W.) comes a little nearer theideal, though there are two divisionsums and not one, and they arerelated by the fact that the six-figurequotient of the first happens to be thedividend of the second. There appearsto be only one solution.Cryptarithm Puzzles 47

151. ALPHABETICAL SUMSThere is a family resemblance between puzzles where an arithmeticalworking has to be reconstructed from a few figures and a number of asterisks,and those in which every digit is represented by a letter, but they are reallyquite different. The resemblance lies in the similarity of the process of solving.Here is a little example of the latter class. It can hardly be called difficult.PR)MTVVR(RSRMVRKKVKMDMVRMVRCan you reconstruct this simple division sum? Every digit is represented bya different letter.

152. ALPHABETICAL ARITHMETICHere is a subtraction puzzle that will keep the reader agreeably employedfor several minutes.Let AB multiplied by C = DE. When DE is taken from FG, the result isHI:FGDEHIEach letter stands for a different figure (1,2,3,4,5, 6, 7, 8, or 9) and 0 isnot allowed.

153. FIGURES FOR LETTERSProfessor Rackbrane the other morning gave his young friends this ratherdifficult problem. He wrote down the letters of the alphabet in this order:ABCD X EFGHI = ACGEFHIBD

48 Arithmetic & Algebraic ProblemsEvery letter, he said, stood for a different digit, I to 9 (0 excluded). Thenumber represented by the first four digits, when multiplied by the numbercontaining five digits, equals the number containing all the nine digits in theorder shown. Can you substitute digits for letters so that it works?

154. THE SHOPKEEPER'S PUZZLEA shopkeeper, for private marking, selects a word of ten letters (all different) such as NIGHTMARES, where each letter stands for one of the figuresI, 2, 3, 4, 5, 6, 7, 8, 9, 0, in their order. So NI stands for $12.00, and$34.00 would be GH. Assuming this little addition sum is in such a privatecode, can you find the man's key word? It is not difficult.GAUNTOILERRGUOEI

155. BEESWAXThe word BEESWAX represents a number in a criminal's secret code, butthe police had no clue until they discovered among his papers the followingsum:EASEBSBSXB PWWKSETQKPE PWEKKQThe detectives assumed that it was an addition sum and utterly failedto solve it. Then one man hit on the brilliant idea that perhaps it was a caseof subtraction. This proved to be correct, and by substituting a different figurefor each letter, so that it worked out correctly, they obtained the secret code.What number does BEESWAX represent?

156. WRONG TO RIGHT"Two wrongs don't make a right," said somebody at the breakfast table."I am not so sure about that," Colonel Crackham remarked. "Take this asCryptarithm Puzzles 49an example. Each letter represents a different digit, and no 0 is allowed."WRONGWRONGRIGHT"If you substitute correct figures the little addition sum will work correctly.There are several ways of doing it."

157. LETTER MULTIPLICATIONIn this little multiplication sum the five letters represent five differentdigits. What are the actual figures? There is no O.SEAMTMEATS

158. THE CONSPIRATORS' CODEA correspondent (G. P.) sends this interesting puzzle. Two conspirators hada secret code. Their letters sometimes contained little arithmetical sumsrelating to some quite plausible discussion, and having an entirely innocentappearance. But in their code each of the ten digits represented a differentletter of the alphabet. Thus, on one occasion, there was a little sum in simpleaddition which, when the letters were substituted for the figures, read asfollows:FL YFORYOURL I FEIt will be found an interesting puzzle to reconstruct the addition sum withthe help of the due that I and 0 stand for the figures I and 0 respectively.

50 Arithmetic & Algebraic Problems

159. LETTER-FIGURE PUZZLEA correspondent (c. E. B.) sends the following. It is not difficult, if properlyattacked:AXB=B, BXC=AC, CXD=BC, DXE=CH, EXF=DK,F X H = C J, H X J = K J, J X K = E, K X L = L, A X L = L. Everyletter represents a different digit, and, of course, AC, BC, etc., are twofigure numbers. Can you find the values in figures of all the letters?

160. THE MILLER'S TOLLHere is a very simple puzzle, yet I have seen people perplexed by it for a fewminutes. A miller was accustomed to take as toll one-tenth of the flour that heground for his customers. How much did he grind for a man who had just onebushel after the toll had been taken?

161. EGG LAYINGThe following is a new variation of an old friend. Though it looks rathercomplicated and difficult, it is absurdly easy if properly attacked. If a hen anda half lays an egg and a half in a day and a half, how many and a half who laybetter by half will lay half a score and a half in a week and a half?

162. THE FLOCKS OF SHEEPFour brothers were comparing the number of sheep that they owned. It wasfound that Claude had ten more sheep than Dan. If Claude gave a quarter ofhis sheep to Ben, then Claude and Adam would together have the same number as Ben and Dan together. If, then, Adam gave one-third to Ben, and Bengave a quarter of what he then held to Claude, who then passed on a fifth ofhis holding to Dan, and Ben then divided one-quarter of the number he thenpossessed equally amongst Adam, Claude, and Dan, they would all have anequal number of sheep.How many sheep did each son possess?

163. SELLING EGGSA woman took a certain number of eggs to market and sold some of them.The next day, through the industry of her hens, the number left over had beenMiscellaneous Puzzles 51doubled, and she sold the same number as the previous day. On the third daythe new remainder was trebled, and she sold the same number as before. Onthe fourth day the remainder was quadrupled, and her sales the same asbefore. On the fifth day what had been left over were quintupled, yet she soldexactly the same as on all the previous occasions and so disposed of herentire stock.What is the smallest number of eggs she could have taken to market thefirst day, and how many did she sell daily?

164. PUSSY AND THE MOUSE"There's a mouse in one of these barrels," said the dog."Which barrel?" asked the cat."Why, the five hundredth barrel.""What do you mean by the five hundredth? There are only five barrels in all.""It's the five hundredth if you count backwards and forwards in this way."And the dog explained that you count like this:

12345

9 8 7 6

10 Il 12 13So that the seventh barrel would be the one marked 3 and the twelfth barrelthe one numbered 4."That will take some time," said the cat, and she began a laborious count.Several times she made a slip and had to begin again."Rats!" exclaimed the dog. "Hurry up or you will be too late!""Confound you! You've put me out again, and I must make a fresh start."

52 Arithmetic & Algebraic ProblemsMeanwhile the mouse, overhearing the conversation, was working madly atenlarging a hole, and just succeeded in escaping as the cat leapt into thecorrect barrel."I knew you would lose it," said the dog. "Your education has been sadlyneglected. A certain amount of arithmetic is necessary to every cat, as it is toevery dog. Bless me! Even some snakes are adders!"Which was the five hundredth barrel? Can you find a quick way of arrivingat the answer without making the actual count?

165. ARMY FIGURESA certain division in an army was composed of a little over twenty thousandmen, made up of five brigades. It was known that one-third of the firstbrigade, two-sevenths of the second brigade, seven-twelfths of the third, ninethirteenths of the fourth, and fifteen-twenty-seconds of the fifth brigade happened in every case to be the same number of men. Can you discover howmany men there were in every brigade?

166. A CRITICAL VOTEA meeting of the Amalgamated Society of Itinerant Askers (better known asthe "Tramps' Union") was held to decide whether the members should strikefor reduced hours and larger donations. It was arranged that during the countthose in favor of the motion should remain standing, and those who votedagainst should sit down."Gentlemen," said the chairman in due course, "I have the pleasure to announce that the motion is carried by a majority equal to exactly a quarter ofthe opposition." (Loud cheers.)"Excuse me, guv'nor," shouted a man at the back, "but some of us overhere couldn't sit down.""Why not?"" 'Cause there ain't enough chairs.""Then perhaps those who wanted to sit down but couldn't will hold up theirhands .... I find there are a dozen of you, so the motion is lost by a majorityof one." (Hisses and disorder.)How many members voted at that meeting?Miscellaneous Puzzles 53

167. THE THREE BROTHERSThe discussion arose before one of the tribunals as to which of a tradesman's three sons could best be spared for service in the Army. "All I knowas to their capacities," said the father, "is this: Arthur and Benjamin can doa certain quantity of work in eight days, which Arthur and Charles will do innine days, and which Benjamin and Charles will take ten days over."Of course, it was at once seen that as more time was taken over the jobwhenever Charles was one of the pair, he must be the slowest worker. Thiswas all they wanted to know, but it is an interesting puzzle to ascertain justhow long each son would require to do that job alone.

168. THE HOUSE NUMBERA man said the house of his friend was in a long street, numbered on hisside one, two, three, and so on, and that all the numbers on one side of himadded up exactly the same as all the numbers on the other side of him. Hesaid he knew there were more than fifty houses on that side of the street, butnot so many as five hundred.Can you discover the number of that house?

169. A NEW STREET PUZZLEBrown lived in a street which contained more than twenty houses, butfewer than five hundred, all numbered one, two, three, four, etc., throughout.Brown discovered that all the numbers from one upwards to his own numberinclusive summed to exactly half the sum of all the numbers from one up to,and including, the last house. What was the number of his house?

170. ANOTHER STREET PUZZLEA long street in Brussels has all the odd numbers of the houses on one sideand all the even numbers on the other-a method of street numbering quitecommon in our own country.(1) If a man lives in an odd-numbered house and all the numbers on oneside of him, added together, equal the numbers on the other side, how manyhouses are there, and what is the number of his house?

54 Arithmetic & Algebraic Problems(2) If a man lives on the even side and all the numbers on one side of himequal those on the other side, how many houses are there, and what is hisnumber?We assume that there are more than fifty houses on each side of the streetand fewer than five hundred.

171. CORRECTING AN ERRORHilda Wilson was given a certain number to multiply by 409, but she madea blunder that is very common with children when learning the elements ofsimple arithmetic: she placed the first figure of her product by 4 below thesecond figure from the right instead of below the third. We have all done thatas youngsters when there has happened to be a 0 in the multiplier. The resultof Hilda's mistake was that her answer was wrong by 328,320, entirelyin consequence of that little slip. What was the multiplicand-the number shewas given to multiply by 409?

172. THE SEVENTEEN HORSES"I suppose you all know this old puzzle," said Jeffries. "A farmer leftseventeen horses to be divided among his three sons in the following proportions: To the eldest, one-half; to the second, one-third; and to the youngest,one-ninth. How should they be divided?""Yes; I think we all know that," said Robinson, "but it can't be done. Theanswer always given is a fallacy.""I suppose you mean," Prodgers suggested, "the answer where they borrowanother chap's horse to make eighteen for the purpose of the division. Thenthe three sons take 9, 6, and 2 respectively, and return the borrowed horse tothe lender.""Exactly," replied Robinson, "and each son receives more than his share.""Stop!" cried Benson. "That can't be right. If each man received morethan his share the total must exceed seventeen horses, but 9, 6, and 2 add upto 17 correctly.""At first sight that certainly looks queer," Robinson admitted, "but theexplanation is that if each man received his true fractional snare, these fractions would add up to less than seventeen. In fact, there would be a fractionleft undistributed. The thing can't really be done."Miscellaneous Puzzles 55"That's just where you are all wrong," said Jeffries. "The terms of the willcan be exactly carried out, without any mutilation of a horse."To their astonishment, he showed how it was possible. How should thehorses be divided in strict accordance with the directions?

173. EQUAL P~RIMETERSRational right-angled triangles have been a fascinating subject for studysince the time of Pythagoras, before the Christian era. Every schoolboyknows that the sides of these, generally expressed in whole numbers, are suchthat the square of the hypotenuse is exactly equal to the sum of the squaresof the other two sides. Thus, in the case of Diagram A, the square of 30 (900),added to the square of 40 (1600), is the square of 50 (2500), and similarlywith Band C. It will be found that the three triangles shown have each thesame perimeter. That is the three sides in every case add up to 120.Can you find six rational right-angled triangles each with a common perimeter, and the smallest possible? It is not a difficult puzzle like my "FourPrinces" (in The Canterbury Puzzles), in which you had to find four such triangles of equal area.

174. COUNTING THE WOUNDEDWhen visiting with a friend one of our hospitals for wounded soldiers,I was informed that exactly two-thirds of the men had lost an eye, threefourths had lost an arm, and four-fifths had lost a leg."Then," I remarked to my friend, "it follows that at least twenty-six of themen must have lost all three-an eye, an arm, and a leg."

56 Arithmetic & Algebraic ProblemsThat being so, can you say exactly how many men were in the hospital? Itis a simple calculation, but I have no doubt it will perplex many readers.

175. A COW'S PROGENY"Supposing," said my friend Farmer Hodge, "that cow of mine to have ashe calf at the age of two years, and supposing she goes on having the likeevery year, and supposing everyone of her young to have a she calf at the ageof two years, and afterwards every year likewise, and so on. Now, how manydo you suppose would spring from that cow and all her descendants in thespace of twenty-five years?" I understood from Hodge that we are to countfrom the birth of the original cow, and it is obvious that the family can produce no feminine beef or veal during the period stated.

176. SUM EQUALS PRODUCT"This is a curious thing," a man said to me. "There are two numberswhose sum equals their product. That is, they give the same result whetheryou add them together or multiply them together. They are 2 and 2, for if youadd them or multiply them, the result is 4." Then he tripped badly, for headded, "These are, I find, the only two numbers that have this peculiarity."I asked him to write down any number, as large as he liked, and I wouldimmediately give him another that would give a like result by additionor multiplication. He selected the number 987,654,321, and I promptly wrotedown the second number. What was it? The fact is, no matter what numberyou may select there is always another to which that peculiarity appliesin combination with it. If this is new to the reader it cannot fail to be interesting to him. He should try to find the rule.

177. SQUARES AND CUBESCan you find two whole numbers, such that the difference of their squaresis a cube and the difference of their cubes is a square? What is the answer inthe smallest possible numbers?

178. CONCERNING A CUBEWhat is the length in feet of the side of a cube when (1) the surface areaequals the cubical contents; (2) when the surface area equals the square ofMiscellaneous Puzzles 57the cubical contents; (3) when the square of the surface area equals thecubical contents?

179. A COMMON DlVlSORHere is a puzzle that has been the subject of frequent inquiries by correspondents, only, of course, the actual figures are varied considerably. Acountry newspaper stated that many schoolmasters have suffered in health intheir attempts to master it! Perhaps this is merely a little journalistic exaggeration, for it is really a simple question if only you have the cunning to hit onthe method of attacking it.This is the question: Find a common divisor for the three numbers, 480,608;

508,8\\; and 723,217 so that the remainder shall be the same in every case.

180. CURIOUS MULTIPLICATIONI have frequently been asked to explain the following, which will doubtlessinterest many readers who have not seen it. If a person can add correctly butis incapable of multiplying or dividing by a number higher than 2, it ispossible to obtain the product of any two numbers in this curious way.Multiply 97 by 23.

97 23

48 (46)

24 (92)

12 (184)

6 (368)

3 736

1,472

2,231In the first column we divide by 2, rejecting the remainders, until 1 isreached. In the second column we multiply 23 by 2 the same number oftimes. If we now strike out those products that are opposite to the evennumbers in the first column (we have enclosed these in parentheses for convenience in printing) and add up the remaining numbers we get 2,231, which isthe correct answer. Why is this?

58 Arithmetic & Algebraic Problems

181. THE REJECfED GUNHere is a little military puzzle that may not give you a moment's difficulty.It is such a simple question that a child can understand it and no knowledgeof artillery is required. Yet some of my readers may find themselves perplexedfor quite five minutes.An inventor offered a new large gun to the committee appointed by ourgovernment for the consideration of such things. He declared that when onceloaded it would fire sixty shots at the rate of a shot a minute. The War Officeput it to the test and found that it fired sixty shots an hour, but declined it,"as it did not fulfill the promised condition.""Absurd," said the inventor, "for you have shown that it clearly does allthat we undertook it should do.""Nothing of the sort," said the experts. "It has failed."Can you explain this extraordinary mystery? Was the inventor, or were theexperts, righ t?

182. TWENTY QUESTIONSI am reminded of an interesting old game I used to playas a bachelor.Somebody thinks of an object-say Big Ben, or the knocker on the front door,or the gong of the clock in the next room, or the top button of his friend'scoat, or Mr. Baldwin's pipe. You have then to discover the object, by puttingnot more than twenty questions, each of which must be answered by "yes" or"no." You have to word your questions discreetly, because if you ask, for example, "Is it animal, vegetable, or mineral?" you might get the unsatisfactoryanswer "Yes," and so waste a question. We found that the expert rarely failedto get an exact solution, and I have known some most remarkably difficultcases solved by the twenty questions.A novel limitation of the game is suggested to me, which will call for someingenuity, and the puzzle will doubtless be attacked in various ways by different persons. It is simply this. I think of a number containing six figures. Canyou discover what it is by putting to me twenty questions, each of which canonly be answered by "yes" or "no"? After the twentieth question you mustgive the number.Miscellaneous Puzzles 59

183. A CARD TRICKTake an ordinary pack of playing cards and regard all the picture cards astens. Now, look at the top card-say it is a seven-place it on the table facedownwards and play more cards on top of it, counting up to twelve. Thus, thebottom card being seven, the next will be eight, the next nine, and so on,making six cards in that pile. Then look again at the top card of the packsay it is a queen-then count 10, II, 12 (three cards in all), and complete thesecond pile. Continue this, always counting up to twelve, and if at last youhave not sufficient cards to complete a pile, put these apart.Now, if I am told how many piles have been made and how many unusedcards remain over, I can at once tell you the sum of all the bottom cards inthe piles. I simply multiply by 13 the number of piles less 4, and add thenumber of cards left over. Thus, if there were 6 piles and 5 cards over, then

13 times 2 (i.e., 6 less 4) added to 5 equals 31, the sum of the bottom cards.Why is this? That is the question.

184. THE QUARRELSOME CHILDRENA man married a widow, and they each already had children. Ten years laterthere was a pitched battle engaging the present family of twelve children. Themother ran to the father and cried, "Come at once! Your children and mychildren are fighting our children!"As the parents now had each nine children of their own, how many wereborn during the ten years?

185. SHARING THE APPLESWhile the Crackhams were having their car filled with gasoline, in a pleasantvillage, eight children on their way to school stopped to look at them. Theyhad a basket containing thirty-two apples, which they were taking into thevillage to sell. Aunt Gertrude, in a generous mood, bought the lot, and saidthe children might divide them among themselves.Dora asked the names of all the children and said, later in the day (thoughshe was drawing a little on her imagination), "Anne got one apple, Marytwo, Jane three, and Kate four. But Ned Smith took as many as his sister,Tom Brown twice as many as his sister, Bill Jones three times as many as his

60 Arithmetic & Algebraic Problemssister, and Jack Robinson four times as many as his sister. Now which of youcan give me the full names of the girls?"

186. BUYING RIBBONHere is a puzzle that appears to bear a strong family resemblance to othersgiven in the past. But it really requires an entirely different method of working. The author is unknown.Four mothers, each with one daughter, went into a shop to buy ribbon.Each mother bought twice as many yards as her daughter, and each personbought as many yards of ribbon as the number of cents she paid for each yard.Mrs. Jones spent 76¢ more than Mrs. White; Nora bought three yards less thanMrs. Brown; Gladys bought two yards more than Hilda, who spent 48¢ lessthan Mrs. Smith. What is the name of Mary's mother?

187. SQUARES AND TRIANGULARSWhat is the third lowest number that is both a triangular number anda square? Of course the numbers 1 and 36 are the two lowest that fulfill theconditions. What is the next number?

188. PERFECT SQUARESFind four numbers such that the sum of every two and the sum of all fourmay be perfect squares.

189. ELEMENTARY ARITHMETICThis is the kind of question that was very popular in Venice and elsewhereabout the middle of the sixteenth century. Nicola Fontana, generally knownas "Tartaglia" (the stammerer) was largely responsible for the invention.If a quarter of twenty is four, what would a third of ten be?

190. TRANSFERRING THE FIGURESIfwe wish to multiply 571,428 by 5 and divide by 4, we need only transferthe 5 from the beginning to the end for the correct answer, 714,285. Can youMiscellaneous Puzzles 61find a number that we can multiply by 4 and then divide the product by 5 inthe same simple manner, by moving the first figure to the end?Of course 714,285,just given, would do if we were allowed to transfer fromthe end to the beginning. But it must be from the beginning to the end.

191. A QUEER ADDITIONColonel Crackham asked the junior members of his household at thebreakfast table to write down five odd figures so that they will add up andmake fourteen. Only one of them did it.

192. SIX SIMPLE QUESTIONS( I) Deduct four thousand eleven hundred and a half from twelve thousandtwelve hundred and twelve.(2) Add 3 to 182, and make the total less than 20.(3) What two numbers multiplied together will produce seven?(4) What three figures multiplied by five will make six?(5) If five times four are 33, what is the fourth of 20?(6) Find a fraction whose numerator is less than its denominator, butwhich, turned upside down, shall remain of the same value.

193. THE THREE DROVERSAs the Crackhams were approaching a certain large town they met and weredelayed in passing first a flock of sheep, then a drove of oxen, and afterwardssome men leading a number of horses. They ascertained that it was a specialmarket day at the town. George seized the occasion to construct the followingpuzzle:"Three drovers with varied flocks met on the highway," he proposed."Said Jack to Jim: 'If I give you six pigs for a horse then you will have twiceas many animals in your drove as I will have in mine.' Said Dan to Jack: 'IfI give you fourteen sheep for a horse, then you'll have three times as manyanimals as I have got.' Said Jim to Dan: 'But if I give you four cows for ahorse, then you'll have six times as many animals as I.' There were no deals;but can you tell me just how many animals there were in the three droves?"

62 Arithmetic & Algebraic Problems

194. PROPORTIONAL REPRESENTATIONWhen stopping at Mangleton-on-the-Bliss the Crackhams found the inhabitants of the town excited over some little local election. There were tennames of candidates on a proportional representation ballot. Voters shouldplace No. I against the candidate of their first choice. They might also placeNo.2 against the candidate of their second choice, and so on until all the tencandidates have numbers placed against their names.The voters must mark their first choice, and any others may be marked ornot as they wish. George proposed that they should discover in how manydifferent ways the ballot might be marked by the voter.

195. A QUESTION OF CUBESProfessor Rackbrane pointed out one morning that the cubes of successivenumbers, starting from I, would sum to a square number. Thus the cubes of

1,2,3 (that is, 1,8,27), add to 36, which is the square of 6. He stated that ifyou are forbidden to use the I, the lowest answer is the cubes of 23, 24, and

25, which together equal 2042• He proposed to seek the next lowest number,using more than three consecutive cubes and as many more as you like, butexcluding I.

196. TWO CUBES"Can you find," Professor Rackbrane asked, "two consecutive cube numbers in integers whose difference shall be a square number? Thus the cube of

3 is 27, and the cube of 2 is 8, but the difference, 19, is not here a squarenumber. What is the smallest possible case?"

197. CUBE DIFFERENCESIf we wanted to find a way of making the number 1,234,567 the differencebetween two squares, we could at once write down 617,284 and 617,283-a half of the number plus'll and minus Ih respectively to be squared. But itwill be found a little more difficult to discover two cubes the differenceof which is 1,234,567.Miscellaneous Puzzles 63

198. ACCOMMODATING SQUARESCan you find two three-figure square numbers (no zeros) that, when puttogether, will form a six-figure square number? Thus, 324 and 900 (the squaresof 18 and 30) make 324,900, the square of 570, only there it happens thereare two zeros. There is only one answer.

199. MAKING SQUARESProfessor Rackbrane asked his young friends the other morning if theycould find three whole numbers in arithmetical progression, the sum of everytwo of which shall be a square.

200. FIND THE SQUARES"What number is that," Colonel Crackham asked, "which, added separatelyto 100 and 164, will make them both perfect square numbers?"

201. FORMING SQUARES"An officer arranged his men in a solid square," said Dora Crackham, "andhad thirty-nine men over. He then started increasing the number of men ona side by one, but found that fifty new men would be needed to complete thenew square. Can you tell me how many men the officer had?"

202. SQUARES AND CUBESFind two different numbers such that the sum of their squares shall equala cube, and the sum of their cubes equal a square.

203. MILK AND CREAMProfessor Rackbrane, when helping himself to cream at the breakfast table,put the following question:"An honest dairyman found that the milk supplied by his cows was 5 percent cream and 95 per cent skimmed milk. He wanted to know how much

64 Arithmetic & Algebraic Problemsskimmed milk he must add to a quart of whole milk to reduce the percentageof cream to 4 per cent."

204. FEEDING THE MONKEYSA man went to the zoo monkey house with a bag of nuts. He found thatif he divided them equally among the eleven monkeys in the first cage hewould have one nut over; if he divided them equally among the thirteenmonkeys in the second cage there would be eight left; if he divided themamong the seventeen monkeys in the last cage three nuts would remain.He also found that ifhe divided them equally among the forty-one monkeysin all three cages, or among the monkeys in any two cages, there wouldalways be some left over.What is the smallest number of nuts that the man could have had in his bag?

205. SHARING THE APPLESDora Crackham the other morning asked her brother this question: "Ifthree boys had a hundred and sixty-nine apples which they shared in theratio of one-half, one-third, and one-fourth, how many apples did eachreceive?"

206. SAWING AND SPLITTINGColonel Crackham, one morning at the breakfast table, said that two menof his acquaintance can saw five cords of wood per day, or they can spliteight cords of wood when sawed. He wanted to know how many cords mustthey saw in order that they may be occupied for the rest of the day insplitting it.

207. THE BAG OF NUTSGeorge Crack ham put five paper bags on the breakfast table. On beingasked what they contained, he said:"Well, I have put a hundred nuts in these five bags. In the first and secondthere are altogether fifty-two nuts; in the second and third there are forty-Miscellaneous Puzzles 65three; in the third and fourth, thirty-four; in the fourth and fifth, thirty."How many nuts are there in each bag?

208. DISTRIBUTING NUTSAunt Martha bought some nuts. She gave Tommy one nut and a quarter ofthe remainder; Bessie then received one nut and a quarter of what were left;Bob, one nut and a quarter of the remainder; and, finally. Jessie received onenut and a quarter of the remainder. It was then noticed that the boys had received exactly 100 nuts more than the girls.How many nuts had Aunt Martha retained for her own use?

209. JUVENILE HIGHWAYMENThree juvenile highwaymen, returning from the market town movie house,called upon an apple woman "to stand and deliver." Tom seized half of theapples, but returned ten to the basket; Ben took one-third of what were left,but returned two that he did not fancy; Jim took half of the remainder, butthrew back one that was worm eaten. The woman was then left with onlytwelve in her basket.How many apples had she before the raid was made?

210. BUYING DOG BISCUITSA salesman packs his dog biscuits (all of one quality) in boxes containing

16, 17, 23, 24, 39, and 40 Ibs. respectively, and he will not sell them in anyother way, or break into a box. A customer asked to be supplied with 100 Ibs.of the biscuits.Could you have carried out the order? If not, how near could you have gotto making up the 100 Ibs.? Of course, he has an ample supply of boxesof each size.

211. THE THREE WORKMEN"Me and Bill," said Casey, "can do the job for you in ten days, but give meAlec instead of Bill, and we can get it done in nine days."

66 Arithmetic & Algebraic Problems"I can do better than that," said Alec. "Let me take Bill as a partner, andwe will do the job for you in eight days."Then how long would each man take over the job alone?

212. WORKING ALONEAlfred and Bill together can do a piece of work in twenty-four days. IfAlfred can do only two-thirds as much as Bill, how long will it take each ofthem to do the work alone?

213. THE FIRST "BOOMERANG" PUZZLEOne of the most ancient forms of arithmetical puzzle is that which I callthe "Boomerang." Everybody has been asked at some time or another to"Think of a number," and, after going through some process of privatecalculation, to state the result, when the questioner promptly tells you thenumber you thought of. There are hundreds of varieties of the puzzle.The oldest recorded example appears to be that given in the Arithmetica ofNicomachus, who died about the year 120. He tells you to think of anywhole number between 1 and 100 and then divide it successively by 3, 5, and

7, telling him the remainder in each case. On receiving this information hepromptly discloses the number you thought of.Can the reader discover a simple method of mentally performing this feat?If not, he will perhaps be interested in seeing how the ancient mathematiciandid it.

214. LONGFELLOW'S BEESWhen Longfellow was Professor of Modern Languages at Harvard Collegehe was accustomed to amuse himself by giving more or less simple arithmeticalpuzzles to the students. Here is an example:If one-fifth of a hive of bees flew to the ladamba flower, one-third flew tothe slandbara, three times the difference of these two numbers flew to an arbor,and one bee continued to fly about, attracted on each side by the fragrantketaki and the malati, what was the number of bees?Miscellaneous Puzzles 67

215. LlLIVATI, 1150 A.D.Here is a little morning problem from Lilivati (1150 A.D.).Beautiful maiden, with beaming eyes, tell me which is the number that,multiplied by 3, then increased by three-fourths of the product, divided by 7,diminished by one-third of the quotient, multiplied by itself, diminished by

52, the square root found, addition of 8, division by 10, gives the number 2?This, like so many of those old things, is absurdly easy if properly attacked.

216. BIBLICAL ARITHMETICIf you multiply the number of Jacob's sons by the number of times whichthe Israelites compassed Jericho on the seventh day, and add to the productthe number of measures of barley which Boaz gave Ruth, divide this by thenumber of Haman's sons, subtract the number of each kind of clean beasts thatwent into the Ark, multiply by the number of men that went to seek Elijahafter he was taken to Heaven, subtract from this Joseph's age at the time hestood before Pharaoh, add the number of stones in David's bag when he killedGoliath, subtract the number of furlongs that Bethany was distant fromJerusalem, divide by the number of anchors cast out when Paul was shipwrecked, subtract the number of persons saved in the Ark, and the answerwill be the number of pupils in a certain Sunday school class.How many pupils are in the class?

217. THE PRINTER'S PROBLEMA printer had an order for 10,000 bill forms per month, but each month thename of the particular month had to be altered: that is, he printed 10,000"JANUARY," 10,000 "FEBRUARY," 10,000 "MARCH," etc.; but as theparticular types with which these words were to be printed had to be speciallyobtained and were expensive, he only purchased just enough movable typesto enable him, by interchanging them, to print in turn the whole of the monthsof the year.How many separate types did he purchase? Of course, the words wereprinted throughout in capital letters, as shown.

68 Arithmetic & Algebraic Problems

218. THE SWARM OF BEESHere is an example of the elegant way in which Bhaskara, in his great work,Lilivati, in 1150, dressed his little puzzles:The square root of half the number of bees in a swarm has flown out upona jessamine bush; eight-ninths of the whole swarm has remained behind; onefemale bee flies about a male that is buzzing within the lotus flower into whichhe was allured in the night by its sweet odor, but is now imprisoned init. Tell me the number of bees.

219. BLINDNESS IN BATSA naturalist, who was trying to pull the leg of Colonel Crackham, said thathe had been investigating the question of blindness in bats."I find," he said, "that their long habit of sleeping in dark corners duringthe day, and only going abroad at night, has really led to a great prevalence ofblindness among them, though some had perfect sight and others could see outof one eye. Two of my bats could see out of the right eye, just three of themcould see out of the left eye, four could not see out of the left eye, andfive could not see out of the right eye."He wanted to know the smallest number of bats that he must have examinedin order to get these results.

220. A MENAGERIEA travelling menagerie contained two freaks of nature-a four-footed birdand a six-legged calf. An attendant was asked how many birds and beaststhere were in the show, and he said:"Well, there are 36 heads and 100 feet altogether. You can work it out foryourself."How many were there?

221. SHEEP STEALINGSome sheep stealers made a raid and carried off one-third of the flock ofsheep and one-third of a sheep. Another party stole one-fourth of whatMiscellaneous Puzzles 69remained and one-fourth of a sheep. Then a third party of raiders carried offone-fifth of the remainder and three-fifths of a sheep, leaving 409 behind. Whatwas the number of sheep in the flock?

222. SHEEP SHARINGA correspondent (c. H. P.) puts the following little question:An Australian farmer dies and leaves his sheep to his three sons. Alfred isto get 20 per cent more than John, and 25 per cent more than Charles. John'sshare is 3,600 sheep. How many sheep does Charles get? Perhaps readers maylike to give this a few moments' consideration.

223. THE ARITHMETICAL CABBYThe driver of the taxicab was wanting in civility, so Mr. Wilkins askecl himfor his number."You want my number, do you?" said the driver. "Well, work it out foryourself. If you divide my number by 2, 3, 4, 5, or 6 you will find there is always lover; but if you divide it by II there ain't no remainder. What's more,there is no other driver with a lower number who can say the same."What was the fellow's number?

224. THE LENGTH OF A LEASE"I happened to be discussing the tenancy of a friend's property," said theColonel, "when he informed me that there was a 99 years' lease. I asked himhow much of this had already expired, and expected a direct answer. But hisreply was that two-thirds of the time past was equal to four-fifths of the timeto come, so I had to work it out for myself."

225. MARCHING AN ARMYA body of soldiers was marching in regular column, with five men more indepth than in front. When the enemy came in sight the front was increasedby 845 men, and the whole was thus drawn up in five lines. How many menwere there in all?

70 Arithmetic & Algebraic Problems

226. THE YEAR 1927A French correspondent sends the following little curiosity. Can you findvalues for p and q so that pq - qP = 1927? To make it perfectly clear, we willgive an example for the year 1844, where p = 3, and q = 7:

37 - 73 = 1844.Can you express 1927 in the same curious way?

227. BOXES OF CORDITECordite charges (writes W. H. 1.) for 6-inch howitzers were served out fromammunition dumps in boxes of IS, 18, and 20."Why the three different sizes of ,?oxes?" I asked the officer on the dump.He answered, "So that we can give any battery the number of charges itneeds without breaking a box."This was ~n excellent system for the delivery of a large number of boxes,but failed in small cases, like 5, 10,25, and 61. What is the biggest number ofcharges that cannot be served out in whole boxes of IS, 18, and 20? It is nota very large number.

228. THE ORCHARD PROBLEMA market gardener was planting a new orchard. The young trees were arranged in rows so as to form a square, and it was found that there were 146trees unplanted. To enlarge the square by an extra row each way he had tobuy 31 additional trees.How many trees were there in the orchard when it was finished?

229. BLOCKS AND SQUARESHere is a curious but not easy puzzle whose author is not traced.Three children each possess a box containing similar cubic blocks, thesame number of blocks in every box. The first girl was able, using all herblocks, to make a hollow square, as indicated by A. The second girl madea larger square, as B. The third girlmade a still larger square, as C, buthad four blocks left over for the corners, as shown. Each girl used all herblocks at each stage. What is thesmallest number of blocks that eachbox could have contained?The diagram must not be taken totruly represent the proportion of thevarious squares.Miscellaneous Puzzles 71

230. FIND THE TRIANGLEThe sides and height of a triangle are four consecutive whole numbers.What is the area of the triangle?

231. COW, GOAT, AND GOOSEA farmer found that his cow and goat would eat all the grass in a certainfield in forty-five days, that the cow and the goose would eat it in sixty days,but that it would take the goat and the goose ninety days to eat it down. Now,if he had turned cow, goat, and goose into the field together, how long wouldit have taken them to eat all the grass?Sir Isaac Newton showed us how to solve a puzzle of this kind with thegrass growing all the time; but, for the sake of greater simplicity, we will assume that the season and conditions were such that the grass was not growing.

232. THE POSTAGE STAMPS PUZZLEA youth who collects postage stamps was asked how many he had in hisalbum, and he replied:"The number, if divided by 2, will give a remainder 1; divided by 3, a remainder 2; divided by 4, a remainder 3; divided by 5, a remainder 4; dividedby 6, a remainder 5; divided by 7, a remainder 6; divided by 8, a remainder

7; divided by 9, a remainder 8; divided by 10, a remainder 9. But there arefewer than 3,000."Can you tell how many stamps there were in the album?

72 Arithmetic & Algebraic Problems

233. MENTAL ARITHMETICTo test their capacities in mental arithmetic, Rackbrane asked his pupils theother morning to do this:Find two whole numbers (each less than 10) such that the sum of theirsquares, added to their product, will make a square.The answer was soon found.

234. SHOOTING BLACKBIRDSTwice four and twenty blackbirdsWere sitting in the rain;I shot and killed a seventh part,How many did remain?

235. THE SIX ZEROSA B CIII III 100

333 333 000

555 500 005

777 077 007

999 090 999

2,775 I,ll I I, IIIWrite down the little addition sum A, which adds up 2,775. Now substitutesix zeros for six of the figures, so that the total sum shall be I, Ill. It will beseen that in the case B five zeros have been substituted, and in case C ninezeros. But the puzzle is to do it with six zeros.

236. MULTIPLICATION DATESIn the year 1928 there were four dates which, written in a well-knownmanner, the day multiplied by the month will equal the year. These are

28/1/28, 14/2/28, 7/4/28, and 4/7/28. How many times in this century1901-2000, inclusive-does this so happen? Or, you can try to find out whichyear in the century gives the largest number of dates that comply with theconditions. There is one year that beats all the others.Miscellaneous Puzzles 73

237. SHORT CUTSWe have from time to time given various short cuts in mental arithmetic.Here is an example that will interest those who are unfamiliar with the process.Can you multiply 993 by 879 mentally?It is remarkable that any two numbers of two figures each, where the tensare the same, and the sum of the units digits make 10, can always be multiplied mentally thus: 97 X 93 = 9,021. Multiply the 7 by 3 and set it down,then add 1 to the 9 and multiply by the other 9, 9 X 10 = 90.This is very useful for squaring any number ending in 5, as 852 = 7,225.With two fractions, when we have the whole numbers the same and the sumof the fractions equal unity, we get an easy rule for multiplying them. Take

7¥.! X 73/4 = 5M16. Multiply the fractions = 1'16, add 1 to one of the 7's, andmultiply by the other, 7 X 8 = 56.

238. MORE CURIOUS MULTIPLICATIONHere is Professor Rackbrane's latest:What number is it that, when multiplied by 18,27,36,45,54,63,72,81, or

99, gives a product in which the first and last figures are the same as those inthe multiplier, but which when multiplied by 90 gives a product in which thelast two figures are the same as those in the multiplier?

239. CROSS-NUMBER PUZZLEOn the following page is a cross-number puzzle on lines similar to those ofthe familiar cross-word puzzle. The puzzle is to place numbers in the spacesacross and down, so as to satisfy the following conditions:Across-I. a square number; 4. a square number; 5. a square number;

8. the digits sum to 35; 11. square root of 39 across; 13. a square number; 14.a square number; 15. square of 36 across; 17. square of half 11 across;

18. three similar figures; 19. product of 4 across and 33 across; 21. a squarenumber; 22. five times 5 across; 23. all digits alike, except the central one;

25. square of 2 down; 27. see 20 down; 28. a fourth power; 29. sum of 18across and 31 across; 31. a triangular number; 33. one more than 4 times 36across; 34. digits sum to 18, and the three middle numbers are 3; 36. an oddnumber; 37. all digits even, except one, and their sum is 29; 39. a fourth power;

40. a cube number; 41. twice a square.

74 Arithmetic & Algebraic ProblemsDown-I. reads both ways alike; 2. square root of 28 across; 3. sum of 17across and 21 across; 4. digits sum to 19; 5. digits sum to 26; 6. sum of 14 acrossand 33 across; 7. a cube number; 9. a cube number; 10. a square number; 12.digits sum to 30; 14. all similar figures; 16. sum of digits is 2 down; 18. allsimilar digits except the first, which is I; 20. sum of 17 across and 27 across;

21. a multiple of 19; 22. a square number; 24. a square number; 26. square of 18across; 28. a fourth power of 4 across; 29. twice 15 across; 30. a triangularnumber; 32. digits sum to 20 and end with 8; 34. six times 21 across; 35. a cubenumber; 37. a square number; 38. a cube number.

240. COUNTING THE LOSSAn officer explained that the force to which he belonged originally consistedof a thousand men, but that it lost heavily in an engagement, and the survivorssurrendered and were marched down to a prisoner of war camp.On the first day's march one-sixth of the survivors escaped; on the secondday one-eighth of the remainder escaped, and one man died; on the third day'smarch one-fourth of the remainder escaped. Arrived in camp, the rest wereset to work in four equal gangs.How many men had been killed in the engagement?Miscellaneous Puzzles 75

241. THE TOWER OF PISA"When I was on a little tour in Italy, collecting material for my book onImprovements in the Cultivation of Macaroni," said the Professor, "I happenedto be one day on the top of the Leaning Tower of Pisa, in company with anAmerican stranger. 'Some lean!' said my companion. 'I guess we can build abit straighter in the States. If one of our skyscrapers bent in this way therewould be a hunt round for the architect.'"I remarked that the point at which we leant over was exactly 179 feet fromthe ground, and he put to me this question: 'If an elastic ball was droppedfrom here, and on each rebound rose exactly one-tenth of the height fromwhich it fell, can you say what distance the ball would travel before it came torest?' I found it a very interesting proposition."

242. A MATCHBOARDING ORDERA man gave an order for 297 ft. of match boarding of usual width and thickness. There were to be sixteen pieces, all measuring an exact number of feet-no fractions of a foot. He required eight pieces of the greatest length, theremaining pieces to be I ft., 2 ft., or 3 ft. shorter than the greatest length.How was the order carried out? Supposing the eight of greatest length were

15 ft. long, then the others must be made up of pieces of 14 ft., 13 ft., or 12 ft.lengths, though every one of these three lengths need not be represented.

243. GEOMETRICAL PROGRESSIONProfessor Rackbrane proposed, one morning, that his friends should writeout a series of three or more different whole numbers in geometrical progression, starting from 1, so that the numbers should add up to a square. Thus,

1 + 2 + 4 + 8 + 16 + 32 = 63.But this is just one short of bemg a square. I am only aware of two answersin whole numbers, and these will be found easy to discover.

244. A PAVEMENT PUZZLETwo square floors had to be paved with stones each I ft. square. The number of stones in both together was 2,120, but each side of one floor was 12 ft.

76 Arithmetic & Algebraic Problemsmore than each side of the other floor. What were the dimensions of the twofloors?

245. THE MUDBURY WAR MEMORIALThe worthy inhabitants of Mudbury-in-the-Marsh recently erected a warmemorial, and they proposed to enclose a piece of ground on which it standswith posts. They found that if they set up the posts I ft. apart they would havetoo few by 150. But if they placed them a yard apart there would be too manyby 70.How many posts had they in hand?

246. MONKEY AND PULLEYHere is a funny tangle. It is a mixture of Lewis Carroll's "Monkey andPulley," Sam Loyd's "How old was Mary?" and some other trifles. But it isquite easy if you have a pretty clear head.A rope is passed over a pulley. It has a weight at one end and a monkey atthe other. There is the same length of rope on either side and equilibrium ismaintained. The rope weighs four ounces per foot. The age of the monkey andthe age of the monkey's mother together total four years. The weight of themonkey is as many pounds as the monkey's mother is years old. The monkey'smother is twice as old as the monkey was when the monkey's mother was halfas old as the monkey will be when the monkey is three times as old as themonkey's mother was when the monkey's mother was three times as old asthe monkey. The weight of the rope and the weight at the end was half as muchagain as the difference in weight between the weight of the weight and theweight and the weight of the monkey. Now, what was the length of the rope?

247. UNLUCKY BREAKDOWNSOn an occasion of great festivities a considerable number of townspeoplebanded together for a day's outing and pleasure. They pressed into servicenearly every wagon in the place, and each wagon was to carry the same number of persons. Half-way ten of these wagons broke down, so it was necessaryfor every remaining wagon to carry one more person. Unfortunately, whenthey started for home, it was found that fifteen more wagons were in such badMiscellaneous Puzzles 77condition that they could not be used; so there were three more persons inevery wagon than when they started out in the morning.How many persons were there in the party?

248. PAT IN AFRICASome years ago ten of a party of explorers fell into the hands of a savagechief, who, after receiving a number of gifts, consented to let them go after halfof them had been flogged by the Chief Medicine Man. There were fiveBritons and five native carriers, and the former planned to make the floggingfall on the five natives. They were all arranged in a circle in the order shownin the illustration, and Pat Murphy (No. I) was given a number to countround and round in the direction he is pointing. When that number fell on aman he was to be taken out for flogging, while the counting went on fromwhere it left off until another man fell out, and so on until the five had beenselected for punishment.If Pat had remembered the number correctly, and had begun at the rightman, the flogging would have fallen upon all the five natives. But poor Patmistook the number and began at the wrong man, with the result that theBritons all got the flogging and the natives escaped.

78 Arithmetic & Algebraic ProblemsCan you find (I) the number the Irishman selected and the man at whomhe began to count, and (2) the number he ought to have used and the manat whom the counting ought to have begun? The smallest possible number isrequired in each case.

249. BLENDING THE TEASA grocer buys two kinds of tea-one at 32¢ per pound, and the other, abetter quality, at 40¢ a pound. He mixes together some of each, which heproposes to sell at 43¢ a pound, and so make a profit of 25 per cent on the cost.How many pounds of each kind must he use to make a mixture of a hundredpounds weight?

250. THE WEIGHT OF THE FISHThe Crackhams had contrived that their tour should include a certain placewhere there was good fishing, as Uncle Jabez was a good angler and theywished to givll him a day's sport. It was a charming spot, and they made apicnic of the occasion. When their uncle landed a fine salmon trout there wassome discussion as to its weight. The Colonel put it into the form of a puzzle,saying:"Let us suppose the tail weighs nine ounces, the head as much as the tailand half the body, and the body weighs as much as the head and tail together.Now, if this were so, what would be the weight of the fish?"

251. CATS AND MICEOne morning, at the breakfast table, Professor Rackbrane's party were discussing organized attempts to exterminate vermin, when the Professor suddenly said:"If a number of cats killed between them 999,919 mice, and every catkilled an equal number of mice, how many cats must there have been?"Somebody suggested that perhaps one cat killed the lot; but Rackbrane replied that he said "cats." Then somebody else suggested that perhaps 999,919cats each killed one mouse, but he protested that he used the word "mice."He added, for their guidance, that each cat killed more mice than there werecats. What is the correct answer?Miscellaneous Puzzles 79

252. THE EGG CABINETA correspondent (T. S.) informs us that a man has a cabinet for holdingbirds' eggs. There are twelve drawers, and all-except the first drawer, whichonly holds the catalog-are divided into cells by intersecting wooden strips,each running the entire width or length of a drawer. The number of cells inany drawer is greater than that of the drawer above. The bottom drawer,No. 12, has twelve times as many cells as strips, No. 11 has eleven times asmany cells as strips, and so on.Can you show how the drawers were divided-how many cells and stripsin each drawer? Give the smallest possible number in each case.

253. THE IRON CHAINTwo pieces of iron chain were picked up on the battlefield. What purposethey had originally served is not certain, and does not immediately concernus. They were formed of circular links (all of the same size) out of metal halfan inch thick. One piece of chain was exactly 3 ft. long, and the other 22 in.in length. Assuming that one piece contained six links more than the other.how many links were there in each piece of chain?

254. LOCATING THE COINS"Do you know this?" said Dora to her brother. "Just put a dime in one ofyour pockets and a nickel in the pocket on the opposite side. Now the dimerepresents 1O¢ and the nickel, 5¢. I want you to triple the value of the coin inyour right pocket, and double the value of the coin in your left pocket. Addthose two products together and tell me whether the result is odd or even."He said the result was even, and she immediately told him that the dimewas in the right pocket and the nickel in the left one. Every time he tried itshe told him correctly how the coins were located. How did she do it?

255. THE THREE SUGAR BASINSThe three basins on the following page each contain the same number oflumps of sugar, and the cups are empty. If we transfer to each cup one-

80 Arithmetic & Algebraic Problems~~~~~~ ce')l ~~~~~rfjeighteenth of the number of lumpsthat each basin contains, we then findthat each basin holds twelve morelumps than each of the cups. Howmany lumps are there in each basinbefore they are removed?

256. A RAIL PROBLEMThere is a garden railing similar to our design. In each division between twouprights there is an equal number of ornamental rails, and a rail is divided inhalves and a portion stuck on each side of every upright, except that the uprights at the ends have not been given half rails. Idly counting the rails fromone end to the other, we found that there were 1,223 rails, counting two halvesas one rail. We also noticed that the number of those divisions was five morethan twice the number of whole rails in a division.How many rails were there in each division?GeometricalProblems
Geometrical Problems

257. MAKING A PENTAGON"I am about to start on making asilk patchwork quilt," said a lady, "allcomposed of pieces in the form of apentagon. How am I to cut out a truepentagon in cardboard, the sides ofwhich must measure exactly an inch?Of course, I can draw a circle andthen by trial with the compass findfive points equidistant on the circumference" (see the illustration), "butunless I know the correct size of mycircle the pentagon is just as it happens, and the sides are always a littlemore, or -a little less, than an exactinch."Could you show her a simple anddirect way of doing it without anytrial?

258. WITH COMPASSES ONLYCan you show how to mark off the four comers of a square, using thecompasses only? You simply use a sheet of paper and the compasses, andthere is no trick, such as folding the paper.

259. LINES AND SQUARESHere is a simple question. With how few straight lines can you makeexactly one hundred squares?Thus, in the first diagram it will be found that with nine straight lines I

83

84 Geometrical ProblemsA 11> e 1)have made twenty squares (twelvewith sides of the length A B, six withsides A C, and two with sides of thelength A D). In the second diagram,although I use one more line, I onlyget seventeen squares. So, you see,everything depends on how the linesare drawn. Remember there must beexactly one hundred squares-neithermore nor fewer.

260. MR. GRINDLE'S GARDEN"My neighbor," said Mr. Grindle,"generously offered me, for a garden,as much land as I could enclose with

9

7 A 8loOfour straight walls measuring 7, 8, 9,and 10 rods in length respectively.""And what was the largest areayou were able to enclose?" asked hisfriend.Perhaps the reader can discoverMr. Grindle's correct answer. Yousee, in the case of three sides the triangle can only enclose one area, butwith four sides it is quite different.For example, it is obvious that thearea of Diagram A is greater thanthat of B, though the sides are thesame.

261. THE GARDEN PATHThis is an old puzzle that I find frequently cropping up. Many find it perplexing, but it is easier than it looks. A man has a rectangular garden, 55 yds.by 40 yds., and he makes a diagonal path, one yard wide, exactly in the man-Triangle, Square, & Other Polygon Puzzlesner indicated in the diagram. Whatis the area of the path?Dimensions for the garden are generally given that only admit of anapproximate answer, but I select figures that will give an answer that isquite exact. The width of the path isexaggerated in the diagram for thesake of clearness.

262. THE GARDEN BED

85Here is quite a simple little puzzle.A man has a triangular lawn of theproportions shown, and he wants tomake the largest possible rectangularflower bed without enclosing the tree.How is he to do it?sible rectangular table top without including a bad knot in the.wood.This will serve to teach the uninitiated a simple rule that may proveuseful on occasion. For example, itwould equally apply to the case of acarpenter who had a triangular boardand wished to cut out the largest pos263. A PROBLEM FOR SURVEYORSThere are tricks in every trade, and the science of numbers contains an infinite number of them. In nearly every vocation of life there are little wrinklesand short cuts that are most useful when known. For example, a man boughta little field, and on page 86 is the scale map (one inch to the rod) that wasgiven to me. I asked my surveyor to tell me the area of the field, but he said

86 Geometrical Problemsit was impossible without some further measurements; the mere length of oneside, seven rods, was insufficient.What was his surprise when I showed him in about two minutes what wasthe area! Can you tell how it is to be done?

264. A FENCE PROBLEMThis is a problem that is very frequently brought to my notice in variousforms. It is generally difficult, but in the form in which I present it it shouldbe easy to the cunning solver. A man has a square field, 60 feet by 60 feet,with other property, adjoining the highway. For some reason he put upa straight fence in the line of the three trees, as shown, and the length offence from the middle tree to the tree on the road was just 91 feet.What is the distance in exact feet from the middle tree to the gate on theroad?...: -o

10Triangle, Square, & Other Polygon Puzzles 87

265. THE FOUR CHECKERSHere is a queer new puzzle that I know will interest my readers considerably. The four checkers are shown exactly as they stood on a squarecheckered board-not necessarily eight squares by eight-but the board wasdrawn with vanishing ink, so that all the diagram except the men has disappeared. How many squares were there in the board and how am I to reconstruct it?@)AD(@I know that each man stood in the middle of a square, one on the edge ofeach side of the board and no man in a corner. It is a real puzzle, until youhit on the method of solution, and then to get the correct answer is absurdlyeasy.

266. A MILITARY PUZZLEAn officer wished to form his men into twelve rows, with eleven men inevery row, so that he could place himself at a point that would be equidistantfrom every row."But there are only one hundred and twenty of us, sir," said one of the men.Was it possible to carry out the order?

267. THE HIDDEN STARThe illustration on page 88 represents a square tablecloth of choice silkpatchwork. This was put together by the members of a family as a little

88 Geometrical Problemsbirthday present for one of its number. One of the contributors supplied aportion in the form of a perfectly symmetrical star, and this has been workedin exactly as it was received. But the triangular pieces so confuse the eye thatit is quite a puzzle to find the hidden star.Can you discover it, so that, if you wished, by merely picking out thestitches, you could extract it from the other portions of the patchwork?

268. A GARDEN PUZZLEThe four sides of a garden are known to be 20, 16, 12, and 10 rods, and ithas the greatest possible area for those sides. What is the area?

269. A TRIANGLE PUZZLEIn the solution to Puzzle No. 230, we said that "there is an infinite numberof rational triangles composed of three consecutive numbers such as 3, 4, 5,and 13, 14, 15." We here show these two triangles. In the first case the area(6) is half of 3 X 4, and in the second case, the height being 12, the area (84)is a half of 12 X 14.Triangle, Square, & Other Polygon Puzzles 89It will be found interesting to discover such a triangle with the smallestpossible three consecutive numbers for its sides, that has an area that may beexactly divided by twenty without remainder.

270. THE DONJON KEEP WINDOWIn The Canterbury Puzzles Sir Hughde Fortibus calls his chief builder and,pointing to a window, says: "Methinks yon window is square, andmeasures, on the inside, one footevery way, and is divided by the narrow bars into four lights, measuringhalf a foot on every side." See ourFigure A. "I desire that another window be made higher up, whose foursides shall also be each one foot, butit shall be divided by bars into eightlights, whose sides shall be all equal."This the craftsman was unable to do,EBA.B

90 Geometrical Problemsso Sir Hugh showed him our Figure B, which is quite correct. But he added,"I did not tell thee that the window must be square, as it is most certainit never could be."Now, an ingenious correspondent, Mr. George Plant, found a flaw in SirHugh's conditions. Something that was understood is not actually stated, andthe window may, as the conditions stand, be perfectly square. How is it to bedone?

271. THE SQUARE WINDOWCrackham told his family that a man had a window a yard square, and itlet in too much light. He blocked up one half of it, and still had a squarewindow a yard high and a yard wide. How did he do it?

272. DIVIDING THE BOARD~[ IIIIAI ,

10 JutA man had a board measuring lO feet in length, 6 inches wide at one end,and 12 inches wide at the other, as shown in the illustration. How far from Bmust the straight cut at A be made in order to divide it into two piecesof equal area?~ ____________ -,·c

110ydS . .D

273. A RUNNING PUZZLEABCD is a square field of fortyacres. The line BE is a straight path,and E is llO yards from D. In a raceAdams runs direct from A to D, butBrown has to start from B, go fromB to E, and thence to D. Each keepsto a uniform speed throughout, andwhen Brown reaches E, Adams is 30yards ahead of him. Which wins therace, and by how much?Triangle, Square, & Other Polygon Puzzles 91

274. THREE TABLECLOTHSMrs. Crackham at the breakfast table recently announced that she had hadas a gift from an old friend three beautiful new tablecloths, each of which isexactly four feet square. She asked the members of her family if they couldtell her the length of the side of the largest square table top that these threecloths will together cover. They might be laid in any way so long as they coverthe surface, and she only wanted the answer to the nearest inch.

275. AN ARTIST'S PUZZLEAn artist wished to obtain a canvas for a painting which would allow forthe picture itself occupying 72 square inches, a margin of 4 inches on top andon bottom, and 2 inches on each side. What are the smallest dimensions possible for such a canvas?

276. IN A GARDEN"My friend Tompkins loves to spring on you little puzzling questions onevery occasion, but they are never very profound," said the Colonel. "I waswalking round his garden with him the other day when he pointed to a rectangular flower bed, and said: 'If I had made that bed 2 feet broader and

3 feet longer it would have been 64 square feet larger; but if it had been 3 feetbroader and 2 feet longer it would then have been 68 square feet larger.What is its length and breadth?'''

277. COUNTING THE TRIANGLESProfessor Rackbrane has just givenme the following puzzle as an exampleof those that interested his party atChristmas. Draw a pentagon, andconnect each point with every otherpoint by straight lines, as in the diagram.How many different triangles arecontained in this figure? To make itquite clear, AFB, AGB, ACB, BFG,AEo t;---+---T---~

92 Geometrical ProblemsBFC, and BGC, are six such triangles. It is not a difficult count if you proceedwith some method, but otherwise you are likely to drop triangles or includesome more than once.

278. A HURDLES PUZZLEThe answers given in the old books to some of the best-known puzzles areoften clearly wrong. Yet nobody ever seems to detect their faults. Here isan example. A farmer had a pen made of fifty hurdles, capable of holding ahundred sheep only. Supposing he wanted to make it sufficiently large tohold double that number, how many additional hurdles must he have?

279. THE ROSE GARDEN"A friend of mine," said ProfessorRackbrane, "has a rectangular garden, and he wants to make exactlyone-half of it into a large bed of roses,with a gravel path of uniform widthround it. Can you find a general rulethat will apply equally to any rectangular garden, no matter what its proportions? All the measurements mustbe made in the garden. A plain ribbon, no shorter than the length of thegarden, is all the material required."

280. CORRECTING A BLUNDERMathematics is an exact science,but first-class mathematicians are apt,like the rest of humanity, to err badlyon occasions. On refering to PeterBarlow's valuable work on The Theoryof Numbers, we hit on this problem:"To find a triangle such that itsthree sides, perpendicular, and theline drawn from one of the angles bisecting the base may be all expressedin rational numbers." He gives as hisanswer the triangle 480, 299, 209,which is wrong and entirely unintelligible.Readers may like to find a correctsolution when we say that all the fivemeasurements may be in whole numbers, and every one of them less thana hundred. It is apparently intendedthat the triangle must not itself beright angled.Triangle, Square, & Other Polygon Puzzles 93

281. THE RUSSIAN MOTORCYCLISTSTwo Army motorcyclists, on the road at Adjbkmlprzll, wish to go toBrczrtwxy, which, for the sake of brevity, are marked in the accompanyingmap as A and B. Now, Pipipoff said: "I shall go to D, which is six miles, and~--------~~--------~then take the straight road to B, another fifteen miles." But Sliponsky thoughthe would try the upper road by way of C. Curiously enough, they found onreference to their cyclometers that the distance either way was exactly thesame. This being so, they ought to have been able easily to answer theGeneral's simple question, "How far is it from A to C?"It can be done in the head in a few moments, if you only know how. Canthe reader state correctly the distance?

282. THOSE RUSSIAN CYCLISTS AGAINHere is another little experience of the two Russian Army motorcyclists thatI described in our last puzzle. In the section from a map given in our illustration on page 94 we are shown three long straight roads, forming a right-angledtriangle. The General asked the two men how far it was from A to B. Pipipoffreplied that all he knew was that in riding right round the triangle, from A to B,from there to C and home to A, his cyclometer registered exactly sixty miles,while Sliponsky could only say that he happened to know that C was exactly

94 Geometrical Problemstwelve miles from the road A to B-that is, to the point D, as shown by thedotted line. Whereupon the General made a very simple calculation in his headand declared that the distance from A to B must be --.Can the reader discover so easily how far it was?

283. THE PRICE OF A GARDENProfessor Rackbrane informed his friends one morning that a neighbortold him that he had been offered a piece of ground for a garden. He said itwas triangular in shape, with the dimensions as shown in our diagram. As theprice proposed was ten dollars per square yard, what will the cost be?

117 yds.Triangle, Square, & Other Polygon Puzzles 95

284. CHOOSING A SITEA man bought an estate enclosed by three straight roads forming an equilateral triangle, as shown in the illustration. Now, he wished to build a housesomewhere on the estate so that if he should have a straight driveway madefrom the front to each of the three roads he might be put to the least expense.The diagram shows one such position. Where should he build the house?

285. THE COUNTER CROSSArrange twenty counters in theform of a cross, in the manner shownin the diagram. Now, in how manydifferent ways can you point out fourcounters that will form a perfectsquare if considered alone? Thus thefour counters composing each arm ofthe cross, and also the four in thecenter, form squares. Squares are alsoformed by the four counters markedA, the four marked B, and so on.How may you remove six countersso that not a single square can be soindicated from those that remain?®o®O O®O@O® ®O®OOO00 o®

96 Geometrical Problems

7 ... \ci···-·~

286. THE TRIANGULARPLANTATIONA man had a plantation of twentyone trees set out in the triangularform shown in our diagram. If hewished to enclose a triangular piece ofground with a tree at each of the threeangles, how many different ways ofdoing it are there from which he mightselect? The dotted lines show threeways of doing it. How many are therealtogether?'1K1. THE CIRCLE AND DISCSDuring a recent visit to a fair wesaw a man with a table, on the oilclothcovering of which was painted a largered circle, and he invited the public tocover this circle entirely with five tindiscs which he provided, and offereda substantial prize to anybody whowas successful. The circular discs wereall of the same size, and each, ofcourse, smaller than the red circle.The diagram, where three discs areshown placed, will make everythingclear.He showed that it was "quite easywhen you know how" by covering upthe circle himself without any apparent difficulty, but many tried overand over again and failed every time.I should explain that it was a condition that when once you had placedany disc you were not allowed to shiftit, otherwise, by sliding them aboutafter they had been placed, it mightbe tolerably easy to do.Let us assume that the red circle issix inches in diameter. What is thesmallest possible diameter (say, tothe nearest half-inch) for the fivediscs in order to make a solutionpossible?Circle Puzzles 97

288. THE THREE FENCES"A man had a circular field," said Crackham, "and he wished to divide itinto four equal parts by three fences, each of the same length. How mightthis be done?""Why did he want them of the same length?" asked Dora."That is not recorded," replied the Colonel, "nor are we told why he wishedto divide the field into four parts, nor whether the fence was of wood or iron,nor whether the field was pasture or arable. I cannot even tell you the man'sname, or the color of his hair. You will find that these points are not essential to the puzzle."

289. SQUARING THE CIRCLEThe problem of squaring the circledepends on finding the ratio of thediameter to the circumference. Thiscannot be found in numbers withexactitude, but we can get it nearenough for all practical purposes.It is equally impossible, by Euclidean geometry, to draw a straight lineequal to the circumference of a givencircle. You can roll a penny carefullyon its edge along a straight line on asheet of paper and get a pretty exactresult, but such a thing as a circulargarden bed cannot be so rolled.The line shown, when straightenedout, is very nearly the exact length ofthe circumference of the accompanying circle. The horizontal part of theline is half the circumference. Couldyou have found it by a simple method,using only pencil, compasses, andruler?

290. THE CIRCLING CARThe outside wheels of a car, running on a circular track, are going twice asfast as the inside ones.What is the length of the circumference described by the outer wheels? Thewheels are five feet apart on the two axles.

98 Geometrical Problems

291. SHARING A GRINDSTONEThree men bought a grindstone twenty inches in diameter. How much musteach grind off so as to share the stone equally, making an allowance of fourinches off the diameter as waste for the aperture? We are not concerned withthe unequal value of the shares for practical use-only with the actual equalquantity of stone each receives.

292. THE WHEELS OF THE CAR"You see, sir," said the automobile salesman, "at present the fore wheel ofthe car I am selling you makes four revolutions more than the rear wheel ingoing 120 yards; but if you have the circumference of each wheel reduced bythree feet, it would make as many as six revolutions more than the rear wheelin the same distance."Why the buyer wished that the difference in the number of revolutions between the two wheels should not be increased does not concern us. Thepuzzle is to discover the circumference of each wheel in the first case. It isquite easy.

293. A WHEEL FALLACYHere is a curious fallacy that I have found to be very perplexing to manypeople. The wheel shown in the illustration makes one complete revolutionin passing from A to B. It is therefore obvious that the line (AB) is exactlyequal in length to the circumference of the wheel. What that length is cannotbe stated with accuracy for any diameter, but we can get it near enough forall practical purposes. Thus, if it is a bicycle wheel with a diameter of 28 inches,we can multiply by 22 and divide by 7, and get the length-88 inches. This isa trifle too much, but if we multiply by 355 and divide by 113 we get 87.9646,which is nearer; or by multiplying by 3.1416 we get 87.9648, which is still morenearly exact. This is just by the way.@.h .... h ... U_ .. h@A ~Circle Puzzles 99Now the inner circle (the large hub in the illustration) also makes onecomplete revolution along the imaginary dotted line (CD) and, since the line(CD) is equal to the line (AB), the circumference of the larger and smallercircles are the same! This is certainly not true, as the merest child can see ata glance. Yet, wherein lies the fallacy?Try to think it out. There can be no question that the hub makes one complete revolution in passing from C to D. Then why does not CD equalin length its circumference?

294. A FAMOUS PARADOXThere is a question that one is perpetually hearing asked, but to whichI have never heard or read an answerthat was satisfactory or really convincing to the ordinary man. It is this,"When a bicycle is in motion, doesthe upper part of each wheel movefaster than the bottom part near theground?" People who are not accustomed to the habit of exact thoughtwill invariably dismiss the subjectwith a laugh, and the reply, "Ofcourse not!" They regard it as tooabsurd for serious consideration. A wheel, they say is a rigid whole, revolvinground a central axis, and if one part went faster than another it would simplybreak in pieces.Then you draw attention of your skeptic to a passing cart and ask him toobserve that, while you can clearly distinguish the spokes as they pass thebottom, and count them as they go by, those at the top are moving so fastthat they are quite indistinguishable. In fact, a wheel in motion looks something like our rough sketch, and artists will draw it in this way. Our friendhas to admit that it is so, but as he cannot explain it he holds to his originalopinion, and probably says, "Well, I suppose it is an optical illusion."I invite the reader to consider the matter: Does the upper part of a wheelmove faster than the lower part?

100 Geometrical Problems

295. ANOTHER WHEEL PARADOXTwo cyclists were resting on a railway bridge somewhere in Sussex, when arailway train went by."That's a London train, going to Brighton," said Henderson."Most of it is," replied Banks, "but parts of it are going direct towardsLondon.""What on earth are you talking about?""I say that if a train is going from London to Brighton, then parts of thattrain are all the time going in the opposite direction-from Brighton toLondon.""You seriously tell me that while I am cycling from Croydon to Eastbourne,parts of my machine are flying back to Croydon?""Steady on, old man," said Banks calmly. "I said nothing about bicycles.My statement was confined to railway trains."Henderson thought it was a mere catch and suggested the smoke or steamof the engine, but his friend pointed out that there might be a strong wind inthe direction. the train was going. Then he tried "the thoughts of the passengers," but here there was no evidence, and these would hardly be parts of thetrain! At last he gave it up. Can the reader explain this curious paradox?

296. A MECHANICAL PARADOXA remarkable mechanical paradox, invented by James Ferguson* aboutthe year 1751, ought to be known by everyone, but, unfortunately, it is not.It was contrived by him as a challenge to a skeptical watchmaker duringa metaphysical controversy. "Suppose," Ferguson said, "I make one wheel asthick as three others and cut teeth in them all, and then put the three wheelsall loose upon one axis and set the thick wheel to turn them, so that its teethmay take into those of the three thin ones. Now, if I turn the thick wheelround, how must it turn the others?" The watchmaker replied that it wasobvious that all three must be turned the contrary way. Then Ferguson produced his simple machine, which anybody can make in a few hours, showingthat, turning the thick wheel which way you would, one of the thin wheels revolved the same way, the second the contrary way, and the third remained[. Ferguson was an eccentnc self-educated Scottish astronomer of the 18th-century, well knownin his day as the "Peasant-Boy Phtlosopher." For a biographtcal sketch, see The EncyclopaediaBritannica, 11th edition.-M. G.]Dividing-the-Plane Puzzles 101stationary. Although the watchmaker took the machine away for carefulexamination, he failed to detect the cause of the strange paradox.

297. THE FOUR HOUSEHOLDERSHere is a square plot of land withfour houses, four trees, a well (W) inthe center, and hedges planted acrosswith four gateways (G).Can you divide the ground so thateach householder shall have an equalportion ofland, one tree, one gateway,an equal length of hedge, and freeaccess to the well without trespass?• 'f\'~~ • 9!•f, ~tft ! 'f

298. THE FIVE FENCESA man owned a large, square,fenced-in field in which were sixteenoak trees, as depicted in the illustration. He wished, for some eccentricreason, to put up five straight fences,so that every tree should be in aseparate enclosure.How did he do it? Just take yourpencil and draw five straight strokesacross the field, so that every tree shallbe fenced off from all the others.

299. THE FARMER'S SONSA farmer once had a square piece of ground on which stood twenty-fourtrees, exactly as shown in the illustration on the following page. He left

102 Geometrical Problemsinstructions in his will that each of his eight sons should receive the sameamount of ground and the same number of trees. How was the land tobe divided in the simplest possible manner?

300. AVOIDING THE MINESHere we have a portion of the North Sea thickly sown with mines by theenemy. A cruiser made a safe passage through them from south to north intwo straight courses, without striking a single mine. Take your pencil and tryto discover how it is done. Go from the bottom of the chart to any point youlike on the chart in a straight line, and then from that point to the top in another straight line without touching a mine.Dividing-the-Plane Puzzles 103Il ~ Q• • Q •Q q Ii qq Ii 12 " <i'q q Q flQ • ~ Q

301. SIX STRAIGHT FENCESA man had a small plantation of thirty-six trees, planted in the form of asquare. Some of these died and had to be cut down in the positions indicatedby the dots in our illustration. How is it possible to put up six straight fencesacross the field, so that everyone of the remaining twenty trees shall be in aseparate enclosure? As a matter of fact, twenty-two trees might be so enclosedby six straight fences if their positions were a little more accommodating, butwe have to deal with the trees as they stand in regular formation, whichmakes all the difference.Just take your pencil and see if you can make six straight lines across thefield so as to leave every tree separately enclosed.

302. DISSECTING THE MOONIn how large a number of piecescan this crescent moon be cut withfive straight cuts of the knife? Thepieces may not be piled or shiftedafter a cut.

104 Geometrical Problems

303. DRAWING A STRAIGHT LINEIf we want to describe a circle we use an instrument that we call a pair ofcompasses, but if we need a straight line we use no such instrument-we employ a ruler or other straight edge. In other words, we first seek a straight lineto produce our required straight line, which is equivalent to using a coin,saucer, or other circular object to draw a circle. Now, imagine yourself to bein such a position that you cannot obtain a straight edge-not even a pieceof thread. Could you devise a simple instrument that would draw yourstraight line, just as the compasses describe a circle?It is an interesting abstract question, but, of course, of no practical value.We shall continue to use the straightedge.

304. DRAWING AN ELLIPSEI suppose a large proportion of my readers are familiar with this trick fordrawing an ellipse. It is very useful if you have to cut a mount for a portraitor to make an oval flower bed. You drive in two pins or nails (or, in the caseof the flower bed, two stakes) and enclose them with an endless bandof thread or string, as shown in our diagram, where the pins are at A and B,and the pencil point, at C, stretches the loop of thread. If you keep the threadtaut and pass the pencil all round until you come back to the starting pointyou will describe the perfect oval shown.But I have sometimes heard the complaint that the method is too haphazard:that it was only by a lot of trials that you can draw an ellipse of the exact dimensions required. This is a delusion, and it will make an interesting littlepuzzle to show at what distance apartthe pins should be placed, and whatlength the string should be, to drawan ellipse, say, twelve inches in lengthby eight inches in breadth.Can you discover the simple rulefor doing this?

305. THE BRICKLAYER'S TASKWhen a man walked in his estate, one of the walls was partly level andpartly over a small rise or hill, precisely as shown in the drawing herewith,Plane Geometry Puzzles 105:. .. _.- ............ _ ..... ----.A cwherein it will be observed that the distance from A to B is the same as fromB to C. Now, the contractor desired and claimed that he should be paid morefor the part that was on the hill than for the part that was level, since(at least, so he held) it demanded the use of more material. But the employerinsisted that he should pay less for that part.It was a nice point, over which they nearly had recourse to the law. Whichof them was in the right?A "----f

306. MEASURING THE RIVERA traveller reaches a river at thepoint A, and wishes to know the widthacross to B. As he has no means ofcrossing the river, what is the easiestway of finding its width?

307. PAT AND HIS p~.r .~, ,~L .. ~ _'.Our diagram represents a field 100 t· ~ ~ "_~~~~1 yards square. Pat and the pig that he iwishes to catch are in opposite corners, I t.

100 yards apart. The pig runs straight t (-for the gateway in the top left-hand t' corner. As the Irishman can run just ' " , 'twice as fast as the pig, you would , ~expect that he would first make , •'lnrightthis is notfo,Pat'sth' g.toway..ofddoingdo"things.it. But{.~ ~ ~ He goes directly for the pig all the it ~~. '11..... ","" ~ '\ -.0;., "'~. • .... ~J time, thus taking a curved course. ~ ~ -~.:- b~""''l;~~~

106 Geometrical ProblemsDoes the pig escape, or does Pat catch it? And if he catches it, exactly howfar does the pig run?

308. THE LADDERThere was some talk at the breakfast table about a ladder that was neededfor some domestic purposes, when Professor Rackbrane interrupted the discussion with this little puzzle:"A ladder was fastened on end against a high wall of a building. A manunfastened it and pulled it out four yards at the bottom. It was then foundthat the top of the ladder had descended just one-fifth of the length of theladder. What was the length of the ladder?"

309. A MA VPOLE PUZZLEDuring a gale a maypole was broken in such a manner that it struck thelevel ground at a distance of twenty feet from the base of the pole, where itentered the earth. It was repaired, and broken by the wind a second time at apoint five feet lower down, and struck the ground at a distance of thirty feetfrom the base.What was the original height of the pole? In neither case did the brokenpart become actually detached.

310. THE BELL ROPEA bell rope, passing through the ceiling above, just touches the belfry floor,and when you pull the rope to the wall, keeping the rope taut, it touches apoint just three inches above the floor, and the wall was four feet from therope when it hung at rest. How long was the rope from floor to ceiling?

311. THE DISPATCH RIDER IN FLANDERSA dispatch rider on horseback, somewhere in Flanders, had to ride with allpossible speed from the position in which he is shown to the spot indicatedby the tent. The distances are marked on the plan. Now, he can ride just twiceas fast over the soft turf (the shaded ground) as he can ride over the loosesand. Can you show what is the quickest possible route for him to take? ThisSolid Geometry Puzzles 107is just one of those practical problems with which the soldier is faced fromday to day when on active service. Important results may hang on the ridertaking the right or the wrong route.SAND*-~----------------~Which way would you have gone? Of course, the turf and the sand extendfor miles to the right and the left with the same respective depths of threemiles and two miles, so there is no trick in the puzzle.

312. THE SIX SUBMARINESReaders may remember a puzzle,to place five pennies so that everypenny shall touch every other penny,that is given in my book, Amusementsin Mathematics, and a correspondenthas suggested that as many as sixcoins can be placed under the conditions if we arrange them as shown inthe upper diagram-that is, with A,B, and C in the form of a triangle,and D, E, and F respectively on thetop of A, B, and C. If we take a section of the coins at XY (see the lower FC

108 Geometrical Problemsdiagram), he held that E and C and also Band F meet at a "mathematicalpoint," and are therefore in contact. But he was wrong, for if E touches C abarrier is set up between Band F. If B touches F, then E cannot touch C. Itis a subtle fallacy that I know will interest my readers. When we say that anumber of things meet at a point (like the spokes of a wheel) only three canbe in contact (each with each) on the same plane.This has led me to propound a new "touching" problem. If five submarines,sunk on the same day, all went down at the same spot where another hadpreviously been sunk, how might they all lie at rest so that everyone of thesix V-boats should touch every other one? To simplify we will say, place sixordinary wooden matches so that every match shall touch every other match.No bending or breaking allowed.

313. ECONOMY IN STRINGOwing to the scarcity of string alady found herself in this dilemma.In making up. a parcel for her son, aprisoner in Germany, she was limitedto using twelve feet of string, exclusive of knots, which passed round theparcel once lengthways and twiceround its girth, as shown in theillustration.What was the largest rectangularparcel that she could make up, subjectto these conditions?

314. THE STONE PEDESTALIn laying the base and cubic pedes- actually used. The base is only a singletal for a certain public memorial, the block in depth.stonemason used cubic blocks ofstone all measuring one foot on everyside. There was exactly the same number of these blocks (all uncut) in thepedestal as in the square base on thecenter of which it stood.Look at the sketch and try to determine the total number of blocksSolid Geometry Puzzles 109

315. A CUBE PARADOXI had two solid cubes of lead, onevery slightly larger than the other,just as shown in the illustration.Through one of them I cut a hole(without destroying the continuity ofits four sides) so that the other cubecould be passed right through it. Onweighing them afterwards it wasfound that the larger cube was stillthe heavier of the two! How was thispossible?

316. THE CARDBOARD BOXReaders must have often remarked on the large number of little things thatone would have expected to have been settled generations ago, and yet neverappear to have been considered. Here is a case that has just occurred to me.If I have a closed cubical cardboard box, by running the penknife alongseven of the twelve edges (it must always be seven) I can lay it out in oneflat piece in various shapes. Thus, in the diagram, if I pass the knife along thedarkened edges and down the invisible edge indicated by the dotted line, Iget the shape A. Another way of cutting produces B or C. It will be seen thatD is simply C turned over, so we will not call that a different shape. Howmany different shapes can be produced?

317. THE AUSTRIAN PRETZELOn the next page is a twisted Vienna bread roll, known as a pretzel. Thetwist, like the curl in a pig's tail, is entirely for ornament. The Wiener pretzel,

110 Geometrical Problemslike some other things, is doomed to of a knife? In what direction wouldbe cut up or broken, and the interest you make the cut?lies in the number of resultant pieces.Suppose you had the pretzel depicted in the illustration lying on thetable before you, what is the greatestnumber of pieces into which youcould cut it with a single straight cut

318. CUTIING THE CHEESEHere is a simple question that will to get an exact answer. I have a piecerequire just a few moments' thought of cheese in the shape of a cube. How

319. THE FLY'S JOURNEYA fly, starting from the point A, cancrawl round the four sides of the baseof this cubical block in four minutes.Can you say how long it will take it tocrawl from A to the opposite uppercorner B?am I to cut it in two pieces with onestraight cut of the knife so that thetwo new surfaces produced by the cutshall each be a perfect hexagon?Of course, if cut in the direction ofthe dotted line the surfaces would besquares. Now produce hexagons.:B

320. THE TANK PUZZLEThe area of the floor of a tank is 6 square feet, the water in it is 9 inchesdeep. How much does the water rise (l) if a I-foot metal cube is put in it;(2) how much farther does it rise if another cube like it is put in by its side?Solid Geometry Puzzles 111

321. THE NOUGAT PUZZLEA block of nougat is 16 inches long, 8 inches wide, and 71f2 inches deep.What is the greatest number of pieces that I can cut from it measuring 5 inchesby 3 inches by 21h inches?

322. AN EASTER EGG PROBLEM"Here is an appropriate Easter egg problem for you," said Professor Rackbrane at the breakfast table. "If I have an egg measuring exactly three inchesin length, and three other eggs all similar in shape, having together the samecontents as the large egg, can you give me exact measurements for the lengthsof the three smaller ones?"

323. THE PEDESTAL PUZZLEAn eccentric man had a block of wood measuring 3 feet by I foot by I foot,which he gave to a wood turner with instructions to turn from it a pedestal,saying that he would pay him a certain sum for every cubic inch of woodtaken from the block in process of turning. The ingenious turner weighed theblock and found it to contain 30 pounds. After he had finished the pedestalit was again weighed, and found to contain 20 pounds. As the original blockcontained 3 cubic feet, and it had lost just one-third of its weight, the turnerasked payment for I cubic foot. But the gentleman objected, saying that theheart of the wood might be heavier or lighter than the outside.How did the ingenious turner contrive to convince his customer thathe had taken not more and not less than I cubic foot from the block?

324. THE SQUIRREL'S CLIMBA squirrel goes spirally up a cylindrical post, making the circuit in four feet.How many feet does it travel to the top if the post is sixteen feet high and threefeet in circumference?

112 Geometrical Problems

325. THE FLY AND THE HONEYI have a cylindrical cup four incheshigh and six inches in circumference.On the inside of the vessel, one inchfrom the top, is a drop of honey, andon the opposi te side of the vessel, oneinch from the bottom on the outside,is a fly. Can you tell exactly how farthe fly must walk to reach the honey?

326. PACKING CIGARETTESA manufacturer sends out his cigarettes in boxes of 160; they are packedin eight rows of 20 each, and exactly fill the box. Could he, by packingdifferently, get more cigarettes than 160 into the box? If so, what is thegreatest number that he could add? At first sight it sounds absurd to expectto get more cigarettes into a box that is already exactly filled, but a moment'sconsideration should give you the key to the paradox.

327. A NEW CUTTING-OUT PUZZLECut the figure into four pieces that will fit together and form a square ...... .o •• :_ .... _ .... ~ ......... ,....... -._-_. -----~- .. -.-, ...... . · "· . •... '0"_:." ___ ,, .......... _ ...... _ .. ;_. __ 0_ · ,.... -.... --,"-'-" · ... --- --~ .. ---_. -," ----- · · ."

328. THE SQUARE TABLE lOPA man had three pieces of beautifulwood, measuring 12 inches, 15 inches,and 16 inches square respectively. Hewanted to cut these into the fewestpieces possible that would fit togetherand form a small square table top

25 inches by 25 inches. How was heto do it? I have found several easysolutions in six pieces, very pretty,but have failed to do it in five pieces.Perhaps the latter is not possible. IDissection Puzzles 11300D 16;,,_know it will interest my readers toexamine the question.

329. THE SQUARES OF VENEERA man has two square pieces of valuable veneer, each measuring 25 inchesby 25 inches. One piece he cut, in the manner shown in our illustration, infour parts that will form two squares, one 20 inches by 20 inches and the other

15 inches by 15 inches. Simply join C to A and D to B. How is he to cut theother square into four pieces that will form again two other squares with sidesin exact inches, but not 20 and 15 as before?

10 5C 10A

10

10 1 5.» .IIID I S

10

114 Geometrical Problems

9

6;IS ~,

3

6

330. DISSECTING THE LETTER ECan you cut this letter E into onlyfive pieces so that they will fit together to form a perfect square? Ihave given all the measurements ininches so that there should be nodoubt as to the correct proportionsof the letter. In this case you are notallowed to turn over any piece.After you have solved the problem,see if you can reduce the number ofpieces to four, with the added freedom to turn over any of the pieces.

331. HEXAGON TO SQUARECan you cut this perfect hexagoninto five pieces that will fit togetherand form a square?

332. SQUARING A STARThis six-pointed star can be cutinto as few as five pieces that will fittogether and form a perfect square.To perform the feat in seven pieces isquite easy, but to do it in five is moredifficult. I introduce the dotted linesmerely to show the true proportionsof the star, which is thus built up oftwelve equilateral triangles.

333. THE MUTILATED CROSSHere is a symmetrical Greek Crossfrom which has been cut a squarepiece exactly equal to one of the armsof the cross. The puzzle is to cut whatremains into four pieces that will fittogether and form a square. This is apleasing but particularly easy cuttingout puzzle.Dissection Puzzles 115

334. THE VICTORIA CROSSWe have shown elsewhere how innumerable puzzles may be devisedon the Greek, St. George, or RedCross, so familiar to us all, composedas it is of five equal squares assembledtogether. Let us now do homage tothe Maltese or Victoria Cross. Cutthe cross shown into seven pieces thatwill fit together and form a perfectsquare. Of course, there must be notrickery or waste of material.In order that the reader may haveno doubt as to the exact proportionsof the cross as given, I have insertedthe dotted lines. As the pieces A andB will fit together to form one of thoselittle squares, it is clear that the areaof the cross is equal to seventeensuch squares.

335. SQUARING THE SWASTIKACut out the swastika on page 116 and then cut it up into four pieces that willfit together and form a squ:ue. There can be no question as to the proportions

116 Geometrical Problems

336. THE MALTESE CROSSCan you cut the star into four piecesand place them inside the frame so asto show a perfect Maltese Cross?of the figure if we regard it as builtup of seventeen equal squares. Youcan divide it up into these seventeensquares with your pencil without difficulty. Now try to cut it into fourpieces to form a single square.

337. THE PIRATES' FLAGHere is a flag taken from a band stripes represented the number of menof pirates on the high seas. The twelve in the band, and when a new man wasadmitted or dropped out a new stripewas added or one removed, as thecase might be. Can you discover howthe flag should be cut into as fewpieces as possible so that they may beput together again and show onlyten stripes? No part of the materialmay be wasted, and the flag must retain its oblong shape.

338. THE CARPENTER'S PUZZLEHere is a well-known puzzle, given in all the old books. A ship's carpenterhad to stop a hole twelve inches square, and the only piece of wood that wasDissection Puzzles 117available measured 9 inches in breadth by 16 inches in length. How did hecut it into only two pieces that would exactly fit the hole? The answer is basedon what I call the "step principle," as shown in the diagram. If you move thepiece marked B up one step to the left, it will exactly fit on A and forma perfect square measuring twelve inches on every side.

12. in. -.----- --- ------ - - -- -- --- ------,:.~ II, (() , 4- in. B I,,,IA ,

16 In.This is very simple and obvious. But nobody has ever attempted to explainthe general law of the thing. As a consequence, the notion seems to have gotabroad that the method will apply to any rectangle where the proportion oflength to breadth is within reasonable limits. This is not so, and I have hadto expose some bad blunders in the case of published puzzles that were supposed to be solved by an application of this step principle, but were reallyimpossible of solution. * Let the reader take different measurements, insteadof 9 in. by 16 in., and see if he can find other cases in which this trick will workin two pieces and form a perfect square.[* See Dudeney's Amusements in Mathematics, Problem 150, where he catches Sam Loyd insuch a blunder.-M. G.)

118 Geometrical Problems

339. THE CRESCENT AND THE STARAHere is a little puzzle on the Crescent and the Star. Look at the illustration, and see if you can determinewhich is the larger, the Crescent orthe Star. If both were cut out of asheet of solid gold, which would bethe more valuable?As it is very difficult to guess bythe eye, I will state that the outer arc,A C B, is a semicircle; the radius ofthe inner arc is equal to the straightline B C; the distance in a straight linefrom A to B is twelve inches; and thepoint of the star, D, contains threesquare inches. It is quite easy to settlethe matter at a glance-when youknow how.

340. THE PATCHWORK QUILTHere is a patchwork quilt that was produced by two young ladies for somecharitable purpose. When they came to join their work it was found that eachlady had contributed a portion of exactly the same size and shape. It is anDissection Puzzles 119amusing puzzle to discover just where these two portions are joined together.Can you divide the quilt into two parts, simply by cutting the stitches, so thatthe portions shall be of the same size and shape? You may think you havesolved it in a few minutes, but-wait and see!

341. THE IMPROVISED CHECKERBOARDSome Englishmen at the front during the Great War wished to pass a restfuJ hour at a game of checkers. They had coins and small stones for the men,but no board. However, one of them found a piece of linoleum as shown inthe illustration, and, as it contained the right number of squares, it was decidedto cut it and fit the pieces together to form a board, blacking some of thesquares afterwards for convenience in playing.An ingenious Scotsman showed how this couJd be done by cutting the stuffin two pieces only, and it is a really good puzzle to discover how he did it.Cut the linoleum along the lines into two pieces that will fit together and formthe board, eight by eight.

342. TESSELLATED PAVEMENTSThe reader must often have noticed, in looking at tessellated pavementsand elsewhere, that a square space had sometimes to be covered with squaretiles under such conditions that a certain number of the tiles have to be

120 Geometrical Problemsa square has been formed with tensquare tiles. As ten is not a squarenumber a certain number of tiles mustbe cut. In this case it is six. It will beseen that the pieces I and I are cutfrom one tile, 2 and 2 from another,and so on.If you had to cover a square spacewith exactly twenty-nine square tilesof equal size, how would you do it?cut in two parts. A familiar exam- What is the smallest number of tilespie is shown in our illustration, where that you need cut in two parts?

343. SQUARE OF SQUARESCutting only along the lines, what is the smallest number of square piecesinto which the diagram can be dissected? The largest number possible is, ofcourse, 169, where all the pieces will be of the same size-one cell-butwe want the smallest number. We might cut away the border on two sides,leaving one square 12 X 12, and cutting the remainder in 25 little squares,making 26 in all. This is better than 169, but considerably more than thefewest possible.Dissection Puzzles 121

344. STARS AND CROSSESThis puzzle calls for a certain amount of ingenuity on account of the awkward position of the cross in the comer.+* *.* * + + +The puzzle is to cut the square into four parts by going along the lines, sothat each part shall be exactly the same size and shape, and each part containa star and a cross.

345. GREEK CROSS PUZZLEHere is a puzzle for our youngerreaders. Cut a square into four piecesin the manner shown, then put thesefour pieces together so as to form asymmetrical Greek cross.

346. SQUARE AND CROSSCut a symmetrical Greek cross intofive pieces, so that one piece shall bea smaller symmetrical Greek cross,entire, and so that the remaining fourpieces will fit together and form aperfect square.

122 Geometrical Problems

347. THREE GREEK CROSSESFROM ONE.r.DIn Amusements in Mathematics(page 168) is given this elegant solution for cutting two symmetricalGreek crosses of the same size froma larger cross. Of course A is cut out

348. MAKING A SQUAREHere is an elegant, but not difficult,little cutting-out puzzle that will interest readers (sent by E. B. E.).*Cut the figure into four pieces, eachof the same size and shape, that willfit together and form a perfect square.entire, and the reader will have nodifficulty in placing the other fourpieces together to form a similarcross. It was then added:"The difficult question now presents itself-how are we to cut threeGreek crosses from one in the fewestpossible pieces? As a matter of factthis problem may be solved in as fewas thirteen pieces; but as I know manyof my readers, advanced geometricians, will be glad to have somethingto work at of which they are notshown the solution, I leave the mystery for the present undisclosed."Only one correspondent has eversucceeded in solving the problem,and his method is exceedingly complex and difficult. Of course the threecrosses must be all of one size.[* Edward B. Escott. See footnote to solution for Problem 332 -M G 1

349. TABLE TOP AND STOOLSMost people are familiar with theold puzzle of the circular table topcut into pieces to form two oval stools,each with a hand hole. The old solution was in eight pieces, but an improved version was given in only sixpieces in Amusements in Mathematics,No. 157.Those who remember the puzzlewill be interested in a solution in asfew as four pieces by the late SamLoyd. Can you cut the circle intofour pieces that will fit together (twoand two) and form two oval stooltops, each with a hand hole?Dv~

351. CHANGING THE SUITYou are asked to cut the spade intothree pieces that will fit together andform a heart. Of course no part of thematerial may be wasted, or it wouldbe absurd, since it would be necessary merely to cut away the stem ofthe spade.Dissection Puzzles J 23

350. TRIANGLE AND SQUARECan you cut each of the equilateraltriangles into three pieces, so that thesix pieces will fit together and form aperfect square?

124 Geometrical Problems

352. PROBLEM OF THE EXTRA CELLHere is a fallacy that is widely known but imperfectly understood. Doubtless many readers will recognize it, and some of them have probably been nota little perplexed. In Figure A the square resembling a chessboard is cut into

8...... 1'3

8..... ...... r--.~JIII Iffi#mtHBAfour pieces along the dark lines, and these four pieces are seen reassembledin diagram B. But in A we have sixty-four of these little square cells, whereasin B we have sixty-five. Where does the additional cell come from?Examine it carefully and see if you can discover how that extra squarecreeps in, and whether it is really possible that you can increase the size ofa slice of bread and butter by merely cutting it in pieces and putting them together again differently.

353. PROBLEM OF THE MISSING CELLCan you arrange the four pieces of the previous puzzle in another way sothat instead of gaining a square we have lost one, the new figure apparentlycontaining only 63 cells?

354. A HORSESHOE PUZZLEHere is an easy little puzzle, but wehave seen people perplexed by it forsome time. Given a paper horseshoe,similar to the one in the illustration,can you cut it into seven pieces, withtwo straight clips of the scissors, sothat each part shall contain a nailhole? There is no objection to yourDissection Puzzles 125shifting the pieces and putting them together after the first cut, only you mustnot bend or fold the paper in any way.

355. SQUARE TABLE TOPFrom a square sheet of paper orcardboard, divided into smallersquares, 7 X 7, as in the diagram, cutout the eight pieces in the manner indicated. The shaded parts are thrownaway. A cabinetmaker had to fittogether these eight pieces of veneerto form a small square table top,

6 X 6, and he stupidly cut that piecenumber eight into three parts.How would you form the squarewithout cutting anyone of the pieces?

356. TWO SQUARES IN ONETwo squares of any relative size can be cut into five pieces, in the mannershown below, that will fit together and form a larger square. But this involvescutting the smaller square. Can you show an easy method of doing it withoutin any way cutting the smaller square?

357. CUTTING THE VENEERA cabinetmaker had a 7 X 7 square checkerboard of beautiful veneer whichhe wished to cut into six pieces to form three separate squares, all differentsizes. How can this be done without any waste, and by cutting only along thelattice lines?

126 Geometrical Problems

8 9 1~

358. IMPROVISED CHESSBOARDA good cutting-out puzzle in as fewas two pieces is not often forthcoming.Here is one that I think cannot fail tointerest readers. Cut this piece ofcheckered linoleum into only twopieces, that will fit together and forma perfect chessboard, without disturbing the checkering of black and white.Of course, it would be easy to cut offthose two overhanging white squaresand put them in the vacant places,but this would be doing it in threepieces.

359. THE PATCHWORK CUSHIONA lady had twenty pieces of silk, allof the same triangular shape and size.She found that four of these would fittogether and form a perfect square, asin the illustration. How was she to fittogether these twenty pieces to form aperfectly square patchwork cushion?There must be no waste, and no allowance need be made for turnings.

360. THE DAMAGED RUGA lady had a valuable Persian rug

12 feet by 9 feet, which was badlydamaged by fire. So she cut from themiddle a strip 8 feet by 1 foot, asshown in the illustration, and thencut the remainder into two pieces thatfitted together and made a perfectlyPaper Folding Puzzles 127square rug 10 feet by 10 feet. How did she do it? Of course, no allowance isto be made for turnings.

361. FOLDING A HEXAGONPaper folding is a branch of puzzledom both instructive and interesting.I do not refer to folding paper into theforms of boxes, boats, frogs, and suchthings, for these are toys rather thanpuzzles, but to the solving of certaingeometrical problems with paper andfingers alone.I will give a comparatively easyexample. Suppose you are given aperfectly square piece of paper, howare you going to fold it so as to indicate by creases a regular hexagon, asshown in the illustration, all ready tobe cut out? Of course, you must useno pencil, measure, or instrument ofany kind whatever. The hexagon maybe in any position in the square.

362. FOLDING A PENTAGONHere is another puzzle in paperfolding of a rather more difficult character than the hexagon example thatwe have considered. If you are givena perfectly square piece of paper, howare you to fold it so as to indicate bycreases a regular pentagon, as in ourillustration all ready to be cut out?Remember that you must use yourfingers alone, without any instrumentor measure whatever.

128 Geometrical Problems

363. FOLDING AN OCTAGONCan you cut the regular octagon from a square piece of paper without usingcompasses or ruler, or anything but scissors? You can fold the paper so as tomake creases.

364. SQUARE AND TRIANGLETake a perfectly square piece of paper, and so fold it as to form the largestpossible equilateral triangle. A triangle in which the sides are the same length\,\\\ \\\. ,\'.0 ..... \,'.'. \, , , .' ...,,.,.,.,. ,,. . IIIIII ,as those of the square, as shown in our diagram, will not be the largest possible. Of course, no markings or measurements may be made except by thecreases themselves.Paper Folding Puzzles 129

365. STRIP TO PENTAGONGiven a ribbon of paper, as in the illustration, of any length-say more thanfour times as long as broad-it can all be folded into a perfect pentagon, withB cJ Ievery part lying within the boundaries of the figure. The only condition is thatthe angle ABC must be the correct angle of two contiguous sides of a regularpentagon. How are you to fold it?

366. A CREASE PROBLEMFold a page, so that the bottom outside corner touches the inside edge andthe crease is the shortest possible. That is about as simple a question as we

2~ __ -:7:C

3>· · • · · ..............could put, but it will puzzle a good many readers to discover just whereto make that fold. I give two examples of folding. It will be seen that the creaseAB is considerably longer than CD, but the latter is not the shortest possible.

130 Geometrical Problems

367. FOLDING POSTAGE STAMPS

1 2 3 -it

5 6 7 8If you have eight postage stamps,

4 by 2, as in the diagram, it is veryinteresting to discover the variousways in which they can be folded sothat they will lie under one stamp, asshown. I will say at once that theycan actually be folded in forty different ways so that number 1 is face upwards and all the others invisiblebeneath it. Numbers 5, 2, 7, and 4will always be face downwards, butyou may arrange for any stamp exceptnumber 6 to lie next to number 1,though there are only two ways eachin which numbers 7 and 8 can bebrought into that position. From alittle law that I discovered, I wasconvinced that they could be foldedin the order 1, 5, 6, 4, 8, 7, 3, 2, andalso 1, 3, 7, 5, 6, 8, 4, 2, with numberI at the top, face upwards, but it puzzled me for some time to discoverhow.Can the reader so fold them without, of course, tearing any of the perforation? Try it with a piece of paperwell creased like the diagram, andnumber the stamps on both sides forconvenience. It is a fascinating puzzle.Do not give it up as impossible!

368. COUNTER SOLITAIREThis simplification of the board of the old game of solitaire lends itself tomany entertaining little pastimes of patience. Copy the simple diagram on asheet of paper or cardboard and use sixteen counters, numbered and placedas shown. The puzzle is to remove all but one counter by a succession of leaps.A counter can leap over another adjoining it to the next square beyond,if vacant, and in making the leap you remove the one jumped over. But noleap can be made in a diagonal direction.® @ ® @ @ I@ @) @CD ® @ ® ® @ (f) ®Moving Counter Puzzles 131The following is a solution in eight moves: 5-13, (6-14,6-5),16-15, (3-11,

3-6),2-10, (8-7,8-16,8-3), (1-9, 1-2, 1-8), (4-12,4-1). This means that 5leaps over 13 and 13 is removed, 6 then leaps over 14 and 14 is removed, andso on. The leaps within parentheses count as one move, because the leaps aremade with the same counter in succession. It will be seen that number 4makes the last leap. Now try to find a solution, in seven moves, in whichnumber I makes the last leap.

369. A NEW LEAP-FROG PUZZLEMake a rough board, as shown,and place seventeen counters on thesquares indicated. The puzzle is to remove all but one by a series of leaping moves, as in checkers or solitaire.A counter can be made to leap overanother to the next square beyond, ifvacant, and you then remove the onejumped over. It will be seen that thefirst leap must be made by the centralcounter, number 9, and one has thechoice of eight directions. A continuous series of leaps with the samecounter will count as a single move.It is required to take off sixteencounters in four moves, leaving thenumber 9 on its original centralsquare. Every play must be a leap.

370. TRANSFERRING THE COUNTERSD EcFDivide a sheet of paper into sixcompartments, as shown in the illustration, and place a pile of fifteencounters, numbered consecutively I,

2,3 ... 15 downwards, in compartment A. The puzzle is to transfer thecomplete pile, in the fewest possible

132 Geometrical Problemsmoves, to compartment F. You can move the counters one at a time to anycompartment, but may never place a counter on one that bears a smallernumber than itself. Thus, if you place I on Band 2 on C, you can then place

1 on 2, but not 2 on 1.

371. MAGIC FIFfEEN PUZZLEThis is Sam Loyd's famous 14-15 puzzle, in which you were asked to getthe 14 and 15 in their proper order by sliding the blocks about in the box. Itwas, of course, impossible of solution. I now propose to slide them about until they shall form a perfect magic square in which the four columns, fourrows, and two diagonals all add up to 30.CD ® @ ®0) ® (j) ®® @ @ @@ @ @It will be found convenient to use numbered counters in place of theblocks. What are your fewest possible moves?

372. TRANSFERRING THE COUNTERSPlace ten counters on the squares of a chessboard as here shown, andtransfer them to the other comer as indicated by the ten crosses. A countermay jump over any counter to the next square beyond, if vacant, eitherhorizontally or vertically, but not diagonally, and there are no captures andno simple moves-only leaps.Not to waste the reader's time it can be conclusively proved that this is impossible. You are now asked to add two more counters so that it may be done.Moving Counter Puzzles 133If you place these, say, on AA, they must, in the end, be found in the corresponding positions BB. Where will you place them?

373. ODDS AND EVENSPlace eight counters in a pile on the middle circle so that they shall be inproper numerical order, with 1 on the top and 8 at the bottom. It is requiredto transfer 1, 3, 5, 7 to the circle marked ODDS, and 2, 4, 6, 8 to the circlemarked EVENS. You can only move one counter at a time from circle tocircle, and you must never place a number on a smaller number, nor an oddand an even number together on the same circle. That is to say, you may

134 Geometrical Problemsplace the 1 on the top of the 3, or the 3 on the top of the 7, or the 2 on the 6,or the 2 on the 4, but you must not place the I on the 2, or the 4 on the 7, asthat would be an odd and even together.What are the fewest possible moves?

374. RAILWAY SHUNTINGHow are the two trains in our illustration to pass one another, and proceedwith their engines in front? The small sidetrack is only large enough to holdone engine or one car at a time, and no tricks, such as ropes and flyingswitches, are allowed. Every reversal-that is, change of direction-of an~-.engine is counted as a move in the solution. What is the smallest number ofmoves necessary?To work on the problem, make a sketch of the track, and on it placea nickel and three pennies (heads up) for the engine and three cars on the left,and a nickel and two pennies (tails up) for the engine and two cars on theright.(j) @ @@ @ ®® @) @® @ ®@ ® @

375. ADJUSTING THE COUNTERS@@@®@@@@@®Place twenty-five counters in asquare in the order shown. Then it isa good puzzle to put them all intoregular order so that the first linereads I 2 3 4 5, and the second 6 7 8 9

10, and so on to the end, by taking upone counter in each hand and makingthem change places. Thus you mighttake up 7 and 1 and replace them as Iand 7. Then take up 24 and 2 andmake them also change places,Moving Counter Puzzles 135when you will have the first two counters properly placed. The puzzle is todetermine the fewest possible exchanges in which this can be done.

376. NINE MEN IN A TRENCHHere are nine men in a trench. Number I is the sergeant, who wishesto place himself at the other end of the line-at point I-all the other menreturning to their proper places as at present. There is no room to pass in thetrench, and for a man to attempt to climb over another would be a dangerousexposure. But it is not difficult with those three recesses, each of which willhold a man.How is it to be done with the fewest possible moves? A man may go anydistance that is possible in a move.

377. BLACK AND WHITEOne morning Rackbrane showed his friends this old puzzle. Place fourwhite and four black counters alternately in a row as here shown. The puzzleis to transfer two contiguous counters to one end then move two contiguouscounters to the vacant space, and so on until in four such moves they form acontinuous line of four black counters followed by four white ones. Remember that two counters moved must always be contiguous ones."Now," he said, "as you know how to do that, try this variation. The conditions are exactly the same, only in moving a contiguous pair you mustmake them change sides. Thus, if you move 5, 6, to the end, you must replacethem in the order 6, 5. How many moves will you now require?"

136 Geometrical Problems

378. THE ANGELICA PUZZLEHere is a little puzzle that will soonbecome quite fascinating if you attempt it. Draw a square with threelines in both directions and place onthe intersecting points eight letteredcounters as shown in our illustration.The puzzle is to move the counters,one at a time, along the lines frompoint to vacant point until you getthem in the order ANGELICA thus:A N GE LC ATry to do this in the fewest possiblemoves. It is quite easy to record yourmoves, as you merely have to writethe letters thus, as an example:A E L N, etc.

379. THE FLANDERS WHEELPlace eight lettered counters on thewheel as shown. Now move them oneat a time along the lines from circle tocircle until the word FLANDERS canbe correctly read round the rim of thewheel as at present, only that the F isin the upper circle now occupied bythe N. Of course two counters cannotbe in a circle at the same time. Findthe fewest possible moves.

380. CATCHING THE PRISONERSMake a rough diagram on a sheet of paper, and use counters to indicate thetwo warders (the men in peaked caps) and the two prisoners. At the beginning the counters must be placed in the squares shown. The first player moveseach of the warders through a doorway to the next cell, in any direction. Thenthe second player moves each prisoner through a doorway to an adjoining cell;Moving Counter Puzzles 137and so on until each warder captures his prisoner. If one warder makesa capture, both he and his captive are out of the game, and the other paircontinue alone.Thus (taking only one side, just for illustration of the moves), the warderon the left may go to F, then the prisoner to D, then the warder to E, thenthe prisoner to A, the warder to B, the prisoner to D, and so on. You maycome to the conclusion that it is a hopeless chase, but it can really be done ifyou use a little cunning.

381. GRASSHOPPERS' QUADRILLEIt is required to make the whitemen change places with the black menin the fewest possible moves. There isno diagonal play or captures. Thewhite men can only move to the rightor downwards, and the black men tothe left or upwards, but they mayleap over one of the opposite color, asin checkers. It is quite easy when onceyou have hit on the method ofsolution.

138 Geometrical Problems

382. THE FOUR PENNIESTake four pennies and arrange themon the table without the assistance ofanother coin or any other means ofmeasurement, so that when a fifthpenny is produced it may be placed inexact contact with each of the four(without moving them) in the manner shown in the illustration. Theshaded circle represents the fifthpenny.If you trust to the eye alone you willprobably fail to get the four in correctposition, but it can be done withabsolute exactitude. How should youproceed?

383. THE SIX PENNIESLay six pennies on the table, andthen arrange them as shown by thesix white circles in the illustration, sothat if a seventh penny (the blackcircle) were produced it could bedropped in the center and exactlytouch each of the six. It is requiredto get it exact, without any dependence on the eye. In this case you arenot allowed to lift any penny off thetable-otherwise there would be nopuzzle at all-nor can any measuringor marking be employed. You requireonly the six pennies.Combinatorial&Topological Problems
Combinatorial & Topological Problems

384. AN IRREGULAR MAGIC SQUAREHere we have a perfect magic squarecomposed of the numbers I to 16 inclusive. The rows, columns, and twolong diagonals all add up 34. Now,supposing you were forbidden to usethe two numbers 2 and 15, butallowed, in their place, to repeat anytwo numbers already used, how wouldyou construct your square so thatrows, columns, and diagonals shouldstill add up 34? Your success will de1 14 I lZIS 4 9 6

10 5 16 3

8 11 2- 13pend on which two numbers you selectas substitutes for the 2 and 15.

385. A MAGIC SQUARE DELUSIONHere is a magic square of the fifth people who have not gone very proorder. I have found that a great many foundly into these things believe that

17 2.4 1

20 5 7

4 6 1~

10 12, 19

11 18 25B

14

20

2.1

2-

15

16

22

3

9

141the central number in all squares ofthis order must be 13. One correspondent who had devoted years toamusing himself with this particularsquare was astounded when I toldhim that any number from I to 25might be in the center. I will showthat this is so. Try to form such amagic square with I in the central cell.

142 Combinatorial & Topological Problems

386. DIFFERENCE SQUARES

4 3 2

7 1 9

6 5 8Can you rearrange the nine digitsin the square so that in all the eightdirections the difference between oneof the digits and the sum of the remaining two shall always be the same?In the example shown it will be foundthat all the rows and columns givethe difference 3; (thus 4 + 2 - 3, andI + 9 - 7, and 6 + 5 - 8, etc.), butthe two diagonals are wrong, because

8 - (4 + I) and 6 - (I + 2) is notallowed: the sum of two must not betaken from the single digit, but thesingle digit from the sum. How manysolutions are there?

387. SWASTIKA MAGIC SQUAREA correspondent sent me this littlecuriosity. It is a magic square, therows, columns, and two diagonals alladding up 65, and all the prime numbers that occur between I and 25(viz., 1,2,3,5,7, II, 13, 17, 19,23) areto be found within the swastika except II. "This number," he says, "inoccult lore is ominous and is associated with the eleven Curses of Ebal,so it is just as well it does not comeinto this potent charm of goodfortune."He is clearly under the impressionthat II cannot be got into the swastikawith the other primes. But in this heis wrong, and the reader may like totry to reconstruct the square so thatthe swastika contains all the ten primenumbers and yet forms a correctmagic square, for it is quite possible.

8 t.2

16 15

388. IS IT VERY EASY?Here is a simple magic square, the three columns, three rows, and twodiagonals adding up 72. The puzzle is to convert it into a multiplying magicsquare, in which the numbers in allthe eight lines if multiplied togethergive the same product in every case.You are not allowed to change, or addto, any of the figures in a cell or useany arithmetical sign whatever! Butyou may shift the two figures withina cell. Thus, you may write 27 as 72,if you like. These simple conditionsmake the puzzle absurdly easy, if youonce hit on the idea; if you miss it, itMagic Square Puzzles 143

27 20 25

22 24 26

23 28 21will appear to be an utter impossibility.

389. MAGIC SQUARE TRICKHere is an advertising trick thatappeared in America many years ago.Place in the empty squares such figures (different in every case, and notwo squares containing the same figure) so that they shall add up 15 in asmany straight directions as possible.A large prize was offered, but nocorrect solution received. Can thereader guess the trick?

390. A FOUR-FIGURE MAGIC SQUAREBecause every cell in this squarecontains the same number, 1234, thethree columns, three rows, and twolong diagonals naturally add up alike.The puzzle is to form and place ninedifferent four-figure numbers (usingthe same figures) so that they alsoshall form a perfect magic square.Remember that the numbers togetherlL34lZ3~

1234

1234 1234

123~ 1234

1Z~4 1234

144 Combinatorial & Topological Problemsmust contain nine of each figure 1, 2, 3, 4, and that they must be four-squarenumbers without fractions or trick of any kind.

391. PROGRESSIVE SQUARESThis is a magic square, adding up 287 in every row, every column, and eachof the two diagonals. If we remove the outer margin of numbers we have another square giving sums of 205. If we again remove the margin there is left amagic square adding up 123. Now fill up the vacant spaces in the diagram withsuch numbers from 1 to 81 inclusive as have not already been given, so thatthere shall be formed a magic square adding up 369 in each oftwenty directions.

20 5S :'0 57 28 71 26

14 31 SO 29 60 35 68

58 46 3S 45 40 36 24

65 33 43 41 39 49 17

64 48 42 37 44 34 IS

10 41 32 53 22 51 72

56 27 52 25 54 11 62

392. CONDITIONAL MAGIC SQUAREThough there is nothing new to be said about the mere construction of aperfect magic square, and the subject has a very large, though scattered,literature of its own, a little variation that has some fresh condition is generally welcome. Here is a not difficult example.Magic Star Puzzles 145Can you form a magic square withall the columns, rows, and two longdiagonals, adding up alike, with thenumbers I to 25 inclusive, placingonly the odd numbers on the shadedsquares in our diagram, and the evennumbers on the other squares? Thereare a good many solutions. Can youfind one of them?

393. THE FIVE-POINTED STARThere is something very fascinatingabout star puzzles. I give an example,taking the case of the simple fivepointed star. It is required to place adifferent number in every circle so thatthe four circles in a line shall add upto 24 in all the five directions. Nosolution is possible with ten consecutive numbers, but you can use anywhole numbers you like.

394. THE SIX-POINTED STARWe have considered the question ofthe five-pointed star. We shall nowfind the six-pointed star even moreinteresting. In this case we can alwaysuse the twelve consecutive numbersI to 12 and the sum of the four numbers in every line will always be 26.The numbers at the six points of thestar may add up to any even numberfrom 24 to 54 inclusive, except 28 and

50, which are impossible. It will beseen that in the example I have given

146 Combinatorial & Topological Problemsthe six points add up to 24. If for every number in its present position yousubstitute its difference from 13 you will get another solution, its complementary, with the points adding up 54, which is 78 less 24. The two complementarytotals will always sum to 78.I will give the total number of different solutions and point out some of thepretty laws which govern the problem, but I will leave the reader this puzzleto solve. There are six arrangements, and six only, in which all the lines offour and the six points also add up to 26. Can you find one or all of them?

395. THE SEVEN-POINTED STARWe have already dealt briefly withstars of five and six points. The caseof the seven-pointed star is particularly interesting. All you have to dois to place the numbers 1, 2, 3, up to

14 in the fourteen disks so that everyline of four disks shall add up to 30.If you make a rough diagram anduse numbered counters, you will soonfind it difficult to break away fromthe fascination of the thing. Possibly,however, not a single reader will hitupon a simple method of solution;his answer, when found, will be obtained by mere patience and luck. Yet,like those of the large majority of thepuzzles given in these pages, the solution is subject to law, if you can unravel it.

396. TWO EIGHT-POINTED STARSThe puzzles of stars with five, six, and seven points that I have given leadus to the eight-pointed star. The star may be formed in two different ways, asshown in our illustration, and the first example is a solution. The numbers 1to 16 are so placed that every straight line of four adds up to 34. If you substitute for every number its difference from 17 you will get the complementarysolution.Magic Star Puzzles 147Let the reader try to discover some of the other solutions, and he will findit a very hard nut, even with this one to help him. But I will present the puzzle in an easy and entertaining form. When you know how, every arrangement in the first star can be transferred to the second one automatically.Every line of four numbers in the one case will appear in the other, only theorder of the numbers will have to be changed. Now, with this informationgiven, it is not a difficult puzzle to find a solution for the second star.

397. FORT GARRISONS Here we have a system of fortifications. It will be seen that there are tenforts, connected by lines of outworks,and the numbers represent thestrength of the small garrisons. TheGeneral wants to dispose these garrisons afresh so that there shall be 100men in everyone of the five lines offour forts. Can you show how it canbe done?The garrisons must be moved bodily-that is to say, you are not allowedto break them up into other numbers.It is quite an entertaining little puzzlewith counters, and not very difficult.

148 Combinatorial & Topological Problems

398. THE CARD PENTAGONMake a rough pentagon on a largesheet of paper. Then throw down theten non-court cards of a suit at theplaces indicated in the illustration, sothat the pips on every row of threecards on the sides of the pentagonshall add up alike. The example willbe found faulty. After you have foundthe rule you will be able to deal thecards into their places without anythought. And there are very few waysof placing them.

400. ROSES, SHAMROCKS,AND THISTLESPlace the numbers I to 12 (onenumber in every design) so that theyshall add up to the same sum in thefollowing seven different ways-viz.,each of the two center columns, eachof the two central rows, the four rosestogether, the four shamrocks together,and the four thistles together.

399. A HEPTAGON PUZZLEUsing the fourteen numbers, 1,2,3,up to 14, place a different number inevery circle so that the three numbersin every one of the seven sides addup to 19.000000GOOO00

402. THE WHEEL PUZZLEPlace the numbers 1 to 19 in the

19 circles, so that wherever there arethree in a straight line they shall addup to 30. It is, of course, very easy.Liquid Pouring Puzzles 149

401. THE MAGIC HEXAGONIn the illustration it will be seenhow we have arranged the numbers

1 to 19 so that all the twelve lines ofthree add up to 23. Six of these linesare, of course, the six sides, and theother six lines radiate from the center.Can you find a different arrangementthat will still add up to 23 in all thetwelve directions? There is only onesuch arrangement to be found.

403. AT THE BROOKIn introducing liquid measuring puzzles in my book Amusements in Mathematics, I have said, "It is the general opinion that puzzles of this class canonly be solved by trial, but I think formulas can be constructed for the solution generally of certain related cases. It is a practically unexplored field forinvestigation."So far as I know, the hint has not been taken and the field is still unexplored,so I recently took advantage of a little unexpected leisure to look into thematter. The result, as I thought probable, was that I struck some new andvery interesting things. For example, let us take the simplest possible case of

150 Combinatorial & Topological Problemsa man who goes to a brook with only two vessels with which to measure agiven quantity of water. When we are dealing, say, with a barrel of wine wemay have complications arising from the barrel being full or empty, from itscapacity and contents being known or unknown, from waste of wine beingpermitted or not permitted, and from pouring back into the barrel being allowed. All these points are eliminated. Is it then possible that any puzzle remains? Let us see.A man goes to the brook with two measures of 15 pints and 16 pints. Howis he to measure exactly 8 pints of water, in the fewest possible transactions?Filling or emptying a vessel or pouring any quantity from one vessel to another counts as a transaction.The puzzle is not difficult, but I think the reader will find it very entertainingand instructive. I need hardly add that no tricks, such as marking or tiltingthe vessels, are allowed.

404. A PROHIBITION POSERLet us now take another step andlook at those cases where we are stillallowed any amount of waste, thoughthe liquid is now limited to a statedquantity.The American prohibition authorities discovered a full barrel of beer,and were about to destroy the liquorby letting it run down a drain whenthe owner pointed to two vesselsstanding by and begged to be allowedto retain in them a small quantity forthe immediate consumption of hishousehold. One vessel was a 7-quartand the other a 5-quart measure. Theofficer was a wag, and, believing itto be impossible, said that if the mancould measure an exact quart intoeach vessel (without any pouringback into the barrel) he might do so.How was it to be done in the fewestpossible transactions without anymarking or other tricks? Perhaps Ishould state that an American barrelof beer contains exactly 120 quarts.Liquid Pouring Puzzles 151

405. PROHIBITION AGAINLet us now try to discover the fewest possible manipulations under the sameconditions as in the last puzzle, except that we may now pour back into thebarrel.

406. THE KEG OF WINEA man had a IO-gailon keg of wine and a jug. One day he drew off a jugfulof wine and filled up the keg with water. Later on, when the wine and waterhad got thoroughly mixed, he drew off another jugful, and again filled up thekeg with water. The keg then contained equal quantities of wine and water.What was the capacity of the jug?

407. WATER MEASUREMENTA maid was sent to the brook with two vessels that exactly measured 7 pintsand II pints respectively. She had to bring back exactly 2 pints of water.What is the smallest possible number of transactions necessary? A "transaction" is filling a vessel, or emptying it, or pouring from one vessel to another.

408. MIXING THE WINEA glass is one-third full of wine, and another glass, with equal capacity, isone-fourth full of wine. Each is filled with water and their contents mixed ina jug. Half of the mixture is poured into one of the glasses. What proportionof this is wine and what part water?

409. THE STOLEN BALSAMThree men robbed a gentleman of a vase containing 24 ounces of balsam.While running away, they met in a forest a glass seller, of whom, in a greathurry, they purchased three vessels. On reaching a place of safety theywished to divide the booty, but they found that their vessels contained 5, II,and 13 ounces respectively. How could they divide the balsam into equalportions?

152 Combinatorial & Topological Problems

410. DELIVERING THE MILKA milkman one morning was driving to his dairy with two lO-gallon cansfull of milk, when he was stopped by two countrywomen, who implored himto sell them a quart of milk each. Mrs. Green had a jug holding exactly 5 pints,and Mrs. Brown a jug holding exactly 4 pints, but the milkman had no measure whatever.How did he manage to put an exact quart into each of the jugs? It was thesecond quart that gave him all the difficulty. But he contrived to do it in asfew as nine transactions-and by a "transaction" we mean the pouring froma can into a jug, or from one jug to another, or from a jug back to the can.How did he do it?

411. THE WAY TO TIPPERARYThe popular bard assures us that"it's a long, long way to Tipperary."Look at the accompanying chart andsee if you can discover the best wayfrom London to "the sweetest girl Iknow." The lines represent stagesfrom town to town, and it is necessaryto get from London to Tipperary inan even number of stages.You will find no difficulty in gettingthere in 3, 5, 7, 9, or 11 stages, butthese are odd numbers and will notdo. The reason they are odd is thatthey all omit the sea passage, a verynecessary stage. If you get to yourdestination in an even number ofstages, it will be because you havecrossed the Irish Sea. Which stage isthe Irish Sea?

412. MARKING A TENNIS COURTThe lines of our tennis court are faint and need remarking. My marker isof such a kind that, though I can start anywhere and finish anywhere, it cannot be lifted off the lines when working without making a mess. I have therefore to go over some of the lines twice.Route & Network Puzzles 153

18 - 18 .....:!l+2-....a .Where should I start and what route should I take, without lifting themarker, to mark the court completely and yet go over the minimum distancetwice? I give the correct proportions of a tennis court in feet. What is thebest route?

413. WATER, GAS, AND ELECTRICITYI think I receive, on an average, about ten letters a month from unknown correspondents respecting this puzzle which I published some years ago under theabove title. They invariably say that someone has shown it to them who didnot know the answer, and they beg me to relieve their minds by telling themwhether there is, or is not, any possible solution. As many of my readers mayhave come across the puzzle and be equally perplexed, I will try to clear upthe mystery for them in a more complete way than I have done in Amusements in Mathematics. First of all here is the thing, as I originally gave it, fortheir consideration.It is required to install water, gas, and electricity from W, G, and E to each

154 Combinatorial & Topological Problemsof the three houses, A, B, and C, without any pipe crossing another. Takeyour pencil and draw lines showing how this should be done. You will soonfind yourself landed in difficulties.

414. CROSSING THE LINESThere is a little puzzle about which, for many years, I have perpetually received enquiries as to its possibility of solution. You are asked to draw thediagram in Figure 1 (exclusive of the little crosses) with three continuousstrokes of the pencil, with out removing the pencil from the paper during astroke, or going over a line twice.As generally understood, it is quite impossible. Wherever I have placed across there is an "odd node," and the law for all such cases is that halfas many lines will be necessary as there are odd nodes-that is, points fromwhich you can depart in an odd number of ways. Here we have, as indicated,~rr I~ I~eight nodes, from each of which you can proceed in three directions (an oddnumber), and, therefore,Jour lines will be required. But, as I have shown inmy book of Amusements, it may be solved by a trick, overriding the conditions as understood. You first fold the paper, and with a thick lead pencildraw CD and EF, in Figure 2, with a single stroke. Then draw the line fromA to B as the second stroke, and GH as the third!During the last few years this puzzle has taken a new form. You are givenRoute & Network Puzzles 155the same diagram and asked to start where you like and try to pass throughevery short line comprising the figure, once and once only, without crossingyour own path. Figure 3 will make quite clear what is meant. It is anattempted solution, but it fails because the line from K to L has not beencrossed. We might have crossed it instead ofKM, but that would be no better.Is it possible? Many who write to me about the puzzle say that though theyhave satisfied themselves as a "pious opinion" that it cannot be done,yet they see no way whatever of proving the impossibility, which is quite another matter. I will show my way of settling the question.

415. THE NINE BRIDGESThe illustration represents the map of a district with a peculiar system ofirrigation. The lines are waterways enclosing the four islands, A, B, C, and D,each with its house, and it will be seen that there are nine bridges available.Whenever Tompkins leaves his house to visit his friend Johnson, who lives inone of the others, he always carries out the eccentric rule of crossing everyone of the bridges once, and once only, before arriving at his destination.How many different routes has he to select from? You may choose any houseyou like as the residence of Tompkins.

156 Combinatorial & Topological Problems

416. SINKING THE FISHING-BOATS~ ~ ~ ~ 0 ~ ~.~ ~ e ~ ~ ~ 6e e lJ ~ lJ 17 6tJ ~ ~ i) ~ <!l ~~ ~ ~ ~ ~ ~ ~~ 6 tJ lJ 6 tJ ~tJ l7 tJ b ~ & &There are forty-nine fishing-boats in the North Sea. How could an enemyram and sink the lot in twelve straight courses, starting from the place shownand finishing up at the same place?

417. GOING TO CHURCHA man living in the house shown in the diagram wants to know what is thegreatest number of different routes by which he can go to the church. TheRoute & Network Puzzles 157possible roads are indicated by the lines, and he always walks either due N,due E, or NE; that is, he goes so that every step brings him nearer tothe church. Can you count the total number of different routes from whichhe may select?

418. THE SUBMARINE NETThe illustration is supposed to represent a portion of a long submarine net,and the puzzle is to make as few cuts as possible from top to bottom, to divide the net into two parts, and so make an opening for a submarine to passthrough.Where would you make the cuts? No cuts can be made through the knots.Remember that the cuts must be made from the top line to the bottom.

419. THE TWENTY-TWO BRIDGESWe have a rough map on page 158 of a district with an elaborate system ofirrigation, as the various waterways and numerous bridges will show. A manset out from one of the lettered departments to pay a visit to a friend living

158 Combinatorial & Topological Problemsin a different department. For thepurpose of pedestrian exercise hecrossed every one of the bridges once,and once only.The puzzle is to show in which twodepartments their houses are situated.It is exceedingly easy if you give it afew moments' thought. You must not,of course, go outside the borders ofthe diagram.

420. FOOTPRINTS IN THE SNOWFour schoolboys, living respectively in the houses A, B, C, and D, attendeddifferent schools. After a snowstorm one morning their footprints were examined, and it was found that no boy had ever crossed the track of anotherboy, or gone outside the square boundary. Take your pencil and continue theirtracks, so that the boy A goes to the school A, the boy B to the school B, andso on, without any line crossing another line.Route & Network Puzzles 159

421. A MONMOUTH TOMBSTONEE I K E R II H o ;] S ~ Ol/tlll E " I EI Ii £ R II H o ;] S £ S :T o H If R £ " J" £ R ,., H 0 J $ E' I E S ~ 0 H II R £ "E R Ii H 0 :T S E I ... I .. S J 0 H II R ER " H 0 ~ $ E I L E ... I IE S T o H tI RtI H (J :r s E I ... e RIEL I E S 7 0 H NIf 0 :r 5 E I " E R E R IE .. I E S ~ o 1/0 :T S E I .. E REX E It E " I E S :r 0If 0 .T S £ I " .. R E R E ... I EO S :r 0 " W If 0 T S E I L E R E ... I IE S :TO H NR IV H 0 T S E I '- E" , E S :T o " N RE R t( H 0 :T S E I .... , E S T 0 H N R E'" E R '"II o 1" s E I E S :r 0 If II R IE 11/, Ii £ R rf Ii 0 of S E S :To If t( It E rf ,E 1 N E- R II If 0 T 8 To H '" R E N I EIn the burial ground attached to St. Mary's Church, Monmouth, is thisarrangement of letters on one of the tombstones. In how many different wayscan these words "HERE LIES JOHN RENIE" be read, starting at thecentral H and always passing from one letter to another that is contiguous?

422. THE FLY'S TOURA fly pitched on the square in thetop left-hand comer of a chessboard,and then proceeded to visit everywhite square. He did this without everentering a black square or ever passing through the same intersection(where a horizontal and a verticalline meet) more than once. Can youshow his route? It can be done inseventeen continuous straight courses.

160 Combinatorial & Topological Problems

423. INSPECTING THE ROADSA man starting from the town A, has to inspect all the roads shown fromtown to town. Their respective lengths, 13, 12, and 5 miles, are all shown.What is the shortest possible route he can adopt, ending his journey whereverhe likes?

424. RAILWAY ROUTESThe diagram below represents a simplified railway system, and we want toknow how many different ways there are of going from A to E, if we nevergo twice along the same line in any journey.Route & Network Puzzles 161This is a very simple proposition, but practically impossible to solve untilyou have hit on some method of recording the routes. You see there are manyways of going, from the short route ABDE, taking one of the large loops, upto the long route ABCDBCDBCDE, which takes you over every line on thesystem and can itself be varied in order in many ways. How many differentways of going are there?

425. A CAR TOURA man started in a car from thetown A, and wished to make a complete tour of these roads, going alongevery one of them once, and onceonly. How many different routes arethere from which he can select? It ispuzzling unless you can devise someingenious method. Every route must Send at the town A, from which youstart, and you must go straight fromtown to town-never turning off atcrossroads.

426. MRS. SIMPER'S HOLIDAY TOURThe illustration represents a plan,very much simplified, of a tour thatmy friend Mrs. Simper proposes tomake next autumn. It will be seen thatthere are twenty towns all connectedby lines of railways. Mrs. Simper livesat the town marked H, and she wantsto visit everyone of the other townsonce, and once only, ending her tourat home.It may interest the reader to knowthat there are just sixty different routesfrom which she may select, counting

162 Combinatorial & Topological Problemsthe reverse way of a route as different. There is a tunnel between Nand 0,and another between Rand S, and the good lady objects very much to goingthrough these. She also wants to delay her visit to D as long as possible in orderto meet the convenience of a friend who resides there. The puzzle is to showMrs. Simper her very best route in these circumstances.

427. SIXTEEN STRAIGHT RUNSA commercial traveller started in his car from the point shown, and wishedto go 76 miles in sixteen straight runs, never going along the same road twice.The dots represent towns and villages, and these are one mile apart. The linesshow the route he selected. It will be seen that he carried out his plan correctly,but six towns or villages were unvisited .• • ••• •Can you show a better route by which he could have gone 76 miles in sixteen straight runs, and left only three towns unvisited?

428. PLANNING TOURSThe illustration represents a map (considerably simplified for our purposes)of a certain district. The circles are towns and villages, and the lines roads.Can you show how five automobile drivers can go from A to A, from B to B,Route & Network Puzzles 163from C to C, from D to D, from E to E, respectively, without one ever crossing the track or going along the same road as another car?Just take your pencil and mark the routes you propose, and you willprobably find it a little puzzling. Of course it makes no difference whichof two similar letters is the startingplace, because we are only concernedwith the routes joining them. You see, if you take the route straight down fromA to A you will have barred out every possible route for the other cars, withthe exception of B to B, because, of course, the drivers are restricted to theroads shown on the map.

429. A MADAM PROBLEMIn how many different ways is itpossible to read the word "MADAM"in the diagram? You may go as youplease, upwards and downwards, forwards and backwards, any way possible along the open paths. But theletters in every case must be contiguous, and you may never pass a letterwithout using it.

164 Combinatorial & Topological Problems

430. THE ENCIRCLED TRIANGLESHere is a little puzzle that will require some patience and judgment tosolve. You have merely to draw thedesign of circle and triangles in asfew continuous strokes as possible.You may go over a line twice if youwish to do so, and begin and endwherever you like.

431. THE SIAMESE SERPENTThe conditions of this puzzle areexceedingly simple. You are asked todraw as much as possible of the serpent in one continuous line. Startingwhere you like and ending where youlike, just see how much of the serpentyou can trace without once takingyour pencil off the paper or goingover the same line twice.An artful person might dodge thecondition of going over a line twiceby claiming that he drew half thewidth of the line going forward andthe other half going back; but he isreminded that a line has no breadth!

432. A BUNCH OF GRAPESHere is a rough conventionalizedsketch of a bunch of grapes. Thepuzzle is to make a copy of it withone continuous stroke of the pencil,never lifting the pencil from the paper,nor going over a line twice throughout. You can first try tracing it withthe pencil until you get some idea ofthe general method.Point Alignment Puzzles 165

433. A HOPSCOTCH PUZZLEWe saw some boys playing the ancient and ever popular game of hopscotch,and we wondered whether the figure that they had marked on the groundcould be drawn with one continuous stroke. We found it to be possible.Can the reader draw the hopscotch figure in the illustration without takinghis pencil off the paper or going along the same line twice? The curved lineis not generally used in the game, but we give the figure just as we saw it.([X] I

434. A WILY PUZZLEAn unscrupulous advertiser offered a hundred dollars for a correct solutionto this puzzle:"A life prisoner appealed to the king for pardon. Not being ready to favorthe appeal, the king proposed a pardon on condition that the prisoner shouldstart from cell A and go in and out of each cell in the prison, coming back tothe cell A without going into any celltwice."Either the advertiser had no answer,and knew he had none, or he was prepared to fall back on some trick orquibble. What is the best answer thereader can devise that may be held tocomply with the advertiser's conditions as given?++++++++++++++++++A+ +++++-

435. A TREE PLANTING PUZZLEA man planted thirteen trees in the manner shown on the following page,and so formed eight straight rows with four trees in every row. But he was notsatisfied with that second tree in the horizontal row. As he quaintly put it, "itwas not doing enough work-seemed to be a sort of loafer." It certainly does

166 Combinatorial & Topological Problemsappear to be somewhat out of thegame, as the only purpose it serves isto complete one row. So he set towork on a better arrangement, and inthe end discovered that he could plantthirteen trees so as to get nine rows offour. Can the reader show how itmight be done?..<it':" : \~//~'~r~\ ~:/~:;;;':.: .. ~ .. : .... ~~ .. :>\J>

436. THE TWENTY PENNIESIf sixteen pennies are arranged in the form of a square there will be thesame number of pennies in every row, column, and each of the two longdiagonals. Can you do the same with twenty pennies?

437. TRANSPLANTING THE TREESIi «i~ Q It Ii.~ ~ ti tlQ ~ ~ ~

4 ti ~ ~i ., It <2A man has a plantation of twenty-two trees arranged in the manner hereshown. How is he to transplant only six of the trees so that they shall thenform twenty rows with four trees in every row?

438. A PEG PUZZLEThe illustration represents a square mahogany board with forty-nine holesin it. There are ten pegs to be placed in the positions shown, and the puzzlePoint Alignment Puzzles 167is to remove only three of these pegs to different holes, so that the ten shallform five rows with four pegs in every row. Which three would you move,and where would you place them?• • • • • • •• • • • • • •• • • • • • •• • • ~ • • •• • ~ • ~ • •• ~ • • • ~ •Qr~ • g; • ~ ~

439. FIVE LINES OF FOURThe illustration shows how ten counters may be placed on the points of thediagram where the lines intersect, so that they form five straight lines withfour counters in every line, as indicated by the dotted lines. Can youfind a second way of doing this?Of course a mere reversal or reflection of the given arrangement is notconsidered different-it must be anew scheme altogether, and of courseyou cannot increase the dimensionsof the diagram or alter its shape.:."

440. DEPLOYING BATILESHIPS. ,:: ::::)- .-: ..... .......Ten battleships were anchored in the form shown on page 168. The puzzleis for four ships to move to new positions (the others remaining where they

168 Combinatorial & Topological Problemsare) until the ten form five straight rows with four ships in every row. Howshould the admiral do it?

441. CONSTELLATION PUZZLEThe arrangement of stars in the illustration is known as "The BritishConstellation." It is not given in any star maps or books, and it is very difficult to find on the clearest night for the simple reason that it is not visible.The 21 stars form seven lines with 5 stars in every line .... : tr ~~:......... : ~ : ". w··· ...... . .... """, ~ ... : ... ..... '. '1.(.... .Joo. "r ..... . '. : .... ~: •••. oJ\.~ W':-.; .... w ./ .... * ~ .. (, .... (f -6 ".: .... ; ... ~ : . . 'Q. ... ~Can you rearrange these 21 stars so that they form eleven straight lineswith 5 stars in every line? There are many solutions. Try to find a symmetrical one.Map Coloring Puzzles 169

442. THE FOUR-COLOR MAP THEOREMFor just about fifty years various mathematicians, including De Morgan,Cayley, Kempe, Heawood, Heffter, Wernicke, Birkhoff, Franklin, and manyothers have attempted to prove the truth of this theorem, and in a long andlearned article in the American Mathematical Monthly for July-August, 1923,Professor Brahana, of the University of Illinois, states that "the problem isstill unsolved." It is simply this, that in coloring any map under the conditionthat no contiguous countries shall be colored alike, not more than four colorscan ever be necessary. Countries only touching at a point, like two Blues andtwo Yellows at a in the diagram, are not contiguous. If the boundary line cahad been, instead, at cb, then the two Yellows would be contiguous, but thatwould simply be a different map, and I should have only to substitute Greenfor that outside Yellow to make it all right. In fact, that Yellow might havebeen Green as the map at present stands.I will give, in condensed form, a suggested proof of my own which severalgood mathematicians to whom I have shown it accept as quite valid. Twoothers, for whose opinion I have great respect, think it fails for a reason thatthe former maintain will not "hold water." The proof is in a form that anybody can understand. It should be remembered that it is one thing to be convinced, as everybody is, that the thing is true, but quite another to givea rigid proof of it.

170 Combinatorial & Topological Problems

443. A SWASTIKLAND MAPHere is a puzzle that the reader willprobably think he has solved at almostthe first glance. But will he be correct?Swastikland is divided in the mannershown in our illustration. The LordHigh Keeper of the Maps was orderedso to color this map of the countrythat there should be a different coloron each side of every boundary line.What was the smallest possible number of colors that he required?

444. COLORING THE MAPColonel Crackham asked his youngson one morning to color all thetwenty-six districts in this map in sucha way that no two contiguous districtsshould be of the same color. The ladlooked at it for a moment, and replied,"I haven't enough colors by one inmy box."This was found to be correct. Howmany colors had he? He was notallowed to use black and whiteonly colors.

445. PICTURE PRESENTATIONA wealthy collector had ten valuable pictures. He proposed to make apresentation to a public gallery, but could not make up his mind as to howmany he would give. So it amused him to work out the exact number of different ways. You see, he could give anyone picture, any two, any three, andso on, or give the whole ten.The reader may think it a long and troublesome calculation, but 1 will givea little rule that will enable him to get the answer in all such cases withoutany difficulty and only trivial labor.Miscellaneous Combinatorial Puzzles 171

446. A GENERAL ELECTIONIn how many different ways maya Parliament of 615 members be electedif there are only four parties: Conservatives, Liberals, Socialists, and Independents? You see you might have C. 310, L. 152, S. 150, I. 3; or C. 0, L. 0,S. 0, I. 615; or C. 205, L. 205, S. 205, I. 0; and so on. The candidates are indistinguishable, as we are only concerned with the party numbers.

447. THE MAGISTERIAL BENCHA friend at Singapore asked me some time ago to give him my solution tothis problem. A bench of magistrates (he does not say where) consists of twoEnglishmen, two Scotsmen, two Welshmen, one Frenchman, one Italian, oneSpaniard, and one American. The Englishmen will not sit beside one another,the Scotsmen will not sit beside one another, and the Welshmen also objectto sitting together.In how many different ways may the ten men sit in a straight line so thatno two men of the same nationality shall ever be next to one another?

448. CROSSING THE FERRYSix persons, all related, have to cross a river in a small boat that will onlyhold two. Mr. Webster, who had to plan the little affair, had quarrelled withhis father-in-law and his son, and, I am sorry to say, Mrs. Webster was noton speaking terms with her own mother or her daughter-in-law. In fact, therelations were so strained that it was not safe to permit any of the belligerentsto pass over together or to remain together on the same side of the river. Andto prevent further discord, no man was to be left with two women ortwo men with three women.How are they to perform the feat in the fewest possible crossings? Notricks, such as making use of a rope or current, or swimming across, areallowed.

449. MISSIONARIES AND CANNIBALSThere is a strange story of three missionaries and three cannibals, who hadto cross a river in a small boat that would only carry two men at a time.

172 Combinatorial & Topological ProblemsBeing acquainted with the peculiar appetites of the cannibals, the missionaries could never allow their companions to be in a majority on either side ofthe river. Only one of the missionaries and one of the cannibals could rowthe boat. How did they manage to get across?

450. CROSSING THE RIVERDuring the Turkish stampede in Thrace, a small detachment found itselfconfronted by a wide and deep river. However, they discovered a boat in whichtwo children were rowing about. It was so small that it would only carry thetwo children, or one grown person.How did the officer get himself and his 357 soldiers across the river and leavethe two children finally in joint possession of their boat? And how many timesneed the boat pass from shore to shore?

451. A GOLF COMPETITION PUZZLEI was asked to construct some schedules for players in American golf competitions. The conditions are:(1) Every player plays every other player once, and once only.(2) There are half as many links as players, and every player plays twiceon every links except one, on which he plays but once.(3) All the players play simultaneously in every round, and the last roundis the one in which every player is playing on a links for the first time.I have written out schedules for a long series of even numbers of players upto twenty-six, but the problem is too difficult for this book except in its mostsimple form-for six players. Can the reader, calling the players A, B, C, D,E, and F, and pairing these in all possible ways, such as AB, CD, EF, AF,BD, CE, etc., complete the above simple little table for six players?ROUNDS

1 Z 3 4 5lOT LINKSZ··L1NKS

3"LlNKSMiscellaneous Combinatorial Puzzles 173

452. FOOTBALL RESULTSNear the close of a football season a correspondent informed me that whenhe was returning from Glasgow after the international match between Scotland and England the following table caught his eye in a newspaper:GoalsPlayed Won Lost Drawn For Against PointsScotland 3 3 0 0 7 I 6England 3 2 3 3Wales 3 I I 3 3 3Ireland 3 0 3 0 6 0As he knew, of course, that Scotland had beaten England by 3--0, it struckhim that it might be possible to find the scores in the other five matches fromthe table. In this he succeeded. Can you discover from it how many goalswere won, drawn, or lost by each side in every match?

453. THE DAMAGED MEASUREHere is a new puzzle that is interesting, and it reminds one, though itis really very different, of the classical problem by Bachet concerning theweight that was broken in pieces which would then allow of any weightin pounds being determined from one pound up to the total weight of all thepieces. In the present case a man has a yardstick from which 3 inches havebeen broken off, so that it is only 33 inches in length. Some of the graduationmarks are also obliterated, so that only eight of these marks are legible; yethe is able to measure any given number of inches from 1 inch up to 33 inches.Where are these marks placed?

13 inches.lu

1 :, 1 6As an example, I give in the illustration the case of a 13-inch rod with fourmarkings. If I want to measure 4 inches, I take 1 and 3; for 8 inches, 6 and 2;for 10 inches, 3, 1, and 6; and so on. Of course, the exact measure must betaken at once on the rod; otherwise the single mark of 1 inch repeateda sufficient number of times would measure any length, which would makethe puzzle absurd.

174 Combinatorial & Topological Problems

454. THE SIX COTTAGESA circular road, twenty-seven mileslong, surrounds a tract of wild anddesolate country, and on this roadare six cottages so placed that onecottage or another is at a distance ofone, two, three up to twenty-six milesinclusive from some other cottage.Thus, Brown may be a mile fromStiggins, Jones two miles from Rogers,Wilson three miles from Jones, andso on. Of course, they can walk ineither direction as required.Can you place the cottages at distances that will fulfill these conditions? The illustration is intended togive no clue as to the relative distances.

455. FOUR IN LINEHere we have a board of thirty-six squares, and four counters are so placedin a straight line that every square of the board is in line horizontally, vertically, or diagonally with at least one counter. In other words, if you regardthem as chess queens, every square on the board is attacked by at least onequeen. The puzzle is to find in how many different ways the four countersmay be placed in a straight line so that every square shall thus be in line witha counter.Every arrangement in which thecounters occupy a different set offoursquares is a different arrangement.Thus, in the case of the example given,they can be moved to the next columnto the right with equal effect, or theymay be transferred to either of thetwo central rows of the board. Thisarrangement, therefore, produces foursolutions by what we call reversals orreflections of the board. RememberMiscellaneous Combinatorial Puzzles 175that the counters must always be disposed in a straight line. It will be foundan entertaining little puzzle.

456. FLIES ON WINDOW PANESHere is a window with eighty-onepanes. There are nine flies on as manypanes, and no fly is in line with another one horizontally, vertically, ordiagonally. Six of these flies are verytorpid and do not move, but each ofthe remaining three goes to an adjoining pane. And yet, after this changeof station, no fly is in line with another.Which are the three lively flies, andto which three panes (at present unoccupied) do they pass?~~~

457. CITY LUNCHEONS~~.f&~~r&The clerks attached to the finn of Pilkins and Popinjay arranged that threeof them would lunch together every day at a particular table so long as theycould avoid the same three men sitting down twice together. The same number of clerks of Messrs. Radson, Robson, and Ross decided to do preciselythe same, only with four men at a time instead of three. On working it outthey found that Radson's staff could keep it up exactly three times as manydays as their neighbors.What is the least number of men there could have been in each staff?

458. THE NECKLACE PROBLEMHow many different necklaces can be made with eight beads, where eachbead may be either black or white, the beads being indistinguishable exceptby color?We may have eight white or eight black, or seven white and one black, orsix white and two black, as in our illustration on page 176. Of course, if

176 Combinatorial & Topological Problemsyou exchange black number 3 with 4,or with 5, or with 6, you get differentnecklaces. But if you exchange 3 with

7 it will be the same as 3 with 5, because it is merely turning the necklaceover. So we have to beware of counting such repetitions as different. Theanswer is a much smaller one thanthe reader may anticipate.

459. AN EFFERVESCENT PUZZLEIn how many different ways can the ever appearing together? Of courseletters in the word EFFERVESCES similar letters, such as FF, have nobe arranged in a line without two E's separate identity, so that to interchange them will make no difference.

5:(svWhen the reader has done this heshould try the case where the lettershave to be arranged differently in acircle, as shown, with no two E's together. We are here, of course, onlyconcerned with the order of the lettersand not with their positions on thecircumference, and you must alwaysread in a clockwise direction, as indicated by the arrow.

460. TESSELLATED TILESHere we have twenty tiles, all colored with the same four colors, and theorder of the coloring is indicated by the shadings: thus, the white may represent white; the black, blue; the striped, red; and the dotted, yellow.The puzzle is to select any sixteen of these tiles that you choose and arrangethem in the form of a square, always placing similar colors together-whiteagainst white, red against red, and so on. It is quite easy to make the squaresMiscellaneous Combinatorial Puzzles 177in paper or cardboard, and color them according to taste, but the order of thecolors must be exactly as shown in the illustration.

461. THE THIRTY-SIXLETTER PUZZLEIf you try to fill up this square byrepeating the letters A, B, C, D, E, F,so that no A shall be in a line, across,downwards or diagonally, with another A, no B with another B, no Cwith another C, and so on, you willfind that it is impossible to get in allthe thirty-six letters under these conditions.The puzzle is to place as many letters as possible. Probably the readerA .B C D E Fwill leave more blank spaces thanthere need be.

462. THE TEN BARRELSA merchant had ten barrels of sugar, which he placed in the form of apyramid, as shown on following page. Every barrel bore a different number,except one, which was not marked. It will be seen that he had accidentally

178 Combinatorial & Topological Problemsarranged them so that the numbers inthe three sides added up alike-thatis, to 16.Can you rearrange them so that thethree sides shall sum to the smallestnumber possible? Of course the central barrel (which happens to be 7 inthe illustration) does not come intothe count.

463. LAMP SIGNALLINGTwo spies on the opposite sides of ariver contrived a method for signallingby night. They each put up a stand,like our illustration, and each possessed three lamps which could showeither white, red, or green light. Theyconstructed a code in which everydifferent signal meant a sentence. Youwill, of course, see that a single lamphung on anyone of the hooks couldonly mean the same thing, that twolamps hung on the upper hooks 1and 2 could not be distinguishedfrom two on 4 and 5. Two red lampson 1 and 5 could be distinguishedfrom two on 1 and 6. And two on

1 and 2 would be different from twoon 1 and 3.Remembering the variations ofcolor as well as of position, what isthe greatest number of signals thatcould be sent?

464. THE HANDCUFFED PRISONERSOnce upon a time there were nine prisoners of particularly dangerouscharacter who had to be carefully watched. Every weekday they were takenout for exercise, handcuffed together, as shown in the sketch made by oneof their guards. On no day in anyone week were the same two men toMiscellaneous Combinatorial Puzzles 179be handcuffed together. It will beseen how they were sent out on Monday. Can you arrange the nine menin triplets for the remaining five days?It will be seen that number I cannot again be handcuffed to number 2(on either side), nor number 2 withnumber 3, but, of course, number Iand number 3 can be put together.Therefore, it is quite a different problem from the old one of the FifteenSchoolgirls, and it will be found tobe a fascinating teaser and amply repay for the leisure time spent on itssolution.

465. SEATING THE PARTYAs the Crackham family were taking their seats on starting out on their tourDora asked in how many different ways they could all be seated, as there weresix of them and six seats-one beside the driver, two with their backs to thedriver, and two behind, facing the driver-if no two of the same sex are everto sit side by side?As the Colonel, Uncle Jabez, and George were the only ones who coulddrive, it required just a little thinking out. Perhaps the reader will like towork out the answer concerning which they were all agreed at the end of theday.

466. QUEER GOLFA certain links had nine holes, 300, 250, 200, 325, 275, 350, 225, 375, and

400 yards apart. If a man could always strike the ball in a perfectly straightline and send it exactly one of two distances, so that it would either gotowards the hole, pass over it, or drop into it, what would the two distancesbe that would carry him in the least number of strokes round the whole course?Two very good distances are 125 and 75, which carry you round in twentyeight strokes, but this is not the correct answer.

180 Combinatorial & Topological Problems

467. THE ARCHERY MATCHOn a target on which the scoringwas 40 for the bull's-eye, and 39, 24,

23, 17, and 16 respectively for therings from the center outwards, asshown in the illustration, three playershad a match with six arrows each.The result was: Miss Dora Talbot, 120points; Reggie Watson, 110 points;Mrs. Finch, 100 points. Every arrowscored, and the bull's-eye was onlyonce hit.Can you, from these facts, determine the exact six hits made by eachcompetitor?

468. TARGET PRACTICEColonel Crackham paid a visit one afternoon by invitation to the Slocombon-Sea Toxophilite Club, where he picked up the following little poser.Miscellaneous Combinatorial Puzzles 181

1bree men in a competition had each six shots at a target, and the result isshown in our illustration, where they all hit the target every time. The bull'seye scores 50, the next ring 25, the next 20, the next 10, the next 5, the next 3,the next 2, and the outside ring scores only 1. It will be seen that the hits onthe target are one bull's-eye, two 25's, three 20's, three lO's, three 1's, and twohits in every other ring. Now the three men tied with an equal score.Next morning the Colonel asked his family to show the exact scoringof each man. Will it take the reader many minutes to find the correct answer?

469. TOM TIDDLER'S GROUNDThe Crackham family were comfortably accommodated at the "Blue Boar"at Puddlebury. Here they had the luck to come upon another guest who wasclearly engaged in solving some sort of puzzle. The Colonel contrived a conversation with him, and learned that the puzzle was called "Tom Tiddler'sGround.""You know," said the stranger, "the words, 'I am on Tom Tiddler's groundpicking up gold and silver.' Here we have a piece of land marked off with 36circular plots, on each of which is deposited a bag containing as many dollars

182 Combinatorial & Topological Problemsas the figures indicate in the diagram. I am allowed to pick up as many bagsas I like, provided I do not take two lying on the same line. What isthe greatest amount of money I can secure?"

470. THE SEVEN CHILDRENFour boys and three girls are seated in a row at random. What are thechances that the two children at the ends of the row will be girls?Game Puzzles
Game Puzzles

471. TIC TAC TOEEvery child knows how to play this ancient game. You make a squareof nine cells, and each of the two players, playing alternately, puts his mark(a zero or a cross, as the case may be) in a cell with the object of getting threein a line. Whichever player gets three in a line wins. I have said in my book"1- 0" )(.000)(.The Canterbury Puzzles that betweentwo players who thoroughly understand the play every game should bedrawn, for neither party could everwin except through the blundering ofhis opponent. Can you prove this?Can you be sure of not losing a gameCROSS HAS WON .". against an expert opponent?

472. THE HORSESHOE GAMEThis little game is an interestingcompanion to tic tac toe. There aretwo players. One has two whitecounters, the other two black. Playingalternately, each places a counter ona vacant point, where he leaves it.When all are played, you slide only,and the player is beaten who is soblocked that he cannot move. In theexample, Black has just placed hislower counter. White now slides hislower one to the center, and wins.Black should have played to thecenter himself and won. Now, whichplayer ought to win at this game?

185

186 Game Puzzles

473. TURNING THE DIEThis is played with a single die.The first player calls any number hechooses, from I to 6, and the secondplayer throws the die at hazard. Thenthey take it in turns to roll over thedie in any direction they choose, butnever giving it more than a quarterturn. The score increases as they proceed, and the player wins who manages to score 25 or force his opponentto score beyond 25. I will give anexample game. A calls 6, and B happens to throw a 3 (as shown in ourillustration), making the score 9. NowA decides to turn up I, scoring 10; Bturns up 3, scoring 13; A turns up 6,scoring 19; B turns up 3, scoring 22;A turns up 1, scoring 23; and B turnsup 2, scoring 25 and winning.What call should A make in orderto have the best chance of winning?Remember that the numbers on opposite sides of a correct die alwayssum to 7, that is, 1-6,2-5,3-4.

474. THE THREE DICEthe beginning of the game. As a matter of fact, Mason selected seven andthirteen, and one of his winningthrows is shown in the illustration.What were his chances of winningMason and Jackson were playing a throw? And what two other numwith three dice. The player won when- bers should Jackson have selected forever the numbers thrown added up his own throws to make his chancesto one of two numbers he selected at of winning exactly equal?

475. THE 37 PUZZLE GAMEHere is a beautiful new puzzlegame, absurdly simple to play butquite fascinating. To most people itwill seem to be practically a game ofchance-equal for both players-butFlr::l~~~ uuuulJGame Puzzles 187there are pretty 3ubtleties in it, and I will show how to win with certainty.Place the five dominoes, 1,2,3,4,5, on the table. There are two players, whoplay alternately. The first player places a coin on any domino, say the 5, whichscores 5; then the second player removes the coin to another domino, say tothe 3, and adds that domino, scoring 8; then the first player removes the coinagain, say to the 1, scoring 9; and so on. The player who scores 37, or forceshis opponent to score more than 37, wins. Remember, the coin must beremoved to a different domino at each play.

476. THE 22 GAMEHere is a variation of our little "Thirty-one Game" (Canterbury Puzzles:No. 79). Layout the sixteen cards as shown. Two players alternately tumdown a card and add it to the common score, and the player who makes thescore of 22, or forces his opponent to go beyond that number, wins. For example, A turns down a 4, B turns down a 3 (counting 7), A turns down a 4(counting 11), B plays a 2 (counting

13), A plays 1 (14), B plays 3 (17),and whatever A does, B scores thewinning 22 next play. Again, supposing the play was 3-1, 1-2, 3-3, 1-2,

1-4, scoring 21, the second playerwould win again, because there is no

1 left and his opponent must gobeyond 22.Which player should always win,and how?

477. THE NINE SQUARES GAMEMake the simple square diagram shown on page 188 and provide a box ofmatches. The side of the large square is three matches in length. The game is,playing one match at a time alternately, to enclose more of those small squaresthan your opponent. For every small square that you enclose you not only scoreone point, but you play again. The illustration shows an illustrative game inprogress. Twelve matches are placed, my opponent and myself having madesix plays each, and, as I had first play, it is now my turn to place a match.

188 Game Puzzles~ <---------'jE. F' :~ ; · . · . · . · . : !: :r----ieo-.' -:i-·------r,;;u.u ... N 8 pWhat is my best line of play in order to win most squares? If I play FG myopponent will play BF and score one point. Then, as he has the right to playagain, he will score another with EF and again with 11, and still again with GK.If he now plays CD, I have nothing better than D H (scoring one), but, as Ihave to play again, I am compelled, whatever I do, to give him all the rest.So he will win by 8 to I-a bad defeat for me.What should I have played instead of that disastrous FG? There is roomfor a lot of skillful play in the game, and it can never end in a draw.

478. THE TEN CARDSPlace any ten playing cards in a row as shown. There are two players. Thefirst player may turn down any single card he chooses. Then the secondplayer can turn down any single card or two adjoining cards. And so on. Theplayer who turns down the last card wins.Remember that the first player must turn down a single, but afterwardseither player can turn down either a single or two adjoining cards, as hepleases. Should the first or second player win?~r.+l~l!!J ~ l!:!JDomino Puzzles
Domino Puzzles

479. THE DOMINO SWASTIKAHere is a little puzzle by Mr. WilfredBailey. Fonn a square frame withtwelve dominoes, as shown in theillustration. Now, with only four extradominoes, fonn within the frame aswastika. The reader may hit on theidea at once, or it may give him considerable trouble. In any case he cannot fail to be pleased with the solution.

480. DOMINO FRACTIONSHere is a new puzzle with dominoes. Taking an ordinary box, discard alldoubles and blanks. Then, substituting figures for the pips, regard the remaining fifteen dominoes as fractions. It will be seen in the illustration that I haveso arranged them that the fractions in every row of five dominoes sum toexactly 2~. But I have only used proper fractions. You are allowed to use asmany improper fractions (such as ~, h, 0/1) as you like, but must makethe five dominoes in every rank sum to 10.

191

192 Domino Puzzles

481. A NEW DOMINO PUZZLE• ,---/-._, ··1It will be seen that I have selectedand placed together two dominoes sothat by taking the pips in unbrokenconjunction I can get all the numbersfrom 1 to 9 inclusive. Thus, 1,2, and

3 can be taken alone; then 1 and 3make 4; 3 and 2 make 5; 3 and 3 make

6; 1,3, and 3 make 7; 3,3, and 2 make

8; and 1,3,3, and 2 make 9. It wouldnot have been allowed to take the

1 and the 2 to make 3, nor to take thefirst 3 and the 2 to make 5. The numbers would not have been in conjunction.Now try to arrange four dominoesso that you can make the pips in thisway sum to any number from 1 to 23inclusive. The dominoes need not beplaced 1 against 1, 2 against 2, and soon, as in play.

482. A DOMINO SQUARE•••• • • • • .1 ••• · . ' , .•, -. I· • • • • • • • • • • • • - - -• ••• • ,-. • -• • I • • • • • ••• • • • • •• • • ••• • , . • I I - •• • • • •• • •• • • • • • • • • • -• - _. • -- •-• -~ . • • • • • • • • • • • • • • -• •Select any eighteen dominoes you please from an ordinary box, and arrangethem any way you like in a square so that no number shall be repeated in anyrow or any column. The example given is imperfect, for it will be seen thatthough no number is repeated in anyone of the columns yet three of the rowsbreak the condition. There are two 4's and two blanks in the first row, two

5's and two 6's in the third row, and two 3's in the fourth row.Can you form an arrangement without such errors? Blank counts as anumber.Domino Puzzles 193

483. A DOMINO STARPlace the twenty-eight dominoes, as shown in the illustration, so as to forma star with alternate rays of four and three dominoes. Every ray must containtwenty-one pips (in the example only one ray contains this number) and thecentral numbers must be 1,2,3,4,5,6, and two blanks, as at present, and thesemay be in any order. In every ray the dominoes must be placed according tothe ordinary rule, six against six, blank against blank, and so on.

484. DOMINO GROUPSI wonder how many of my readers know that if you layout, the twentyeight dominoes in line according to the ordinary rule-six against six, twoagainst two, blank against blank, and so on-the last number must always bethe same as the first, so that they will really always form a circle. It is a veryancient trick to conceal one domino (but do not take a double) and then askhim to arrange all the others in line without your seeing. It will astonish himwhen you tell him, after he has succeeded, what the two end numbers are. Theymust be those on the domino that you have withdrawn, for that domino com-

194 Domino Puzzlespletes the circle. If the dominoes are laid out in the manner shown in ourillustration and I then break the line into four lengths of seven dominoes each,it will be found th<.t the sum of the pips in the first group is 49, in the second

34, in the third 46, and in the fourth 39.I want to play them out so that all the four groups of seven when the lineis broken shall contain the same number of pips. Can you find a way ofdoing it?

485. LES QUADRILLESThis old French puzzle will, I think, be found very interesting. It is requiredto arrange a complete set of twenty-eight dominoes so as to form the figure'--r--;-•· · I • • . I I • · • • • ••· • · · · • - . _. •• .-. .-• · • • • • · · • · • • · • · · · · • · ....... . ••• 1.-.:: ••• 1 · ... I• •.-. . .. -. ... ... ' . . · • · · • • I · · • • . • · • · · · · · · . - ••• ... - -. • I • • • . ..... . • I • .. ..... .I--t--;-•. . • • 1 •••••• 1 • . .Domino Puzzles 195shown in our illustration, with all the numbers forming a series of squares.Thus, in the upper two rows we have a square of blanks, and a square of four

3's, and a square of 4's, and a square of l's; in the third and fourth rows wehave squares of 5, 6, and blank, and so on. This is, in fact, a perfect solutionunder the conditions usually imposed, but what I now ask for is an arrangement with no blanks anywhere on the outer edge. At present every numberfrom blank to 6 inclusive will be found somewhere on the margin. Can youconstruct an arrangement with all the blanks inside?

486. DOMINO FRAMESTake an ordinary set of twenty-eight dominoes and return double 3, double

4, double 5, and double 6 to the box as not wanted. Now, with the remainderform three square frames, in themanner shown, so that the pips inevery side shall add up alike. In theexample given the sides sum to 15. Ifthis were to stand, the sides of the twoother frames must also sum to 15.But you can take any number youlike, and it will be seen that it is notrequired to place 6 against 6,5 against

5, and so on, as in play.••-••r;-;-ee-•• •• • · ·. ~I· .. • I. • I• •••• • -• •• • •-.----.-·· •• --• - • I • 1-· • •• -1 •

487. DOMINO HOLLOW SQUARESEvery game lends itself to the propounding of interesting little puzzles.• • •••• I• • ••• -r • • •• • • - • • • •• , • • • • • •Let us, as an example, take the following poser, devised from an ordinarybox of twenty-eight dominoes. It isrequired with these twenty-eight toform seven hollow squares, all similarto the example given, so that the pipsin the four sides of every square shalladd up alike. All these seven squaresneed not have the same sum, and, ofcourse, the example given need notbe one of your set.

196 Domino PuzzlesThe reader will probably find it easy to form six of the squares correctly,in many ways, but the trouble generally begins when you come to make theseventh square with the four remaining dominoes.

488. DOMINO SEQUENCESA boy who had a complete set of dominoes, up to double 9, was trying toarrange them all in sequence, in the usual way-6 against 6, 3 against 3, blankagainst blank, etc. His father said to him, "You are attempting an impossibility, but if you will let me pick out four dominoes it can then be done. And thosethat I take shall contain the smallest total number of pips possible in thecircumstances. "Now, which dominoes might the father have selected? Remember that thedominoes in common use in this country stop at double 6, but we are here usinga set up to double 9.

489. TWO DOMINO SQUARESArrange the twenty-eight dominoes as shown in the diagram to form twosquares so that the pips in everyone of the eight sides shall add up alike. Thedominoes being on the table one day recently, we set ourselves the above task,and found it very interesting.The constant addition must be within limits to make the puzzle possible, andit will be found interesting to find these limits. Of course, the dominoes neednot be laid according to the rule, 6 against 6, blank against blank, and so on., , I I I I I- - , I I- - - -- f-- I-- -- - - -- - - -- - - - I I I- -I I I I I I IDomino Puzzles 197

490. DOMINO MULTIPLICATIONWe have received the following entertaining puzzle from W. D. W., ofPhiladelphia, Pa.:Four dominoes may be so placed as to form a simple multiplication sum ifwe regard the pips as figures. The example here shown will make everythingperfectly clear. Now, the puzzle is, using all the twenty-eight dominoes to arrange them so as to form seven such little sums in multiplication.

1_· •• . . • I· • .. • •It will be found comparatively easy to construct six such groups, while thefour dominoes left over are impossible of arrangement. But it can be done, andthe quest will be found amusing. No blank may be placed at the left end ofthe multiplicand or product.

491. DOMINO RECTANGLEHere is a new domino puzzle that I hope will be found entertaining. Arrange

11

11

21

21

198 Domino Puzzlesthe twenty-eight dominoes exactly as shown in the illustration, where the pipsare omitted, so that the pips in everyone of the seven columns shall sum to

24, and the pips in everyone of the eight rows to 21. The dominoes need notbe 6 against 6, 4 against 4, and so on.• • • • • , • • • • 492. THE DOMINO COLUMN• • • • , • • • •• ,• • • • •• • • • - • • · • • • • • • • • •••• • • ••• 1. • •Arrange the twenty-eight dominoesin a column so that the three sets ofpips, taken anywhere, shall add upalike on the left side and on the right.Such a column has been started in thediagram. It will be seen that the topthree add up to 9 on both sides, thenext three add up to 7 on both sides,and so on. This is merely an example,so you can start afresh if you like. • , • • •

493. ARRANGING THE DOMINOESSomebody reminded Professor Rackbrane one morning at the breakfasttable that he had promised to tell them in how many different ways the set oftwenty-eight dominoes may be arranged in a straight line, in accordance withthe original rule of the game, left to right and right to left, in any arrangementcounting as different ways. Later on he told them that the answer was

7,959,229,931,520 different ways. He said that it was an exceedingly difficultproblem.He then proposed that they should themselves find out in how manydifferent ways the fifteen smaller dominoes (after discarding all those bearinga 5 or a 6) may similarly be arranged in a line. Of course, you always place

1 against 1, 6 against 6, and so on, the two directions counting different.Match Puzzles
Match Puzzles

494. A NEW MATCH PUZZLEI have a box of matches. I find that I can form with them any given pair ofthese four regular figures, using all the matches every time. Thus, if there wereeleven matches, I could form with them, as shown, the triangle and pentagonor the pentagon and hexagon, or the square and triangle (by using only threematches in the triangle); but could not with eleven matches form the triangleL[]OOand hexagon, or the square and pentagon, or the square and hexagon. Ofcourse there must be the same number of matches in every side of a figure.What is the smallest number of matches I can have in the box?

495. HURDLES AND SHEEPThis is a little puzzle that you can try with matches. A farmer says that fourof his hurdles will form a square enclosure just sufficient for one sheep. That00

201

202 Match Puzzlesbeing so, what is the smallest number of hurdles that he will require for enclosing ten sheep? Everything depends on the shape of your enclosure. Theonly other way of placing the four matches (or hurdles) in A is to form adiamond-shaped figure, and the more attenuated this diamond becomes thesmaller will be its area, until the sides meet, when there will be no area enclosedat all.If you place six matches, as in B, you will have room for two sheep. But ifyou place them as in C, you will only have room for one sheep, for seventenths of a sheep will only exist as mutton. And if you place them as in D, youcan still only accommodate two sheep, which is the maximum for six hurdles.How many hurdles do you require for ten sheep?

496. THE TWENTY MATCHESThe illustration shows how twenty matches, divided into two groups offourteen and six, may form two enclosures so that one space enclosed isexactly three times as large as the other. Now divide the twenty matches intotwo groups of thirteen and seven, and with them again make two enclosures,one exactly three times as large as the other.m [ I] ] J] I

497. A MATCH PUZZLEThe sixteen squares of a chessboard are enclosed by sixteen matches. It isrequired to place an odd number of matches inside the square so as to en-close four groups of four squares each.There are obvious and easy ways ofdoing it with 8, 10, or 12 matches, butthese are even numbers.It may only take the reader a fewmoments to discover the four distinctive ways (mere reversals and reflections not counting as different) ofdoing it with an odd number ofmatches. Of course no duplicatedmatches are allowed.IVIIMatch Puzzles 203

498. AN INGENIOUSMATCH PUZZLEPlace six matches as shown, andthen shift one match without touchingthe others so that the new arrangement shall represent an arithmeticalfraction equal to I. The match forming the horizontal fraction bar mustnot be the one moved.

499. FIFTY-SEVEN TO NOTHINGAfter the last puzzle, this one should be easy.It will be seen that we have arranged six cigarettes (matches will do just aswell) so as to represent the number 57. The puzzle is to remove any two of them

204 Match Puzzlesyou like (without disturbing any of the others) and so replace them as torepresent 0, or nothing. Remember that you can only shift two cigarettesThere are two entirely different solutions. Can you find one or both?

500. THE FIVE SQUARESHere is a new little match puzzle though they will smile when they seethat will perplex a good many readers, the answer. It will be seen that thetwelve matches are so arranged thatthey form four squares. Can you rearrange the same number of matches(all lying flat on the table) so thatthey enclose five squares?Every square must be entirely"empty" or the illustration itself willshow five squares if we were allowedto count the large square forming theboundary. No duplicated match orloose ends are allowed.

501. A MATCH TRICKWe pulled open a box of matches the other day, and showed some friendsthat there were only about twelve matches in it. When opened at that end nohead was visible. The heads were all at the other end of the box. We told themafter we had closed the box in front of them we would give it a shake, and, onreopening, they would find a match turned round with its head visible. Theyafterwards examined it to see that the matches were all sound. How did wedo it?

502. THREE TIMES THE SIZELayout 20 matches in the way shown in our illustration. You will see thatthe two groups of 6 and 14 matches form two enclosures, so that one spaceenclosed is exactly three times as large as the other.Match Puzzles 205Now transfer I match from the larger to the smaller group, and with the 7and 13 enclose two spaces again, one exactly three times as large as the other.· i[ • 1 • .. •Twelve of the matches must remain unmoved from their present positionsand there must be no duplicated matches or loose ends. The dotted lines aresimply to indicate the respective areas.

503. A SIX-SIDED FIGUREHere are 6 matches arranged so asto form a regular hexagon. Can youtake 3 more matches and so arrangethe 9 as to show another regular sixsided figure? No duplicated matchesor loose ends allowed./ \\-========-I

504. TWENTY -SIX MA TeHESMake a rough square diagram, like the one shown on page 206, where the sideof every little square is the length of a match, and put the stars and crosses intheir given positions. It is required to place 26 matches along the lines so as toenclose two parts of exactly the same size and shape, one part containing twostars, and the other two crosses.

206 Match PuzzlesIn the example given, each part is correctly of the same size and shape, andeach part contains either two stars or two crosses; but, unfortunately, only 20matches have been used. So it is not a solution. Can you do it with 26 matches?-------~------'------~: , :-- ----: '+ : .: ' :. 's . -- ----'.: . , I, . I " •• • I •, I +' I : ' I tt • I • . , ,. :~--'---'-==r=: * ' " .~I, " • _____ , ____ ••• ____ _, ,,

505. THE THREE MATCHESCan you place 3 matches on the table, and support the matchbox on them,without allowing the heads of the matches to touch the table, to touchone another, or to touch the box?

506. EQUILATERAL TRIANGLESHere is a little puzzle for youngreaders:Place 16 matches, as shown, to formeight equilateral triangles. Now takeaway 4 matches so as to leave onlyfour equal triangles. No superfluousmatches or loose ends to be left.CD

6

508. HEXAGON TO DIAMONDSHere is another match puzzle foryoung readers. With 6 matches forma hexagon, as here shown. Now, bymoving only 2 matches and adding 1more, can you form two diamonds?Match Puzzles 207

507. SQUARES WITH MATCHESArrange 12 matches on the table, asshown in the illustration. Now it isrequired to remove 6 of these matchesand replace them so as to form fivesquares. Of course 6 matches mustremain unmoved, and there must beno duplicated matches or loose ends.j' \\~,/.

509. QUEER ARITHMETICCan you show with matches how to take seven-tenths from five so thatexactly four remains?

510. COUNTING THE MATCHESA friend writes to say he bought a little box of midget matches, each one inchin length. He found that he could arrange them all in the form of a trianglewhose area was just as many square inches as there were matches. He then usedup six of the matches, and found that with the remainder he could again construct another triangle whose area was just as many square inches as there

208 Match Puzzleswere matches. And using another six matches he could again do precisely thesame.How many matches were there in the box originally? The number is lessthan forty.Unclassified Puzzles
Unclassified Puzzles

511. A PUZZLE WITH CARDSTake from the pack the thirteen cards forming the suit of diamonds andarrange them in this order face downwards with the 3 at the top and 5 at thebottom: 3,8,7, ace, queen, 6, 4, 2,jack, king, 10,9,5. Now play them out ina row on the table in this way. As you spell "ace" transfer for each letter a cardfrom the top to the bottom of the pack-A-C-E-and play the fourth card onto the table. Then spell T-W-O, while transferring three more cards to thebottom, and place the next card on the table. Then spell T-H-R-E-E, whiletransferring five to the bottom, and so on until all are laid out in a row, andyou will find they will be all in regular order. Of course, you will spell out theknave as J-A-C-K.Can you arrange the whole pack so that they will play out correctly inorder, first all the diamonds, then the hearts, then the spades, and lastly theclubs?

512. CARD SHUFFLINGThe rudimentary method of shuffling a pack of cards is to take the pack facedownwards in the left hand and then transfer them one by one to the righthand, putting the second on top of the first, the third under, the fourth above,and so on until all are transferred. If you do this with any even number ofcards and keep on repeating the shuffle in the same way, the cards will in duetime return to their original order.Try with 4 cards, and you will find the order is restored in three shuffles. Infact, where the number of cards is 2, 4, 8, 16, 32, 64, the number of shufflesrequired to get them back to the original arrangement is 2, 3, 4, 5, 6, 7respectively. How many shuffles are necessary in the case of 14 cards?

513. A CHAIN PUZZLEA man has eighty links of old chain in thirteen fragments, as shown on thefollowing page. It will cost him l¢ to open a link and 2¢ to weld one together

2lJ

212 Unclassified Puzzlesagain. What is the lowest price it must cost him to join all the pieces togetherso as to form an endless chain?A new chain will cost him 36¢. What is the cheapest method of procedure?Remember that the large and small links must run alternately.

514. A SQUARE WITH FOUR PENNIESCan you place four English pennies together so as to show a square? Theymust all lie fiat on the table.

515. SIMPLE ADDITIONCan you show that four added to six will make eleven?

516. A CALENDAR PUZZLEI have stated in my book Amusements in Mathematics that, under ourpresent calendar rules, the first day of a century can never fall on a Sunday ora Wednesday or a Friday. As I have not given the proof, I am frequently askedthe reason why. I will try to explain the mystery in as simple a way as possible.

517. THE FLY'S TOURI had a ribbon of paper, divided into squares on each side, as shown in theillustration. I joined the two ends together to make a ring, which I threw onUnclassified Puzzles 213I J r I I J I I I I I ,the table. Later I noticed that a fly pitched on the ring and walked in a lineover every one of the squares on both sides, returning to the point from whichit started, without ever passing over the edge of the paper! Its course passedthrough the centers of the squares all the time. How was this possible?

518. A MUSICAL ENIGMAHere is an old musical enigma that has been pretty well known in Germanyfor some years.~~ rphU \"\. 1.1oJ~~ 1

519. SURPRISING RELATIONSHIPANGELINA: You say that Mr. Tomkins is your uncle?EDWIN: Yes, and I am his uncle!c.IlA "ANGELINA: Then-let me see-you must be nephew to each other, ofcourse! Funny, isn't it?Can you say quite simply how this might be, without any breach of themarriage law or disregard of the Table of Affinity?

214 Unclassified Puzzles

520. AN EPITAPH (A.D. 1538)Two grandmothers, with their two granddaughters;Two husbands, with their two wives;Two fathers, with their two daughters;Two mothers, with their two sons;Two maidens, with their two mothers;Two sisters, with their two brothers;Yet only six in all lie buried here;All born legitimate, from incest clear.How might this happen?

521. THE ENGINEER'S NAMEThree business men-Smith, Robinson, and Jones-all live in the LeedsSheffield district. Three railwaymen of similar names live in the same district.The business man Robinson and the guard live at Sheffield, the business manJones and the stoker live at Leeds, while the business man Smith and therailway engineer live half-way between Leeds and Sheffield. The guard'snamesake earns $10,000.00 per annum, and the engineer earns exactly onethird of the business man living nearest to him. Finally, the railwaymanSmith beats the stoker at billiards.What is the engineer's name?

522. STEPPING STONESThe illustration represents eightstepping stones across a certainstream. The puzzle is to start from thelower bank and land twice on the upper bank (stopping there), havingreturned once to the lower bank. Butyou must be careful to use each stepping stone the same number of times.In how few steps can you make thecrossing?Make the steps with two fingers inthe diagram, and you will see what ao0°0000o BArI "Unclassified Puzzles 215very simple matter it is. Yet it is more than likely that you will at first take agreat many more steps than are necessary.

523. AN AWKWARD TIME"When I told a man the other morning," said Colonel Crackham at thebreakfast table, "that I had to catch the 12:50 train, he surprised me by saying that it was a very awkward time for any train to start. I asked him to explain why. Can you guess his answer?"

524. CRYPTIC ADDITION

31JO)'+81']Can you prove that the above addition sum is correct?

525. THE TWO SNAKESWe have been asked this question:Suppose that two snakes start swallowing one another simultaneously, each

216 Unclassified Puzzlesgetting the tail of the other in its mouth, so that the circle formed by the snakesbecomes smaller and smaller, what will eventually happen?

526. TWO PARADOXESA child may ask a question that will profoundly perplex a learned philosopher, and we are often meeting with paradoxes that demand a little thoughtbefore we can explain them in simple language. The following was put to us:"Imagine a man going to the North Pole. The points of the compass are,as every one knows:NW ES"He reaches the Pole and, having passed over it, must turn about to lookNorth. East is now on his left-hand side, West on his right-hand side, and thepoints of the compass thereforeNE WSwhich is absurd. What is the explanation?Unclassified Puzzles 217"We were standing with a child in front of a large mirror that reflected thewhole body. 'Why is it,' asked the intelligent youngster, 'that I am turned rightround in the mirror, so that right is left and left right, and yet top is not bottom and bottom top? If it reverses sideways, why does it not reverse lengthways? Why am I not shown standing on my head?' "

527. COIN AND HOLEWe have before us a specimen of every American coin from a penny to adollar. And we have a small sheet of paper with a circular hole cut in it ofoexactly the size shown. (It was made by tracing around the rim of a penny.)What is the largest coin I can pass through that hole without tearing the paper?

528. A LEAP YEAR PUZZLEThe month of February in 1928 contained five Wednesdays. There is,of course, nothing remarkable in this fact, but it will be found interesting todiscover the last year prior to 1928 and the first year after 1928 that had fiveWednesdays in February.

529. BLOWING OUT THE CANDLECandles were lighted on Colonel Crackham's breakfast table one foggymorning. When the fog lifted, the Colonel rolled a sheet of paper into the formof a hollow cone, like a megaphone. He then challenged his young friends to

218 Unclassified Puzzlesuse it in blowing out the candles. They failed, until he showed them the trick.Of course, you must blow through the small end..' .... .... _ ... - f\?c -./

530. RELEASING THE STICKHere is a puzzle that will often cause a good deal of bewilderment amongstyour friends, though it is not so generally known as it deserves to be. I think--:it was invented by Sam Loyd, the American chess and puzzle genius. At anyrate, it was he who first showed it to us more than a quarter of a century ago.It is simply a loop of string passed through one end of a stick as here shown,Unclassified Puzzles 219but not long enough to pass round the other end. The puzzle is to suspend itin the manner shown from the top hole of a man's coat, and then get it freeagain.

531. THE KEYS AND RINGColonel Crackham the other dayproduced a ring and two keys, as hereshown, cut out of a solid piece of cardboard, without a break or join anywhere. Perhaps it will puzzle thereader more than it puzzled George,who promptly cut them out.

532. THE ENTANGLED SCISSORSHere is an old puzzle that manyreaders, who have forgotten how toput on the string, will be glad to seeagain. If you start on the loop at thebottom, the string can readily be gotinto position. The puzzle is, of course,to let some one hold the two ends ofthe string until you disengage thescissors. A good length of stringshould be used to give you free play.We would also advise the use of alarge pair of scissors and thick cordthat will slide easily.

533. INTELLIGENCE TESTSNowadays the schools are all out to give their children "Intelligence Tests,"and in The New Examiner, by Dr. Ballard, there is a fine collection. I have

220 Unclassified Puzzlesincluded one here which, although not new, is really worthy of honorablemention.An English officer, after a gruesome experience during the Boxer rebellionin China some years ago, fell asleep in church during the sermon. He wasdreaming that the executioner was approaching him to cut off his head, andjust as the sword was descending on the officer's unhappy neck his wife lightlytouched her husband on the back of his neck with her fan to awaken him. Theshock was too great, and the officer fell forward dead. Now, there is somethingwrong with this. What is it?Another good question on similar lines for the scientific boy would be:Ifwe sell apples by the cubic inch, how can we really find the exact numberof cubic inches in, say, a dozen dozen apples?

534. AT THE MOUNTAIN TOP"When I was in Italy I was taken to the top of a mountain and shown thata mug would hold less liquor at the top of this mountain than in the valleybeneath. Can you tell me," asked Professor Rackbrane, at the breakfast table,"what mountain this might be that has so strange a property?"

535. CUPID'S ARITHMETICDora Crackham one morning produced a slip of paper bearing the jumbleof figures shown in our illustration. She said that a young mathematician hadthis poser presented to him by his betrothed when she was in a playful mood./)122 fffG q601LG2/"What am I to do with it?" asked George."Just interpret its meaning," she replied. "If it is properly regarded itshould not be difficult to decipher."Unclassified Puzzles 221

536. TANGRAMSThose readers who were interested in the article on "Tangrams" in Amusements in Mathematics may be glad to have a further collection of examplesof the strikingly realistic figures and designs that can be produced by combining these curious shaped pieces. In diagram I the square is shown cut into theseven pieces. If you mark the point B, midway between A and C, on one sideof a square of any size, and D, midway between C and E, on an adjoining side,the direction of the cuts is obvious. In the examples given below and on thefollowing page, two complete sets of seven pieces have been combined.Diagram 2 represents a man riding a bicycle; 3, a man pushing a wheelbarrow; 4, a boy riding a donkey; 5, a motor-car; 6, a house; 7, a dog;

8, a horse; 9, the British lion.As will be seen, the possibility with these two sets combined is infinite, andmany interesting subjects may be produced very successfully.

1 ~: . .

222 Unclassified Puzzles[The reader interested in tangrams will enjoy two excellent new paperbacks:Tangrams: 330 Puzzles, by Ronald C. Read (Dover, 1965), and Tangrams:Picture-making Puzzle Game, by Peter Van Note (Tuttle, 1966).-M. G.)Answers
Answers

1. CONCERNING A CHECKThe amount must have been $31.63. He received $63.31. After he hadspent a nickel there would remain the sum of $63.26, which is twice theamount of the check.

2. DOLLARS AND CENTSThe man must have entered the store with $99.98 in his pocket.

3. LOOSE CASHThe largest sum is $1.19, composed ofa half dollar, quarter, four dimes, andfour pennies.

4. GENEROUS GIFTSAt first there were twenty persons, and each received $6.00. Then fifteenpersons (five fewer) would have received $8.00 each. But twenty-four (fourmore) appeared and only received $5.00 each. The amount distributed weeklywas thus $120.00.

5. BUYING BUNSThere must have been three boys and three girls, each of whom receivedtwo buns at three for a penny and one bun at two for a penny, the cost of whichwould be exactly 7¢.

6. UNREWARDED LABORWeary Willie must have worked 1673 days and idled l3Y.! days. Thus theformer time, at $8.00 a day, amounts to exactly the same as the latter at $10.00a day.

225

226 Answers

7. THE PERPLEXED BANKERThe contents of the ten bags (in dollar bills) should be as follows: $1,2,4,

8, 16,32,64, 128,256, 4S9. The first nine numbers are in geometrical progression, and their sum, deducted from 1,000, gives the contents of the tenth bag.

8. A WEIRD GAMEThe seven men, A, B, C, D, E, F, and G, had respectively in their pocketsbefore play the following sums: $4.49, $2.25, $1.13, 57¢, 29¢, 15¢, and 8¢. Theanswer may be found by laboriously working backwards, but a simplermethod is as follows: 7 + 1 = 8; 2 X 7 + 1 = 15; 4 X 7 + 1 = 29; and soon, where the multiplier increases in powers of 2, that is, 2, 4, 8, 16, 32, and 64.

9. DIGGING A DITCHA. should receive one-third of two dollars, and B. two-thirds. Say B. can digall in 2 hours and shovel all in 4 hours; then A. can dig all in 4 hours and shovelall in 8 hours. That is, their ratio of digging is as 2 to 4 and their ratio ofshovelling as 4 to 8 (the same ratio), and A. can dig in the same time that B.can shovel (4 hours), while B. can dig in a quarter of the time that A. canshovel. Any other figures will do that fill these conditions and give two similar ratios for their working ability. Therefore, A. takes one-third and B. twiceas much-two-thirds.

10. NAME THEIR WIVESAs it is evident that Catherine, Jane, and Mary received respectively $122.00,$132.00, and $142.00, making together the $396.00 left to the three wives, ifJohn Smith receives as much as his wife Catherine, $122.00; Henry Snookshalf as much again as his wife Jane, $198.00; and Tom Crowe twice as muchas his wife Mary, $284.00, we have correctly paired these married couples andexactly accounted for the $1,000.00.

11. MARKET TRANSACTIONSThe man bought 19 cows for $950.00, 1 sheep for $10.00, and 80 rabbits for$40.00, making together 100 animals at a cost of $1,000.00.Answers 227A purely arithmetical solution is not difficult by a method of averages, theaverage cost per animal being the same as the cost of a sheep.By algebra we proceed as follows, working in dollars: Since x + y + z =

100, then fu + ~y + 1hz = 50.

50x + lOy + ~z = 1,000~x + ~y + 1hz = 50

49~x + 9~y 950by subtraction, or 99x + 19y = 1900. We have therefore to solve this indeterminate equation. The only answer is x = 19, Y = l. Then, to makeup the 100 animals, z must equal 80.

12. THE SEVEN APPLEWOMENEach woman sold her apples at seven for I¢, and 3¢ each for the odd onesover. Thus, each received the same amount, 20¢. Without questioning the ingenuity of the thing, I have always thought the solution unsatisfactory, becausereally indeterminate, even if we admit that such an eccentric way of sellingmay be fairly termed a "price." It would seemjust as fair if they sold them atdifferent rates and afterwards divided the money; or sold at a single rate withdifferent discounts allowed; or sold different kinds of apples at differentvalues; or sold the same rate per basketful; or sold by weight, the apples beingof different sizes; or sold by rates diminishing with the age of the apples; andso on. That is why I have never held a high opinion of this old puzzle.In a general way, we can say that n women, possessing an + (n - 1),n(a + b) + (n - 2), n(a + 2b) + (n - 3), ... ,n[a + b(n - 1) 1 apples respectively, can sell at n for a penny and b pennies for each odd one over andeach receive a + b(n - I) pennies. In the case of our puzzle a = 2, b = 3,andn = 7.

13. A LEGACY PUZZLEThe legacy to the first son was $55.00, to the second son $275.00, to thethird son $385.00, and to the hospital $605.00, making $1,320.00 in all.

228 Answers

14. PUZZLING LEGACIESThe answer is $1,464.oo-a little less than $1,500.00. The legacies, in order,were $1,296.00, $72.00, $38.00, $34.00, and $18.00. The lawyer's fee would be$6.00.

15. DIVIDING THE LEGACYThe two legacies were $24.00 and $76.00, for if 8 (one-third of 24) be takenfrom 19 (one-fourth of76) the remainder will be II.

16. A NEW PARTNERWe must take it for granted that the sum Rogers paid, $2,500.00, was onethird of the value of the business, which was consequently worth $7,500.00before he entered. Smugg's interest in the business had therefore been$4,500.00 (l\-2 times as much as Williamson), and Williamson's $3,000.00. Aseach is now to have an equal interest, Smugg will receive $2,000.00 of Rogers'scontribution, and Williamson $500.00.

17. POCKET MONEYWhen he left home Tompkins must have had $2.10 in his pocket.

18. DISTRIBUTIONThe smallest number originally held by one person will be (in cents) onemore than the number of persons. The others can be obtained by continuallydoubling and deducting one. So we get their holdings as 10, 19,37, 73, 145,

289,577, 1,153, and 2,305. Let the largest holder start the payment and workbackwards, when the number of cents in the end held by each person will be

29 or 512-that is, $5.12.

19. REDUCTIONS IN PRICEIt is evident that the salesman's rule was to take off three-eighths of the priceat every reduction. Therefore, to be consistent, the motorcycle should beoffered at $156.25 after the next reduction.Answers 229

20. HORSES AND BULLOCKSWe have to solve the indeterminate equation 344x = 265y + 33. This iseasy enough if you know how, but we cannot go into the matter here. Thus xis 252, and y is 327, so that if he buys 252 horses for $344.00 apiece, and 327bullocks for $265.00 apiece, the horses will cost him in all $33.00 more thanthe bullocks.

21. BUYING TURKEYSThe man bought 75 turkeys at 80¢ each, making $60.00. After retaining 15he sold the remaining 60 at 90¢ each, making $54.00, as stated. He thus madea profit of 1O¢ each on the 60 birds he resold.

22. THE THRIFTY GROCERHe must have had 168 each of dollar bills, half dollars, and quarters,making a total of $294.00. In each of the six bags there would be 28 of eachkind; in each of the seven bags 24 of each kind; and in each of the eight bags,

21 of each kind.

23. THE MISSING PENNYThe explanation is simply this. The two ways of selling are only identicalwhen the number of apples sold at three for a penny and two for a penny isin the proportion of three to two.Thus, if the first woman had handed over 36 apples, and the second woman

24, they would have fetched 24¢, whether sold separately or at five for 2¢.But if they each held the same number of apples there would be a loss whensold together of l¢ in every 60 apples. So if they had 60 each there would bea loss of2¢. If there were 180 apples (90 each) they would lose 3¢, and so on.The missing penny in the case of 60 arises from the fact that the three apenny woman gains 2¢, and the two a penny woman loses 3¢.Perhaps the fairest practical division of the 24¢ would be that the firstwoman receives 9l6¢ and the second woman 14'12, so that each loses l6¢on the transaction.

230 Answers

24. THE RED DEATH LEAGUEThe total amount of subscriptions, reduced to cents, is 300,737, whichis the product of 311 and 967, each of which is a prime. As we know that theR.D.L. had fewer than 500 members, it follows that there were 311, and theyeach paid as a subscription 967¢, or $9.67. This is the only possible answer.

25. A POULTRY POSERThe price of a chicken was $2.00, for a duck $4.00, and for a goose $5.00.

26. BOYS AND GIRLSEvery boy at the start possessed 12¢, and he gave 1¢ to every girl. Everygirl held 36¢, of which she gave 3¢ to every boy. Then every child wouldhave 18¢.

27. THE COST OF A SUITThe cost of Melville's suit was $150.00, the coat costing $75.00, the trousers$50.00, and the vest $25.00.

28. A QUEER SETTLING UPRichard had $4.00, and John had $2.50.

29. APPLE TRANSACTIONSThe apples cost 96¢ per 100.

30. PROSPEROUS BUSINESSHe possessed $22,781.25.

31. THE BANKER AND THE COUNTERFEIT BILLSince the identical counterfeit bill can be traced through all the transactions,these are all invalid. Therefore everybody stands in relation to his debtor justwhere he was before the banker picked up the note, except that the butcherowes, in addition, $5.00 to the farmer for the calf received.Answers 231

32. THEIR AGESTom's age was seven years and Mary's thirteen years.

33. MRS. WILSON'S FAMILYThe ages must be as follows: Mrs. Wilson, 39; Edgar, 21; James, 18; John,

18; Ethel, 12; Daisy, 9. It is clear that James and John were twins.

34. DE MORGAN AND ANOTHERDe Morgan was born in 1806. When he was 43, the year was the square ofhis age-1849. Jenkins was born in 1860. He was 52 + 62 (61) in the year

54 + 64 (1921). Also he was 2 X 31 (62) in the year 2 X 31 2 (1922). Again,he was 3 X 5 (15) in the year 3 X 54 (1875).

35. "SIMPLE" ARITHMETICTheir ages were, respectively, sixty-four and twenty.

36. ANCIENT PROBLEMDemochares must be just sixty years of age.

37. FAMILY AGESThe father and mother were both of the same age, thirty-six years old, andthe three children were triplets of six years of age. Thus the sum of all theirages is ninety years, and all the other conditions are correctly fulfilled.

38. MIKE'S AGEMike's present age is 101~1 years, Pat's is 291~1, and Biddy's is 242~1 years.When the sty was built (79121 years ago), Mike was 3~1, Pat was 22~1, andBiddy was 171\6 I. In 1 P \61 years Mike will be 22~ 1 (as old as Pat was whenhe built the sty), Pat will be 41~1, and Biddy will be 361%1, making 100 yearstogether.

232 Answers

39. THEIR AGESThirty years and twelve years, respectively.

40. BROTHER AND SISTERThe boy's age was ten and his sister's four.

41. A SQUARE FAMILYThe ages of the nine children were respectively 2, 5, 8, II, 14, 17,20,23,26,and the age of the father was 48.

42. IN THE YEAR 1900The man was born in 1856 and died in 1920, aged 64 years. Let x = age atdeath. Then 29x = date of birth. The date of birth + age = date of death,so that 29x + x = 30x, or date of death. Now, from the question he was clearlyalive in 1900, and dead by 1930. So death occurred during or between thosedates, and as the date is 30x, it is divisible by 30. The date can only be 1920,which, divided by 30, gives 64. So in 1900 he was 44 years of age.

43. FINDING A BIRTHDAYThe man must have been born at midday on February 19, 1873, and atmidday on November 11, 1928, he had lived 10,17616 days in each century.Of course, the century ended at midnight on December 31, 1900, which wasnot a leap year, and his age on November 11, 1928, was 55 years and nearly

9 months.

44. THE BIRTH OF BOADICEAThere were 129 years between the birth of Oeopatra and the death ofBoadicea; but, as their united ages amounted to 100 years only, there musthave been 29 years when neither existed-that is, between the death ofOeopatra and the birth of Boadicea. Therefore Boadicea must have beenborn 29 years after the death of Cleopatra in 30 B.C., which would be in theyear 1 B.C.Answers 233

45. ROBINSON'S AGERobinson's age must have been 32, his brother's 34, his sister's 38, and hismother's 52.

46. A DREAMLAND CLOCKThe hour indicated would be exactly 23%3 minutes after four oclock. Butbecause the minute hand moved in the opposite direction, the real time wouldbe 361'13 minutes after four. You must deduct the number of minutes indicatedfrom 60 to get the real time.

47. WHAT IS THE TIME?The time is 63,.~ minutes past IX, when the hour hand is 450/16 (the squareof 6*) minutes past XII. If we allow fractions less than a minute point, thereis also the solution, five seconds (one-twelfth of a minute) past XII oclock.

48. THE AMBIGUOUS CWCKThe first time would be 55;'43 minutes past twelve, which might also (thehands being similar) indicate 6o/i43 minutes past one oclock.

49. THE BROKEN CLOCK FACEIf the clock face be broken orcracked in the manner shown in ourillustration, the numerals on each ofthe four parts will sum to 20. Thecunning reader will at once have seenthat, as three X's (in IX, X, and XI)are adjacent, two of these must bealone contained in one piece. Therefore there are only two possible casesfor trial.[The first edition of Puzzles andCurious Problems, in which this puz-

234 Answerszle appeared, gave an inferior solution in which it was necessary to view thehour IX upside down and interpret it as XI. (Note that it is so interpreted inthe illustration accompanying the statement of the problem.) When a revisedsecond edition of the book was issued, this blemish was removed and thesolution shown here was substituted. Actually, there are twelve more legitimatesolutions. The reader is invited to see if he can find all of them without checking my Scientific American columns for May and June, 1966, in whichall thirteen can be found.It is assumed that the Roman numerals are permanently attached to therim of the clock face. Lines of breakage may separate the parts of an hour, asshown in Dudeney's solution, but must not go around any numerals, separating them from the rim.-M. G.)so. WHEN DID mE DANCING BEGIN?The dancing must have begun at 598 ¥I43 minutes past ten, and the handswere noticed to have changed places at 54l31}l43 minutes past eleven.

51. MISTAKING THE HANDSThe time must have been 5¥I I minutes past two oclock.

52. EQUAL DISTANCESAt 23 Vl3 minutes past three oclock.

53. RIGHT AND LEFTIt must be 41Vl3 minutes past three oclock.

54. AT RIGHT ANGLESTo be at right angles the minute hand must always be exactly fifteen minuteseither behind or ahead of the hour hand. Each case would happen eleven timesin the twelve hours-i.e., every 1 hour 5¥Il minutes. Starting from nine oclock,the eighth addition will give the case 5 hours 43rlI minutes. In the other case,starting from three oclock, the second addition gives 5 hours lOHy! I minutes.Answers 235These are the two cases between five and six, and the latter will, of course, bethe sooner.

55. WESTMINSTER CLOCKThe times were 8 hours 237 lt\43 minutes, and 4 hours 41 13 VI43 minutes. Weare always allowed to assume that these fractional times can be indicated inclock puzzles.

56. HILL CLIMBINGIt must have been 63/.1 miles to the top of the hill. He would go up in

4~ hours and descend in I ~ hours.

57. TIMING THE CARAs the man can walk 27 steps while the car goes 162, the car is clearly goingsix times as fast as the man. The man walks 3 ~ miles an hour: therefore thecar was going at 21 miles an hour.

58. THE STAIRCASE RACEIf the staircase were such that each man would reach the top in a certainnumber of full leaps, without taking a reduced number at his last leap, thesmallest possible number of steps would, of course, be 60 (that is, 3 X 4 X 5).But the sketch showed us that A. taking three steps at a leap, has oneodd step at the end; B. taking four at a leap, will have three only at the end;and C. taking five at a leap, will have four only at the finish. Therefore, wehave to find the smallest number that, when divided by 3, leaves a remainder I,when divided by 4 leaves 3, and when divided by 5 leaves a remainder 4. Thisnumber is 19. So there were 19 steps in all, only 4 being left out in the sketch.

59. A WALKING PUZZLEIt will be found (and it is the key to the solution) that the man fromB. can walk 7 miles while the man from A. can walk 5 miles. Say the distancebetween the towns is 24 miles, then the point of meeting would be 14 miles

236 Answersfrom A. and the man from A. walked 3¥7 miles per hour, while the man fromB. walked 4"Ys miles per hour. They both arrived at 7 P.M. exactly.

60. RIDING IN THE WINDHe could ride one mile in 3¥7 minutes, or 7124 mile per minute. The windwould help or retard him to the extent of ~4 mile per minute. Therefore, withthe wind he could ride 1}24 mile per minute and against the wind, ~4 mile perminute; so that is I mile in 3 minutes or 4 minutes, respectively, as stated.

61. A ROWING PUZZLEThe correct answer is 3o/i 7 minutes. The crew can row Vs of the distance perminute on still water, and the stream does y\2 of the distance per minute. Thedifference and sum of these two fractions are ~o and 1 ~o. Therefore, againstthe stream would take 61)7 minutes (or 817 minutes), and with the stream 6iJ'b(or 3o/i7 minutes).

62. THE ESCALATORIf I walk 26 steps I require 30 seconds, and if I walk 34 steps I require only

18 seconds. Multiply 30 by 34 and 26 by 18 and we get 1,020 and 468,the difference between which is 552. Divide this by the difference between 30and 18 (that is, by 12) and the answer is 46, the number of steps in the stairway, which descends at the rate of 1 step in 1 ~ seconds. The speed at whichI walk on the stairs does not affect the question, as the step from whichI alight will reach the bottom at a given moment, whatever I do in the meantime.

63. SHARING A BICYCLELet Anderson ride 11 ¥I miles, drop the bicycle, and walk the rest of the way.Brown will walk until he picks up the bicycle and then rides to their destination, getting there at exactly the same time as Anderson. The journey takesthem 3 hours 20 minutes. Or you can divide the 20 miles into nine stages of

2~ miles each, and drop the machine at every stage, only you must makeAnswers 237Anderson ride at the start. Anderson will then ride each of his five stages in% hour and walk each of his four stages in % hour, making rus total time 3 Y.!hours. Brown will ride each of his four stages in ~18 hour and walk each of hisfive stages in '!-9 hour, making his total time also 3 Y.! hours. The distances thatAnderson and Brown ride respectively must be in the proportion of 5 to 4;the distances they walk in the proportion of 4 to 5.

64. MORE BICYCLINGA. rides 71 Y:17 miles, B. rides l 13b miles, and C. rides I Ph7 miles, makingthe 20 miles in all. They may ride in any order, only each man should complete his ride in one mount and the second rider must always walk both beforeand after riding. They will each take 3% hours on the journey, and thereforewill all arrive together.

65. A SIDECAR PROBLEMAtkins takes Clarke 40 miles in his car and leaves him to walk the remaining 12 miles. He then rides back and picks up Baldwin at a point 16 milesfrom the start and takes him to their destination. All three arrive in exactly

5 hours. Or Atkins might take Baldwin 36 miles and return for Clarke, whowill have walked his 12 miles. The sidecar goes 100 miles in all, with no passenger for 24 miles.

66. THE DISPATCH RIDERThe answer is the square root of twice the square of 40, added to 40. Thisis 96.568 miles, or, roughly, 96~ miles.

67. THE TWO TRAINSIn 5 seconds both trains (together) go 600 feet, or 81'¥t1 miles per hour. In

15 seconds the faster train gains 600 feet, or 273ft 1 miles per hour. From thiswe get 540/11 miles per hour as the rate of the faster train; and it is clear that

273ft 1 miles per hour is the rate of the other.

238 Answers

68. PICKLEMINSTER TO QUlCKVILLEThere are two possible distances that will fit the conditions-21O miles and

144 miles, only I barred the latter by the words, "at an ordinary rate." With

144 miles A would run 140 miles, while Band D ran 4 miles; so if the latterwent 2 miles per hour, the former would have to go 70 miles per hour-rateswhich are certainly not "ordinary"! With 210 miles B and D go half thespeed of A, and C goes three-quarters the speed of A, so you can give themreasonable rates.

69. THE DAMAGED ENGINEThe distance from Angiechester to Clinkerton must be 200 miles. The trainwent 50 miles at 50 m.p.h. and 150 miles at 30 m.p.h. If the accident had occurred 50 miles farther on, it would have gone 100 miles at 50 m.p.h. and

100 miles at 30 m.p.h.

70. THE PUZZLE OF THE RUNNERSWhile Brown has only run ~ or 'Y.!4 of the course, Tompkins has run the remainder %, less JA!, or 1~4. Therefore Tompkins's pace is IV-! times that ofBrown. Brown has now % of the course to run, whereas Tompkins has only ~.Therefore Brown must go five times as fast as Tompkins, or increase his ownspeed to five times I V-!, that is, s¥.I times as fast as he went at first. But thequestion was not how many times as fast, but "how much faster," and S¥.Itimes as fast is equal to s\I.i times faster than Brown's original speed. The correct answer is therefore 20\l.i times faster, though in practice probably impossible.

71. THE TWO SHIPSThe error lies in assuming that the average speeds are equal. They are not.The first ship does a mile in VI2 hour outwards and in JA! hour homewards.Half of the sum of these fractions is ¥.Is. Therefore the ship's average speed forthe 400 miles is a mile in ¥.Is hour. The average speed of the second ship is amile in YIO hour.Answers 239

72. FIND THE DISTANCEThe distance between the two places must have been 18 miles. The meetingpoints were 10 miles from A and 12 miles from B . Simplymultiply 10 (the first distance) by 3 and deduct the second distance, 12.Could anything be simpler? Try other distances for the meeting points (takingcare that the first meeting distance is more than two-thirds of the second) andyou will find the little rule will always work.

73. THE MAN AND THE DOGThe dog's speed was 16 miles per hour. The following facts will give thereader clues to the general solution. The distance remaining to be walked sideby side with the dog was 81 feet, the fourth power of 3 (for the dog returnedfour times), and the distance to the end of the road was 625 feet, the fourthpower of 5. Then the difference between the speeds (in miles per hour)of man and dog (that is, 12) and the sum of the speeds (20) must be in thesame ratio, 3 to 5, as is the case.

74. BAXTER'S DOGIt is obvious that Baxter will overtake Anderson in one hour, for each willbe 4 miles from the hotel in the same direction. Then, as the dog has beenrunning uniformly at 10 miles an hour during that hour, he must have run 10miles! When a friend put this problem before a French professor of mathematics, he exclaimed: "Man Dieu, queUe serle!" quite overlooking the simplemanner of solution.

75. THE RUNNER'S REFRESHMENTAs the radius is t, the diameter is 2t. The diameter multiplied by 'TT (theGreek letter pi) gives us the circumference, 2t'TT miles. As he goes roundn times, 2t'TTn equals the number of miles run, and, as he drinks s quarts permile, he consumes 2t'TTns quarts. We can put the factors in any order: therefore the answer is 2'TTnts (two pints) or one quart.

240 Answers

76. EXPLORING THE DESERTThe nine men, A, B, C, D, E, F, G, H, J, all go 40 miles together on the Igal. in their engine tanks, when A transfers I gal. to each of the other eightand has I gal. left to return home. The eight go another 40 miles, whenB transfers I gal. to each of the other seven and has 2 gals. to take him home.The seven go another 40 miles, when C transfers I gal. to each ofthe six othersand returns home on the remaining 3 gals. The six go another 40 miles, whenD gives each of five I gal. and returns home. The five go 40 miles, when E giveseach of four I gal. and returns home. The four go another 40 miles, when Fgives each of three I gal. and returns home. The three go 40 miles, whenG gives each of two I gal. and returns home. The two go 40 miles, when Hgives I gal. to J and returns home. Finally, the last man, J, goes another

40 miles and then has 9 gals. to take him home. Thus J has gone 360 miles outand home, the greatest distance in a straight line that could be reached underthe conditions.

77. EXPLORING MOUNT NEVERESTDump 5 rations at 90-mile pointand return to base (5 days). Dump Iat 85 and return to 90 (I day). DumpI at 80 and return to 90 (l day).Dump I at 80, return to 85, pick up Iand dump at 80 (l day). Dump I at

70 and return to 80 (I day). Return tobase (l day). We have thus left I

60 50

40.s"s£.ration at 70 and I at 90. Dump I at 5and return to base (l day). Ifhe mustwalk 20 miles he can do so by goingto 10 and returning to base. Dump 4at 10 and return to base (4 days).Dump I at 10 and return to 5; pick upI and dump at 10 (I day). Dump 2 at

20 and return to 10 (2 days). DumpI at 25 and return to 20 (l day).Dump I at 30, return to 25, pick up Iand dump at 30 (I day). March to 70(2 days). March to base (l ~ days).Total, 23 ~ days.A few claims have been made forfewer than 23 ~ days, but they wereall based on tricks that were actually,or in spirit, forbidden, such as dumping only portions of the sealed boxes,making the man take a long marchAnswers 241fasting, or making him eat his day's ration before starting so as to carry anddump two rations. In the last case he would reaJIy be carrying three rations,one inside and two on his back!If the route had been in a straight march across a desert, the shortest timenecessary would be 86 days, as follows:Dump 42 at 10, return to base (42 days). Dump I at 15, return to 10 (I day).Dump 20 at 20 and return to 10 (20 days). Dump I at 20 and return to

15, pick up I and dump at 20 (I day). Dump 10 at 30 and return to 20 (10days). Dump I at 35, return to 30 (I day). Dump 4 at 40 and return to 30(4 days). Dump I at 40 and return to 35. Pick up I and dump at 40 (I day).Dump 2 at 50 and return to 40 (2 days). Dump I at 55 and return to 50 (Iday). Dump I at 60 and return to 55. Pick up I and dump at 60 (I day).March to base (2 days). Total, 86 days.

78. THE BATH CHAIRIf the man leaving A goes I~ miles at 5 miles per hour, he will take 20minutes; and the return journey at 4 miles per hour will take them 25 minutes.He thus overtakes the invalid at 12:35, when the latter has gone ~ milein 35 minutes, and his rate is I Y7 miles per hour.

79. THE PEDESTRIAN PASSENGERAssume that the train runs for an hour, and that it is itself of the absurdlength of 3 miles. Then, as in the diagram, the train will have gone from B toC = 60 miles, but the passenger will have gone from A to C, or 63 miles. Onthe other hand, if he walks from the front to the rear of the train, the train willhave gone from B to C (again 60 miles), while the passenger will have gonefrom B to D = 57 miles. So that in our first case the man would be travellingover the permanent way at the rate of 63 miles per hour, and in the secondcase, 57 miles per hour.

60 c

242 AnswersSO. MEETING TRAINSAs the journey takes five hours, divide the route into five equal distances.Now, when the lady leaves Wurzletown there are four trains on the way anda fifth just starting. Each of these five she will meet. Also, when she has goneone-fifth of the distance, another will start; at two-fifths, another; at threefifths, another; at four-fifths, another; and when she arrives at Mudville a trainwill be just on the point of starting. If we assume, as we must, that she doesnot meet this one "on the way," or meet the one that arrived at Wurzletownjust as she left, she will have met altogether nine trains on the journey.

81. CARRYING BAGSLet the boy continue to carry one bag for I \-) miles; then hand it to thegentleman, who will carry it to the station. Also let the man carry his bag 27'3miles and then deliver it to the boy, who will carry it for the remainingdistance. Then each of the three persons will have carried one bag 27'3 miles-an equal division of labor.

82. THE ESCALATORLet n equal the number of steps in staircase, and take as the unit of time thetime taken by one step to disappear at bottom.The second man takes 75 steps in n - 75 units of time, or (dividing by 25)

3 steps in n 2575 units. Therefore first man takes I step in n 2575 units. Butfir al 50 . 50' f' I . n - 50 . st man so takes steps ill n - uruts 0 tlllle, or step ill ---so umtsof time. Therefore n 5050 = n 2575 and n = lOO-the required answer.

83. THE FOUR CYCLISTSA, B, C, D could ride one mile in Ii, Y9, VI2, and llI5 of an hour respectively.They could, therefore, ride once round in Y18, ~7, \-)6, and \45 of an hour, andconsequently in V9 of an hour (that is, 67'3 minutes) they would meet for thefirst time. Four times in 67'3 minutes is 267'3 minutes. So that they would complete their task in 26 minutes 40 seconds past noon.Answers 243

84. THE DONKEY CARTThe journey took IOS,41 hours. Atkins walked 53¥.!I miles at the end of hisjourney, Brown walked 1327,41 at the beginning, and Cranby's donkey wentaltogether 804%1 miles. I hope the ass had a good rest after performingthe feat.

85. THE THREE CARSB would pass C in 613 minutes.

86. THE FLY AND THE CARS(1) The fly meets car B in 1 hour 48 minutes. (2) There is no necessity hereto work out the distance that the fly travels-a difficult series for the novice.We simply find that the cars meet in exactly 2 hours. The fly really travels inmiles:

270 270 270 270 'nfini d h f h' . 1 d -1- + W + 100 + 1,000' .. to 1 ty, an t e sum 0 t 1S geometnca ecreasing series is exactly 300 miles.

87. THE SUBWAY STAIRSThe least common multiple of 2,3,4,5,6, and 7 is 420. Deduct 1, and 419is a possible number for the stairs, and every addition of 420 will also work.So the number of steps might be either 419, 839, 1,259, 1,679, etc., onlywe were told that there were fewer than 1,000, and that there was anotherstairway on the line with fewer steps that had the same peculiarity. Thereforethere must have been 839 steps in the Curling Street stairway.

88. THE BUS RIDEThey can ride three times as fast as they can walk, therefore three-quartersof their time must have been spent in walking, and only a quarter in riding.Therefore they rode for 2 hours, going 18 miles, and walked back in 6 hours,thus exactly filling their 8 ho~rs.

244 Answers

89. A QUESTION OF TRANSPORTThe car should take four men 12 miles, and drop them 8 miles from theirdestination; then return 8 miles and pick up four men from the eight who havewalked to that point; then proceed 12 miles, and drop the second four 4 milesfrom destination; then return 8 miles and pick up the last four, who will be

8 miles from the starting point; then drive these 12 miles to their destination,where all will arrive at the same time. The car has thus gone 52 miles, whichwould take 23/s hours. The time of arrival was, therefore, 2:36 P.M.

90. HOW FAR WAS IT?The distance must have been 300 miles.

91. OUT AND HOMEThe distance must be 13 Ii miles, so that he walked into the town in 2%hours and returned in 4% hours, making 7 hours, as stated.

92. THE MEETING CARSThe distance from London to Bugleminster must be 72 miles.

93. A BICYCLE RACEThey will come together 12 minutes from the start.

94. A LITTLE TRAIN PUZZLEThere is no necessity for any algebraical working in the solution of thisproblem, nor need we know the distance between the two stations. Whereverthey meet, just send the two trains back for an hour's journey at their respective rates. One will obviously go 60 miles and the other 40 miles, so they were

100 miles (60 added to 40) apart an hour before they met!

95. AN IRISH JAUNTAfter travelling 20 minutes Pat said we had gone just half as far as the re-Answers 245maining distance to Pig town, so it is clear that from Boghooley to Pigtowntook one hour.Five miles beyond Pigtown we were at a point half as far from Ballyfoyneas from Pigtown. Then we reached Ballyfoyne in an hour, from which we knowthat we took 3 hours from Pigtown to Ballyfoyne, and therefore the time ofthe complete journey was four hours. Now we find that the 5-mile run took

2 hours, so in the 4 hours we must have done 10 miles, which is the correctdistance from Boghooley to Ballyfoyne.

96. A WALKING PROBLEMThe second man, on seeing his friend turn and walk towards him, walkedbackwards 200 yards. It was an eccentric thing to do, but he did it, and it isthe only answer to the puzzle. They could thus have their faces towards eachother and be going in a direct line.

97. THE FALSE SCALESIf the scales had been false on account of the pans being unequally weighted,then the true weight of the pudding would be 154 oz., and it would haveweighed 130 oz. in one pan and 178 oz. the other. Half the sum of the apparentweights (the arithmetic mean) equals 154. But the illustration showed that thepans weighed evenly and that the error was in the unequal lengths of the armsof the balance. Therefore, the apparent weights were 121 oz. and 169 oz., andthe real weight 143 oz. Multiply the apparent weights together and we get thesquare of 143-the geometric mean. The lengths of the arms were in the ratioII to 13.If we call the true weight x in each case, then we get the equations:(-iTx + 4) + (-iTx + 52) = x, and x = 154.

2j({lX + 4) X ({IX + 52) = x, and x = 143.

98. WEIGHING THE GOODSSince one canister weighs an ounce, the first illustration shows that in one

246 Answerspan eight packets equal 3 OZ., and, therefore, one packet will weigh % oz. Thesecond illustration shows that in the other pan one packet equals 6 oz. Multiply % by 6 and we get %, the square root of which is 3h, or I ~ oz. as the realweight of one packet. Therefore, eight packets weigh 12 OZ., which is the correct answer.

99. WEIGHING THE BABYIt is important to notice that the man, baby, and dog weigh together 180 lb.,as recorded on the dial in the illustration. Now, the difference between 180 and

162 is 18, which equals twice the weight of the dog, whose weight is 9 lb.Therefore the baby weighs 30 lb., since 30 less 70 per cent is 9.

100. FRESH FRUITSSince the lower scales tell us that one apple and six plums equal in weightone pear, we can substitute one apple and six plums for the pear in the upperscales without disturbing the balance. Then we can remove six plums fromeach pan in the upper scales, and find that four apples equal four plums.Consequently, one apple equals one plum, and if we substitute a plum for theapple in the lower scales, as they originally stood, we see that seven plumsequal one pear in weight. So the old book says Q.E.D. (quite easily done!).

101. WEIGHING THE TEA(1) With the 5 lb. and 9 lb. weights in different pans weigh 4 lb. (2) Withthe 4 lb. weigh second 4 lb. (3) Weigh third 4 lb. (4) Weigh fourth 4 lb., andthe remainder will also be 4 lb. (5), (6), (7), (8), (9) Divide each portionof 4 lb. in turn equally on the two sides of the scales.

102. AN EXCEPTIONAL NUMBER

1 3452. The successive numbers are 1, 2, 3, 4, 5; 13 X 4 = 52. [VictorMeally supplies the following solution for a six-digit number: 947,658.

94 X 7 = 658.-M. G.)Answers 247

103. THE FIVE CARDSThe number is either 3 9 1 5 7 or 5 7 1 3 9. In either case the productof the two pairs 39 and 57, minus 1, results in 2222.

104. SQUARES AND DIGITSIf a square number terminates in similar digits, those digits must be 4, asin the case of 144, the square of 12. But there cannot be more than three equaldigits, and therefore the smallest answer is 1444, the square of 38.

105. THE TWO ADDITIONSArrange the figures in the following way:

173 85

4 92

177 177and both sums add up alike.

106. THE REPEATED QUARTETTEMultiply 273863 by 365 and the product is 99959995. Working the problembackwards, any number whatever consisting of eight figures with the first fourrepeated is divisible by 73 (and by 137) without remainder, and if it ends in

5 or 0 is divisible by 365 (or by 50005). Knowing this, the highest possibleproduct can be written down at once.

107. EASY DIVISIONAll we need do is to proceed as follows:

7--7101449275362318840579

1014492753623188405797Divide 7 by 7, carry the I to the top line, divide again by 7 and carry the 0 to

248 Answersthe top line, and so on until we come to a 7 in the bottom line without anyremainder. Then stop, for the correct number is found. A solution maybe found for any divisor and with any given figure at the beginning. A general examination of the question is very interesting.If we take the divisor 2 we get the number 2-1 0-5 2-6-3 1 5 7 8-9 4-7 3

6-8-4-.This is a complete circuit. The hyphens are at the places where there is noremainder when divided by 2. Note that the successive figures immediatelyfollowing a hyphen are 1,5,6,3,9, 7, 8, 4, 2, so ifI want my number to begin with an 8, I start 8 4 2 1 0 5, etc., taking the 8 that follows a hyphen.Where there is a complete circuit, as in this case, and in the case of thedivisors 3, 6, and 11, the number of figures in the number will always be 10times the divisor less 2. If you try the divisor 4, there are five broken circuits.Thus, 4-1 025 6- will give you numbers beginning with 4 or I; or 2 0-5 I 2-8-with 2,5, or 8; or 7 1 7948- beginning with 7; or 3076-92- with 9 or 3;or 6 I 5 3 8 4- for a number beginning with 6.Some divisors, like 5 and 9, though producing broken circuits, require thesame total of figures as if they were complete circuits. Our divisor 7 givesthree broken circuits: the one shown above for the initial figures 7, 1, or 4,another for 5, 8, or 2, and a third for 6, 9, or 3.

108. A MISUNDERSTANDINGWe may divide 8 5 7 I 4 2 by 3 by simply transferring the 2 from the end tothe beginning. Or 4 2 8 5 7 1, by transferring the 1.

109. THE TWO FOURSThis is how 64 may be expressed by the use of only two fours with arithmetical signs:The process of simplification shown should make everything quite clear.j (fl4t= J( ~)24 = yI2i2 = y4096 = 64[Interest in the "Four Fours" problem has had periodic revivals since itsAnswers 249first mention in the December 30, 1881, issue of Knowledge, a magazineof popular science edited by the astronomer Richard Proctor. For a recentdiscussion of the problem, see my Scientific American column for January 1964,and the answer section of the column for the following month. A table givingthe integers from 1 to 100, each expressed with four fours, will be foundin L. Harwood Clarke, Fun With Figures (William Heinemann Ltd., 1954),pp. 51-53, and Angela Dunn, Mathematical Bafflers (McGraw-Hill, 1964),pp.5-8.The number 64 is easily expressed with four fours: (4 + 4)(4 + 4), andwith three fours: 4 X 4 X 4. In Recreational Mathematics Magazine, No. 14(1964), M. Bicknell and V. E. Hoggatt show 64 ways of expressing 64 withfour fours, and add a good list of references on the Four Fours problem.D. E. Knuth, in his article "Representing Numbers Using Only One 4,"Mathematics Magazine, Vol. 37, November-December, 1964, pp. 308-310,shows how to represent 64 by using only one four and three kinds of symbols:the square root sign, the factorial sign, and brackets. To express 64 by thismethod requires 57 square root signs, nine factorials, and 18 brackets. A computer program, Knuth reports, found that all positive integers less than 208could be represented in ~ similar manner, and he conjectures that the methodapplies to all positive integers.Dudeney is partially correct in his assertion about the number 113. So faras I know, no one has found a way to represent it without adopting highlyunorthodox symbols or complicated procedures such as Knuth's.-M. G.]

110. THE TWO DIGITSOf course it is entirely a matter of individual taste what arithmetical formsand signs are admissible, but I should personally draw the line at expressionsintroducing "log" and "antilog."A few solutions are as follows:

25 = 52

36 = 6 X 3!

64 = y'46 or (V4)6

25 = 5 + .2

250 Answers

111. DIGITAL COINCIDENCESIf we multiply 497 by 2 we get the product 994. If we add together 497 and

2 we get 499. The figures are the same in both cases. Also 263 multiplied by 2and added to 2 will give 526 and 265 respectively.[Harry Lindgren points out that by inserting 9's one obtains two answerswith any desired number of digits: 4997 + 2 = 4999; 4997 X 2 = 9994;

2963 + 2 = 2965; 2963 X 2 = 5926; and similarly for 49997 + or X 2;

29963 + or X 2; and so on.-M. G.]

112. PALINDROMIC SQUARE NUMBERSThe square of 836 is 698896, which contains an even number of digits andreads backwards and forwards alike. There is no smaller square number containing an even number of figures that is a palindrome.

113. FACTORIZINGIf the number of noughts enclosed by the two ones is 2 added to any multiple of 3, two factors can always be written down at once in this curious way.

1001 = 11 X 91; 1000001 = 101 X 9901; 1000000001 = 1001 X 999001;

1000000000001 = 10001 X 99990001. The last is our required answer, and

10001 = 73 X 137. The multiple of 3 in 11 is 3: therefore we insert 3 noughtsin each factor and one more 9.If our number contained 10 1 noughts, as I suggested, then the multiple of 3is 33 and the factors will contain 33 noughts and 34 nines in the form shown.If the number of noughts in the number be even you can get two factors inthis way: 1001 = 11 X 91; 100001 = 11 X 9091; 10000001 = 11 X 909091,and so on.

114. FIND THE FACTORSThe factors of 1234567890 are 2 X 3 X 3 X 5 X 3607 X 3803. If we multiply 3607 by 10 and 3803 by 9 we get two composite factors 36070 and

34227, which multiplied together produce 1234567890 and have the least possible difference between them.Answers 251

115. DIVIDING BY ELEVENTo be divisible by 11, four of the alternate digits must sum to 17 and theremaining five to 28, or four to 28 and five to 17. Thus, in the example I gave(482539761),4,2,3, 7, 1 sum to 17, and 8, 5, 9, 6 to 28. Now four digitswill sum to 17 in 9 different ways and five to 17 in 2 ways, making 11 together.In each of the 11 cases 4 may be permuted in 24 ways and 5 in 120 ways, ortogether in 2880 ways. So that 2880 X 11 = 31,680 ways. As the nine digitscan be permuted in 362,880 ways, the chances are just 115 to 11 againsta haphazard arrangement being divisible by 11.

116. DIVIDING BY 37Write beneath the number successively, from right to left, the numbers 1,

10, 11, as follows:

4 9 293 082 3

10 11 10 11 10 1 11 10Now, regarding the lower figures as multipliers, add together all the productsof 1 and 10 and deduct all the products by 11. This is the same as adding 13,08,29, and 49 together (99) and deducting eleven times 2 plus 3 plus 1 (66).The difference, 33, will be the remainder when the large number is dividedby 37.Here is the key. If we divide 1, 10, 100, 1000, etc., by 37 we get successivelythe remainders 1, 10, 26, but for convenience we deduct the 26 from 37 andcall it minus 11. If you try 49,629,708,213 you will find the minus or negativetotal 165, or in excess of the positive 99. The difference is 66. Deduct 37 andyou get 29. But as the result is minus, deduct it from 37 and you have 8 ascorrect answer. You can now find the method for other prime divisors. Thecases of 7 and 13 are easy. In the former case you write 1, 3, 2 (I, 3, 2), 1, 3,

2, etc., from right to left, the bracketed numbers being minus. In the lattercase, 1 (3, 4, 1), 3, 4, 1 (3,4, 1), etc.

117. ANOTHER 37 DMSIONCall the required numbers ABCABCABC. If the sum of the A digits, the Bdigits, and the C digits respectively are:

252 AnswersA B C

18 19 8

15 15 15

12 11 22

19 8 18

22 12 11

8 18 19

11 22 12then in the first three groups 11A - lOB = C. In the next two groupsllA - lOB - C = III (3 X 37); and in the last two groups lOB + CllA = Ill. It does not matter what the figures are, but if they comply withthese conditions we can always divide by 37. Here is an example of the firstcaseABCABCABC,

984763251,where the 3 A's sum to 18, the 3 B's to 19, and the 3 C's to 8.You will find 22 cases with the first equation, 10 with the second, and 10 withthe third, making 42 fundamental cases in all. But in every case the A figuresmay be permuted in 6 ways, and the B figures in 6 ways, and the C figures in

6 ways, making 6 X 6 X 6 = 216, which multiplied by 42 gives the answer

9072 ways divisible by 37. As the 9 digits may be permuted in 362880 waysth h 9072 I 39 I . d·· ·bil· e c ances are 362880 or 40 or to agamst lVlSl lty.

118. A DIGITAL DIFFICULTYThere are four solutions, as follows: 2,438,195,760; 3,785,942,160;

4,753,869,120; 4,876,391,520. The last figure must be zero. Any arrangementwith an even figure next to the zero will be divisible by 2, 3, 4, 5, 6, 9, 10, 12,

15, and 18. We have therefore only to consider 7, 11, 13, 16, and 17. To be divisible by 11 the odd digits must sum to 28 and the even to 17, or vice versa.To be divisible by 7 X 11 X 13 = 1001, if we ignore the zero, the numbersformed by the first three and the last three digits must sum to the middlethree. (Note that the third case above is really 474-1386-912, with the 1Answers 253carried forward and added to the 4.) But, we cannot do better than take thelowest multiple (82) of the lowest common multiple of the divisors(12,252,240), which gives ten figures (this is 1,004,683,680), and keep onadding that lowest common multiple until all digits are different.The 199th multiple will give us the first answer, 309 the second, 388 thethird, and 398 the fourth. The work can be considerably shortened by leapingover groups where figures will obviously be repeated, and all the answers maybe obtained in about twenty minutes by the use of a calculating machine.

119. THREES AND SEVENSThe smallest number possible is 3,333,377,733, which is divisible by 3 andby 7, and the sum of its digits (42) also divisible by 3 and by 7. There mustbe at fewest three 7's and seven 3's, and the 7's must be placed as far to theright as possible.

120. ROOT EXTRACTIONThe only other numbers are 5832, 17,576, and 19,683, the cube roots ofwhich may be correctly obtained by merely adding the digits, which come to

18,26, and 27 respectively.

121. QUEER DIVISIONThe smallest number that fulfills the conditions is 35,641,667,749. Othernumbers that will serve may be obtained by adding 46,895,573,610 or anymultiple of it.

122. THREE DIFFERENT DIGITSThe numbers are 162,243,324,392,405,512,605,648,810, and 972.These, we think, are all the cases that exist.

123. DIGITS AND CUBESThere are three solutions. They are 56,169 (the square of 237), where

56 + 69 = 125 (thecubeof5); 63,001 (the square of251), where 63 + 01 =

254 Answers

64 (the cube of 4); 23, I 04 (the square of 152), where 23 + 04 = 27 (the cubeof3).

124. REVERSING THE DIGITS

989,010,989 multiplied by 123,456,789 produces 122,100,120,987,654,321,where the last nine digits are in the reverse order.

125. DIGITAL PROGRESSIONThe Professor's answer was:

297

291

237

231

564

564

564

564

831

837

891

897where the common differences are respectively 267, 273, 327, and 333. Hepointed out that the three digits in the central number may be arrangedin any of the six possible ways, and a solution may be found.[Victor Meally tells me that Victor Thebault, in Parmi les Nombres Curieux,page 140, shows that there are 760 such progressions. In addition to 456 andits permutations, the middle number may be any of the permutations of thefollowing four sets of three digits: 258,267,348 and 357.-M. G.]

126. FORMING WHOLE NUMBERSIf you multiply 6,666 by the sum of the four given digits you will get thecorrect answer. As I, 2, 3, 4 sum to 10, then 6,666 multiplied by 10 gives us

66,660 as our answer. Taking all possible selections of four different digits, theanswer is 16,798,320, or 6,666 X 2,520.

127. SUMMING THE DIGITSThere are several ways of attacking this puzzle, and the answer is

201,599,999,798,400. The sum of the digits is 45 and

45 X 8! = 1,814,400Now write down-

18144

18144

18144

18144Answers 255to nine places, add up and put 00 at the end, and there is the answer.

128. SQUARING THE DIGITSIn four moves 73,34,48,25, we can get 157,326,849, which is the squareof 12,543. But the correct solution is I 5, 84, 46, which gives us the number

523,814,769, the square of 22,887, which is in three moves only.

129. DIGITS AND SQUARES(1) 1,026,753,849 (the square of 32,043); (2) 9,814,072,356 (the square of

99,066).

130. DIGITAL SQUARESThe only two solutions are 567, with its square, 321,489; and 854, with itssquare, 729,316. We need only examine cases where the digits in the rootnumber sum to 9, 18, or 27; or 8, 17, or 26, and it can never be a lower sumthan 317 to form the necessary six figures.

131. FINDING A SQUARETaking the six numbers in their order, the sums of their digits are:

46 31 42 34 25 34

146 7 7 7Again adding, where necessary, the digits until we reach a single figure, weget the second row of numbers, which we call the digital roots. These may becombined in different triplets in eight different ways:

146 147 167 177 467 477 677 777

2 3 5 6 8 9 2 3

256 Answersagain giving the digital roots shown. Now, as shown in Amusements inMathematics, the digital root of every square number must be either 1, 4, 7,or 9, so that the required numbers must have the roots 4, 7, 7, to be a square.The two 7's may be selected in three different ways. But if the fifth number isincluded, the total of the three will end in 189 or 389, which is impossible fora square, as the 89 must be preceded by an even figure or O. Therefore therequired numbers must be: 2,494,651 + 1,385,287 + 9,406,087 = 13,286,025,which is the square of 3,645.As illustrating the value of this new method we may be allowed to quotefrom the late Professor W. W. Rouse Ball:"This application is original on Mr. Dudeney's part. Digital properties arebut little known to mathematicians, and we hope his example may serveto direct attention to the method .... In a certain class of arithmetical problems it is of great assistance."

132. JUGGLING WITH DIGITS

7 + 1 = 8; 9 - 6 = 3; 4 X 5 = 20.

133. EQUAL FRACTIONSThe five answers are as follows:

2 3 79 3 7 29. 3 9 27.2 3 _ 58 .. 2 _ .6 _ 97"4 = (; = 158; 6" = 14 = 58' 6" = 18 = 54' 6" = 9" - 174' 1- 3" - 485'

134. DIGITS AND PRIMESThe 4, 6, and 8 must come in the tens place, as no prime number can endwith one of these, and 2 and 5 can only appear in the units place if alone.When those facts are noted the rest is easy, as here shown:

47

61

89

2

3

5

207Answers 257

135. A SQUARE OF DIGITSIn every one of the following eight sums all the nine digits are used once,and the difference between the successive totals is, throughout, 9:

243 341 154 317 216 215 318 235

675 586 782 628 738 748 654 746

918 927 936 945 954 963 972 981

136. THE NINE DIGITSThe number 94,857,312, multiplied by 6, gives the product 569,143,872, thenine digits being used once, and once only, in each case.[Victor Meally supplies two other solutions: 89,745,321 X 6 = 538,471,926,and 98,745,231 X 6 = 592,471,386.-M. G.)

137. EXPRESSING TWENTY-FOURThe following is a simple solution (by G. P. E.) for three 7's:From this we obtain the answer for three l's by substituting I for 7 in everycase, and putting plus instead of minus.[Dudeney does not give solutions for the remaining digits, but VictorMeally has supplied them:(4+4-4)!(5 - %)!(~- 6)!

8+8+8(V9 + ~)!-M.G.)

258 Answers

138. THE NINE BARRELSThere are forty-two different arrangements. The positions of the I and 9 arefixed. Always place the 2 beneath the 1. Then, if the 3 be beneath the 2 thereare five arrangements. If the 3 be to the right of the I there are five arrangements with 4 under the 2, five with 5 under the 2, four with 6 under 2, two with

7 under 2. We have thus twenty-one arrangements in all. But the 2 might havebeen always to the right of I, instead of beneath, and then we get twenty-onereversed and reflected arrangements (practically similar), making forty-twoin all. Either the 4, 5, or 6 must always be in the center.

139. THE EIGHT CARDSYou need only make the 8 and 9 change places, first turning the 9 round soas to change it to a 6. Then each column will add up 18.

140. FIND THE NUMBERSThe two numbers composed of I's that sum and multiply alike are II and

1.1. In both cases the result is 12.1.

141. MULTIPLYING THE NINE DIGITSDora was not to be caught by George's question. She, of course, immediately gave the correct answer, O. •

142. CURIOUS MULTIPLICANDThe number is 142,857. This is, of course, the recurring decimal fraction ofone-seventh.

143. ADDING THEIR CUBESThe required number is 153. The cubes of I, 5, and 3 are respectivelyI, 125, and 27, and these added together make 153.[Dudeney overlooked a fourth number: 371. Aside from I, these are theonly four numbers that are the sums of the cubes of their digits. For the moreAnswers 259general problem, of finding numbers that are the sums of the nth powers oftheir digits, see Joseph S. Madachy, Mathematics on Vacation (Scribner's,

1966, pp. 163-165.-M. G.]

144. THE SOLITARY SEVENThe restored simple division sum is as follows:

124)12128316(97809

1 1 16

968

868

1003

992

1 1 1 6

1 1 1 6[When Dudeney first published this in The Strand Magazine, a reader,Harold Revell, of Sussex, sent him a formal proof that the solution is indeedunique. Mr. Revell sent me a copy of his proof in 1964, but it is too lengthyto give here.-M. G.]

145. A COMPLETE SKELETON

625)631938(1011.1008

625

693

625

688

625

630

625

5000

5000The three D's that must occur at the bottom show that the divisor is a submultiple of 1,000. The factors therefore can only be 5, 5, 5,2,2,2, x, where x

260 Answersis less than 10. To form the three-figure divisor, one factor at least must be 5,and therefore the last figure must be 5 or O. The subtraction from the single 0near the bottom shows that it is a 5 and at once gives us the 5,000. The factor

2 being excluded from the divisor (or it could not end in a 5), the final figurein the quotient must be 8 (2 X 2 X 2), and the divisor 625, making x afourth 5. The rest is quite easy.Here is an answer:

146. SIMPLE MULTIPLICATION

4539281706

2

9078563412If you divide the first number into pairs-45, 39, etc.-these can be arrangedin any order so long as the 06 is not at the beginning or the 45 at the end.

147. AN ABSOLUTE SKELETONIt can soon be discovered that the divisor must be 312, that 9 cannot be inthe quotient because nine times the divisor contains a repeated figure. Wetherefore know that the quotient contains aJl the figures 1 to 8 once, and therest is comparatively easy. We shall find that there are four cases to try, andthat the only one that avoids repeated figures is the following:

312) 10,114,626,600(32,418,675.

148. ODDS AND EVENS

249)764,752,206(3,071,294.

249)767,242,206(3,081,294.

245)999,916,785 (4,081,293.

245)997,466,785 (4,071,293.

248)764,160,912(3,081,294.

248) 761,680,912(3,071,294.Answers 261If the reader will work out each of these little sums in simple division hewill find that they fulfill all the conditions required by the asterisks and O'sand E's.

149. SIMPLE DIVISIONDivide 4,971,636,104 by 124,972, and the quotient is 39,782. The reader cannow work out the little sum for himself, and he will find that all the conditionsare fulfilled. If we were allowed additional 7's in the dividend, an answerwould be 7,471,076,104 divided by 124,972 equals 59,782.

150. A COMPLETE SKELETONThe first division sum is:

333) 100,007,892(300,324and the second:

29)300,324( 10,356.

151. ALPHABETICAL SUMSThe answer is as follows:

35)19775(565

175

227

210

175

175It is clear that R cannot be I: it must therefore be either 5 or 6 to produce theR in the second line. Then D must be 0 to give the V in the fifth line. Also Mmust be 1,2, 3, or 4, if R is 5, but may be 5 if R is 6. Again, S must be an evennumber if R is 5, to make D a 0, and if R is 6, then S must be 5. When we havediscovered and noted these facts, only a little trial is necessary.

262 Answers

152. ALPHABETICAL ARITHMETIC

17 93X 4 -68

68 25

153. FIGURES FOR LETTERS

6,543 X 98,271 = 642,987,153.

154. THE SHOPKEEPER'S PUZZLEThe only word (not a jumble of letters) that will fit the conditions isREGULATION. Used in the way explained, the actual sum was:The key is as follows:from which we get

36,407+98,521

134,928

155. BEESWAX

1 2 345 6 7 8 9 0ATQBKXSWEP

917947476-408857923

509089553and BEESWAX represents the number 4,997,816.

156. WRONG TO RIGHT

25938+25938

51876

157. LETIER MULTIPLICATION

4973X 8

39784

158. THE CONSPIRATORS' CODE

598

507

8047

9152

159. LETIER-FIGURE PUZZLEAnswers 263It is clear that A must be I, and that B and C must be either 6 and 2 or 3and 5, and that in the third equation they are shown to be 3 and 5, since Dmust be 7. Then E must be 8 in order that D X E should show C = 5. Thenthe rest is easy, and we find the answer as follows: A = I, B = 3, C = 5,D=~E=~F=~H=~J=~K=~L=n

160. THE MILLER'S TOLLThere must have been one bushel and one-ninth, which after taking theone-tenth as toll, would leave exactly one bushel.

161. EGG LAYINGThe answer is half a hen and a half hen; that is, one hen. If one and a halfhens lay one and a half eggs in one and a half days, one hen will lay one eggin one and a half days. And a hen who lays better by half will lay one and ahalf eggs in one and a half days, or one egg per day. So she will lay ten anda half (half a score and a half) in ten and a half days (a week and a half).

162. THE FLOCKS OF SHEEPAdam must have possessed 60 sheep, Ben 50, Claude 40, and Dan 30.If the distributions described had taken place, each brother would have thenhad 45 sheep.

264 Answers

163. SELLING EGGSThe smallest possible number of eggs is 103, and the woman sold 60 everyday. Any multiple of these two numbers will work. Thus, she might havestarted with 206 eggs and sold 120 daily; or with 309 and sold 180 daily. Butwe required the smallest possible number.

164. PUSSY AND THE MOUSEYou have simply to divide the given number by 8. If there be no remainder,then it is the second barrel. If the remainder be 1, 2, 3, 4, or 5, then that remainder indicates the number of the barrel. If you get a remainder greater than

5, just deduct it from 10 and you have the required barrel. Now 500 dividedby 8 leaves the remainder 4, so that the barrel marked 4 was the one thatcontained the mouse.

165. ARMY FIGURESThe five brigades contained respectively 5,670; 6,615; 3,240; 2,730; and

2,772 men. Represent all the fractions with the common denominator 12,012,and the numerators will be 4,004; 3,432; 7,007; 8,316; and 8,190. Combiningall the different factors contained in these numbers, we get 7,567,560, which,divided by each number in turn, gives us 1,890; 2,205; 1,080; 910; and 924.To fulfill the condition that the division contained a "little over 20,000 men,"we multiply these by 3 and have the correct total-21,027.

166. A CRITICAL VOTEThere must have been 207 voters in all. At first 115 voted for the motion and

92 against, the majority of 23 being just a quarter of92. But when the 12 whocould not sit down were transferred to the other side, 103 voted for themotion and 104 against. So it was defeated by I vote.

167. THE THREE BROTHERSArthur could do the work in 143%9 days, Benjamin in 172%1 days, andCharles in 23'l31 days.Answers 265

168. THE HOUSE NUMBERSThe numbers of the houses on each side will add up alike if the number ofthe house be 1 and there are no other houses; if the number be 6, with

8 houses in all; if 35, with 49 houses; if 204, with 288 houses; if 1,189, with

1,681 houses; and so on. But it was known that there were more than 50 andfewer than 500 houses, so we are limited to a single case, and the number ofthe house must have been 204.X2 + x Find the integral solutions of --2- = y2. Then we get the answers:x = Number of houses. y = Number of particular house.

1. . . . . . . . . . . . .. . . . . . . . . 1

8 ...................... 6

49. . . . . . . . . . . . . . . . . . . . .. 35

288 ...................... 204

1,681 ...................... 1,189and so on.

169. A NEW STREET PUZZLEBrown's number must have been 84, and there were 119 houses. The numbers from 1 to 84 sum to 3,570 and those from 1 to 119 to 7,140, which is justdouble, as stated.Write out the successive solutions to the Pellian equation (explained onpage 164 in my book Amusements in Mathematics) 2x2 - 1 = y2, thus:x ~

1

5 7

29 41

169 239

985 1,393and so on. Then the integral half of any value of x will give you the housenumber and the integral half of y the total number of houses. Thus (ignoringthe values 0-0) we get 2-3, 14-20,84-119,492-696, etc.

266 Answers

170. ANOTHER STREET PUZZLEOn the odd side of the street the house must have been No. 239, and therewere 169 houses on that side. On the even side of the street the house musthave been No. 408, and there were 288 houses.In the first case, find integral solution of 2x2 - 1 = y2. Then we get theanswers:x = Number of houses. y = Number of particular house.

1......................... 1

5......................... 7

29 ......................... 41

169 ......................... 239

985 ......................... 1,393and so on.In the second case, find integral solution of 2(X2 + x) = y2.Then we get the answers:x = Number of houses. y = Number of particular house.

1........................ 2

8 ........................ 12

49........................ 70

288 ........................ 408

1,681. ....................... 2,378and so on.These two cases, and the two previous puzzles, all involve the well-knownPellian equation and are related.

171. CORRECTING AN ERRORHilda's blunder amounted to multiplying by 49, instead of by 409. Dividethe error by the difference (328,320 by 360) and you will get the requirednumber-912.

172. THE SEVENTEEN HORSESThe farmer's seventeen horses were to be divided in the proportions 1.-2, \-3, ~.Answers 267It was not stated that the sons were to receive those fractions of seventeen.The proportions are thus o/1s, o/1S, and ¥is, so if the sons receive respectively

9, 6, and 2 horses each, the terms of the legacy will be exactly carriedout. Therefore, the ridiculous old method described does happen to givea correct solution.A correspondent suggested to me the ingenious solution:~, i.e., 2 and lover = 3Y.!, i.e., 3 and lover = 4~, i.e., 9 and lover = 10

17

173. EQUAL PERIMETERSThe six right-angled triangles having each the same, and the smallest possible, perimeter (720), are the following: 180, 240, 300; 120, 288, 312; 144,

270, 306; 72, 320, 328; 45, 336, 339; 80, 315, 325.

174. COUNTING THE WOUNDEDThe three fractions are respectively 4%0, 4~;o, and 4%0. Add together 40, 45,and 48, and deduct twice 60. The result is 13, as the minimum numberfor every 60 patients. Therefore as the minimum (who could have each lostan eye, an arm, and a leg) was 26, the number of patients must have been 120.

175. A COW'S PROGENYNote the following series of numbers, first considered by Leonardo Fibonacci(born at Pisa in 1175), who practically introduced into Christian Europe ourArabic numerals:0, 1, 1,2,3,5, 8, 13,21,34, ... , 46,368.Each successive number is the sum of the two preceding it. The sum ofall numbers from the beginning will always equal 1 less than the next but oneterm. Twice any term, added to the preceding term, equals the next but oneterm. Now, there would be 0 calf in the first year, 1 in the second, 1 inthe third, 2 in the fourth, and so on, as in the series. The twenty-fifth term is

268 Answers

46,368, and if we add all the twenty-five terms or years together we get theresult, 121,392, as the correct answer. But we need not do that addition.When we have the twenty-fourth and twenty-fifth terms we simply say (46,368multiplied by 2) plus 28,657 equals 121,393, from which we deduct 1.

176. SUM EQUALS PRODUCTIf you take any number in combination with 1 and a fraction whose numerator is 1 and its denominator 1 less than the given number, then the sum andproduct will always be the same. Thus, 3 and 1 Y.J, 4 and 1 Y.J, 5 and 114, and soon. Therefore, when I was given 987654321, I immediately wrote down

1 9876~4320 and their sum and product is 987654322 9876~4320Now, the reason why 2 and 2 are often regarded as an exceptional case isthat the denominator is 1, thus 2 and I VI, which happens to make the seconda whole number, 2. But it will be seen that it is really subject to the universalrule. A number may be fractional as well as whole, and I did not make it acondition that we must find a whole number, for that would be impossibleexcept in the case of 2 and 2. Of course, decimal fractions may be used, suchas 6 and 1.2, or 11 and 1.1, or 26 and 1.04.In short, the general solution is n added to or multiplied byn I --I = (n + 1) + --1·n- n177. SQUARES AND CUBESThe solution in the smallest possible numbers appears to be this:

102 - 62 = 100 - 36 = 64 = 43•

103 - 63 = 1,000 - 216 = 784 = 282•

178. CONCERNING A CUBE(1) 6 feet. (2) 1.57 feet nearly. (3) Y36 foot.Answers 269

179. A COMMON DIVISORSince the numbers have a common factor plus the same remainder, if thenumbers are subtracted from one another in the manner shown below theresults must contain the common factor without the remainder.

508,811- 480,608

28,203

723,217- 508,811

214,406Here the prime factors of 28,203 are 3, 7, 17, 79, and those of 214,406 are 2,

23, 59, 79. And the only factor common to both is 79. Therefore the requireddivisor is 79, and the common remainder will be found to be 51. Simple,is it not?

180. CURIOUS MULTIPLICATIONIn the first column write in the successive remainders, which are 1 0000 1 1,or reversed, 1 1 0000 1. This is 97 in the binary scale of notation, or 1 plus

25 plus 26. In the second column (after rejecting the numbers opposite to theremainder 0) we add together 23 X I, 23 X 25, 23 X 26, equals 2,231. Thewhole effect of the process is now obvious. It is merely an operation in thebinary scale.

181. THE REJECTED GUNThe experts were right. The gun ought to have fired 60 shots in 59 minutesif it really fired a shot a minute. The time counts from the first shot, so thatthe second would be fired at the close of the first minute, the third at the closeof the second minute, and so on. In the same way, if you put up 60 posts ina straight line, a yard apart, they will extend a length of 59 yards, not 60.

182. TWENTY QUESTIONSThere are various ways of solving this puzzle, but the simplest is, I think,the following. Suppose the six-figure number to be 843,712. (l) If you divideit by 2, is there any remainder? No. (2) If you divide the quotient by 2, is

270 Answersthere any remainder? No. (3) If you divide again by 2, is there any remainder?No. Your twenty questions will be all the same, and writing from right to left,you put down a zero for the answer "No," and 1 for the answer "Yes." The result will be that after the twentieth question you will get 11001101111111000000.This is 843,712 written in the binary scale. Dropping the final 0 in the unitsplace, the first 1 is the sixth figure backwards. Add together the 6th, 7th, 8th,

9th, 10th, lith, 12th, 14th, 15th, 18th, and 19th powers of 2 and you will get

843,712 in our denary scale.If the number is a low one like 100,000, seventeen questions would be sufficient if only you knew that the 0 had been reached in the quotient, but thethree final questions will merely add three noughts to the left of your binarynumber. But to prevent quibbles as to infinities, etc., it is best to state beforebeginning your questions that zero divided by 2 is understood to mean zerowith no remainder.

183. A CARD TRICKEvery pile must contain 13 cards, less the value of the bottom card. Therefore, 13 times the number of piles less the sum of the bottom cards, and plusthe number of cards left over, must equal 52, the number in the pack. Thus

13 times the number of piles plus number of cards left over, less 52, mustequal sum of bottom cards. Or, which is the same thing, the number of pilesless 4, multiplied by 13, and plus the cards left over gives the answer asstated. The algebraically inclined reader can easily express this in terms of hisfamiliar symbols.

184. THE QUARRELSOME CHILDRENEach parent had three children when they married, and six were bornafterward.

185. SHARING THE APPLESNed Smith and his sister Jane took 3 and 3 respectively, Tom and KateBrown took 8 and 4 respectively, Bill and Anne Jones took 3 and 1 respectively, and Jack and Mary Robinson took 8 and 2 respectively. This accountsfor the 32 apples.

186. BUYING RIBBONMary's mother was Mrs. Jones. Now:Daughters.Hilda bought 4 yds. for $ .16Gladys bought 6 yds. for .36Nora bought 9 yds. for .81Mary bought 10 yds. for 1.00Mothers.Mrs. Smith 8 yds. for $ .64Mrs. Brown 12 yds. for 1.44Mrs. White 18 yds. for 3.24Mrs. Jones 20 yds. for 4.00

187. SQUARE AND TRIANGULARSAnswers 271To find numbers that are both square and triangular, one has to solve thePellian equation, 8 times a square plus 1 equals another square. The successive numbers for the first square are 1,6, 35, etc., and for the relative secondsquares 3, 17, 99, etc. Our answer is therefore 1,225 (352), which is both asquare and a triangular number.

188. PERFECT SQUARESSeveral answers can, of course, be found for this problem, but we think thesmallest numbers that satisfy the conditions are:a = 10,430, b = 3,970, c = 2,114, d = 386.a + b = 10,430 + 3,970 = 14,400 = 1202.a + c = 10,430 + 2,114 = 12,544 = 1122.a + d = 10,430 + 386 = 10,816 = 1042.b + c = 3,970 + 2,114 = 6,084 = 782.b + d = 3,970 + 386 = 4,356 = 662.c + d= 2,114 + 386 = 2,500 = 502.a + b + c + d = 10,430 + 3,970 + 2,114 + 386 = 16,900 = 1302.

272 AnswersThe general solution depends on the fact that every prime number of theform 4m + I is the sum of two squares. Readers will probably like to work outthe solution in full.

189. ELEMENTARY ARITHMETICThe answer must be 273. It is merely a sum in simple proportion: If 5 be 4,then 3Yl will be 273.

190. TRANSFERRING THE FIGURESThe required number is:

2,173,913,043,478,260,869,565,which may be multiplied by 4 and the product divided by 5 by simply movingthe 2 from the beginning to the end.

191. A QUEER ADDITIONWrite the following four numbers, composed of five odd figures, in the formof an addition sum, II, I, I, I, and they will add up to 14.

192. SIX SIMPLE QUESTIONS(I) 8, II I Y.J; (2) 1873; (3) 7 and I; (4) 1\-3; (5) 8\4; (6) %.

193. THE THREE DROVERSJack had II animals, Jim 7, and Dan 21 animals, making 39 animals in all.

194. PROPORTIONAL REPRESENTATIONThe number of different ways in which the ballot may be marked is 9,864,100.

195. A QUESTION OF CUBESThe cubes of 14, 15, up to 25 inclusive (twelve in all) add up to 97,344, whichAnswers 273is the square of 312. The next lowest answer is the five cubes of 25,26, 27,28,and 29, which together equa1315 2•

196. TWO CUBESThe cube of 7 is 343, and the cube of 8 is 512; the difference, 169, isthe square of 13.

197. CUBE DIFFERENCESThe cube of 642 is 264,609,288, and the cube of 641 is 263,374,721, the difference being 1,234,567, as required.

198. ACCOMMODATING SQUARESThe number is 225,625 (the squares of 15 and 25), making the square of 475.

199. MAKING SQUARESAn answer is as follows: 482, 3,362, 6,242, which have a common differenceof 2,880. The first and second numbers sum to 622, the first and third to 822,and the second and third to 982•

200. FIND THE SQUARESIf you add 125 to 100 and also to 164, you get two square numbers, 225 and

289, the squares of 15 and 17 respectively.

201. FORMING SQUARESThe officer must have had 1,975 men. When he formed a square 44 X 44 hewould have 39 men over, and when he attempted to form a square 45 X 45he would be 50 men short.

202. SQUARES AND CUBESIf we make one number 625m6, and the other number double the first, we

274 Answerscan get any number of solutions of a particular series. Thus, if we makem = 1, we get the answer 6252 + 1,2502 = 1253, and 6253 + 1,2503 = 46,8752.

203. MILK AND CREAMHalf a pint of skimmed milk must be added.

204. FEEDING THE MONKEYSThe smallest number of nuts is 2,179. The best way of solving this is to dealfirst with the first two cases, and find that 34 (or 34 added to 143 or anymultiple of it) will satisfy the case for 11 and 13 monkeys. You then have tofind the lowest number of this form that will satisfy the condition for the 17monkeys.

2OS. SHARING THE APPLESThe ratio is clearly 6, 4, and 3, which sum to 13. Therefore the boys receive~'iJ, t'iJ, and :Yl3, or 78, 52, and 39 apples.

206. SAWING AND SPLITTINGThe men must saw 3 Yl3 cords of wood.

207. THE BAG OF NUTSThe five bags contained respectively 27, 25, 18, 16, 14 nuts. Each bagcan be found by subtracting the other two pairs together from 100. Thus,

100 - (52 + 30) = 18, the third bag.

208. DISTRIBUTING NUTSThere were originally 1,021 nuts. Tommy received 256; Bessie, 192; Bob,

144; and Jessie, 108. Thus the girls received 300 and the boys 400, or 100 more,and Aunt Martha retained 321.Answers 275

209. JUVENILE HIGHWAYMENThe woman must have had 40 apples in her basket. Tom left her 30, Bobleft 22, and Jim left 12.

210. BUYING DOG BISCUITSThe salesman supplied four boxes of 17 Ibs. each, and two boxes of 16 Ibs.each, which would make exactly the 100 Ibs. required.

211. THE THREE WORKMENAlec could do the work in 143%9 days; Bill in 1723/.11 days; and Caseyin 23V.ll days.

212. WORKING ALONESixty days and forty days.

213. THE FIRST "BOOMERANG" PUZZLEWhen you are given the remainder after dividing by 3 multiply it by 70, theremainder by 5 multiply by 21, and the remainder by 7 multiply by 15. Addthese results together and they will give you either the number thought of, orthat number increased by some multiple of 105. Thus, if the number thoughtof was 79, then the remainder I multiplied by 70, the remainder 4 multipliedby 21, and the remainder 2 multiplied by 15, added together makes 184.Deduct 105, and you get 79-the number thought of.

214. LONGFELLOW'S BEESThe number of bees must have been 15.

215. LILiV A TI, 1150 A.D.The answer is 28. The trick lies in reversing the whole process-multiplying

2 X 10, deducting 8, squaring the result, and so on. Remember, for example,

276 Answersthat to increase by three-fourths of the product is to take seven-fourths. Andin the reverse process, at this step, you take four-sevenths.

216. BIBLICAL ARITHMETICThere were seven in the Sunday School class. The successive numbers required by the questions in their order are as follows: 12, 7, 6, 10, 7, 50, 30, 5,

15,4,8.

217. THE PRINTER'S PROBLEMThe printer must have purchased the following twenty-seven types:AABCDEEEFGHlJLMNOOPRRSTUUVY

218. THE SWARM OF BEESThere were seventy-two bees.

219. BLINDNESS IN BATSThe fewest possible would be 7, and this might happen in either of threeways:(I) 2 with perfect sight, 1 blind only in the right eye, and 4 totally blind.(2) 1 with perfect sight, I blind in the left eye only, 2 blind in the right eyeonly, and 3 totally blind.(3) 2 blind in the left eye only, 3 blind in the right eye only, and 2 totallyblind.

220. THE MENAGERIEAs the menagerie contained two monstrosities-the four-footed bird andthe six-legged calf-there must have been 24 birds and 12 beasts in all.

221. SHEEP STEALINGThe number of sheep in the flock must have been 1,025. It will be foundthat no mutilation of any sheep was necessary.Answers 277

222. SHEEP SHARINGThe share of Charles is 3,456 sheep. Probably some readers will first havefound Alfred's share, and then subtracted 25 per cent, but this will, of course,be wrong.

223. THE ARITHMETICAL CABBYThe driver's number must have been 121.

224. THE LENGTH OF A LEASEThe number of years of the lease that had expired was 54.

225. MARCHING AN ARMYThere must have been 4,550 men. At first they were placed with 65 in frontand 70 in depth; afterwards 910 in front and 5 in depth.

226. THE YEAR 1927

211 - 1}2 = 1927.

227. BOXES OF CORDITEThe dump officer should give boxes of 18 until the remainder is a multipleof 5. Then, unless this is 5, 10, or 25, he gives this remainder in 15's and 20's.The biggest number for which this system breaks down is 72 plus 25, or 97.Of course, in the case of higher numbers, such as 133, where 108, in six boxesof 18, leaves 25, he would give only one box of 18, leaving 115, which he woulddeliver in one box of 15 and five boxes of 20. But in the case of 97, 72 is thefirst and only case leaving a multiple of 5-that is, 25.

228. THE ORCHARD PROBLEMAt first he had 7,890 trees, which formed a square 88 X 88, and left 146 treesover; but the additional 31 trees made it possible to plant a square 89 X 89,or a total of7,921 trees.

278 Answers

229. BLOCKS AND SQUARESThe smallest number of blocks in each box appears to be 1,344. Surrounding the hollow square 342, the first girl makes her square 502, the second girl

622, and the third girl 722, with her four blocks left over for the corners.

230. FIND THE TRIANGLEThe sides of the triangle are 13, 14, and 15, making 14 the base, the height

12, and the area 84. There is an infinite number of rational triangles composedof three consecutive numbers, like 3, 4, and 5, and 13, 14, and 15, but thereis no other case in which the height will comply with our conditions.The triangles having three consecutive numbers for their sides, and havingan integral area, are:

3

13

51

193

723They are found very simply:

4

14

52

194

724

5

15

53

195

725, etc.

52 = 4 X 14 - 4

194 = 4 X 52 - 14

724 = 4 X 194 - 52,or generally U" = 4 U,,_l - Un-2, or the general mathematical formula: Findx, so that 3(x2 - 1) = a perfect square, where 2x, 2x + 1, 2x - 1 are thesides of the triangle.

231. COW, GOAT, AND GOOSEAs cow and goat eat ¥.is in a day, cow and goose Y60 in a day, and goat andgoose ~o in a day, we soon find that the cow eats 51160 in a day, the goat 1160in a day, and the goose 1!J60 in a day. Therefore, together they will eat %60 ina day, or ¥.io. So they will eat all the grass in the field in 40 days, since thereis no growth of grass in the meantime.Answers 279

232. THE POSTAGE STAMPS PUZZLEThe number of postage stamps in the album must have been 2,519.

233. MENTAL ARITHMETICThere are two solutions with numbers less than ten: 3 and 5, and 7 and 8.The general solution to this problem is as follows:Calling the numbers a and b, we have:a2 + b2 + ab = 0 = /a - mbj2 = a2 - 2amb + b2m2 •. . . b + a = - 2am + bm2,. b- a(2m+ I) .. - m2 _1in which m may be any whole number greater than I, and a is chosento make b rational. The general values are a = m2 - I and b = 2m + 1.

234. SHOOTING BLACKBIRDSTwice 4 added to 20 is 28. Four of these (a seventh part) were killed, andthese were those that remained, for the others flew away.

235. THE SIX ZEROS

100

330

505077099

1,111

236. MULTIPLICATION DATESThere are 215 different dates in this century complying with the conditions,if we include such cases as 25/4/00. The most fruitful year was 1924, when we

280 Answersget the seven cases: 24/1/24, 12/2/24, 2/12/24, 8/3/24, 3/8/24, 6/4/24,

4/6/24. One has only to seek the years containing as many factors as possible.

237. SHORT CUTSTo multiply 993 by 879, proceed as follows: Transfer 7 from 879 to 993,and we get 872 and 1,000, which, multiplied together, produce 872,000. And

993 less 872 is 121, which, multiplied by the 7, will produce 847. Add the tworesults together, and we get 872,847, the correct answer.

238. MORE CURIOUS MULTIPLICATIONThe number is 987,654,321, which, when multiplied by 18, gives

17,777,777,778, with 1 and 8 at the beginning and end. And so on with theother multipliers, except 90, where the product is 88,888,888,890, with 90 atthe end.[Dudeney overlooked such numbers as 1,001; 10,101; and 100,101 (madeup of I's and O's, with l's at the ends and no two consecutive I's), all of whichprovide other answers.-M. G.)

239. CROSS-NUMBER PUZZLEThe difficulty is to know where to start, and one method may be suggestedhere. In reading the clues across, the most promising seems to be 18 across.The three similar figures may be Ill, 222, 333, and so on. 26 down is thesquare of 18 across, and therefore 18 across must be either III or 222, as thesquares of 333, 444, etc., have all more than five figures. From 34 across welearn that the middle figure of 26 down is 3, and this gives us 26 down as thesquare of Ill, i.e., 12321.We now have 18 across, and this gives us 14 down and 14 across. Next wefind 7 down. It is a four-figure cube number ending in 61, and this is sufficient to determine it. Next consider 31 across. It is a triangular number-thatis, a number obtained by summing 1,2,3,4,5, etc. 210 is the only triangularnumber that has one as its middle figure. This settles 31 across, 18 down, 21down, and 23 across. We can now get 29 across, and this gives us 30 down.Answers 281From 29 down we can obtain the first two figures of 15 across, and can complete 15 across and 29 down. The remainder can now be worked out.

240. COUNTING THE LOSSThe number killed was 472. If the reader checks the figures for himself hewill find that there were 72 men in each of the four gangs set to work in the end.The general solution of this is obtained from the indeterminate equation

35x - 48

768which must be an integer, where x = number of survivors. Solving in theusual way, we get x = 528. Therefore, the number killed is 1,000 - 528 = 472.

241. THE TOWER OF PISAThe ball would come to rest after travelling 218 ft. 9Y.! in.

242. A MATCHBOARDING ORDERThe answer is: 8 pieces of20 ft., 1 piece of 18 ft., and 7 pieces of 17 ft. Thus

282 Answersthere are 16 pieces in all, measurmg together 297 ft., in accordance with theconditions.

243. GEOMETRICAL PROGRESSIONI + 3 + 9 + 27 + 81 = 121 = 112,andI + 7 + 49 + 343 = 400 = 202•

244. A PAVEMENT PUZZLEOne floor was 38 ft. (1,444 stones) and the other 26 ft. square (676 stones).

245. THE MUDBURY WAR MEMORIALThe number of posts in hand must have been 180, and the length of theenclosing line 330 ft. Then, at a foot apart, they would require 150 more, butat a yard apart 110 would suffice, and they would have 70 too many.

246. MONKEY AND PULLEYWe find the age of the monkey works out at 116 years, and the age of themother 216 years, the monkey therefore weighing 216 lb., and the weight thesame. Then we soon discover that the rope weighed I y.. lb., or 20 oz.; and, asa foot weighed 4 oz., the length of the rope was 5 ft.

247. UNLUCKY BREAKDOWNSThere must have been 900 persons in all. One hundred wagons started offwith 9 persons in each wagon. After 10 wagons had broken down, therewould be 10 persons in every wagon-"one more." As 15 more wagons hadto be withdrawn on the home journey, each of the remaining 75 wagonswould carry 12 persons-"three more than when they started out in themorning."

248. PAT IN AFRICAPat said, "Begorra, one number's as good as another and a little better, so,Answers 283as there are ten of us and myself, sure I'll take eleven, and it's myself that I'llbegin the count at." Of course, the first count fell on himself. Eleven, startingat No.1, is thus the smallest number to count out all the Britons. He was, infact, told to count twenty-nine and begin at No.9. This would have countedout all the natives. These are the smallest numbers.

249. BLENDING THE TEASThe grocer must mix 70 lb. of the 32¢ tea with 30 lb. of the 40¢ tea.

250. THE WEIGHT OF THE FISHThe fish must have weighed 72 oz. or 4~ lb. The tail weighed 9 oz., thebody 36 oz., and the head 27 oz.

251. CATS AND MICEI t is clear that 999,919 cannot be a prime number, and that if there is to beonly one answer it can have only two factors. As a matter of fact these are

991 and 1,009, both of which are primes, and as each cat killed more mice thanthere were cats, the correct answer is clearly that 991 cats each killed 1,009mice.

252. THE EGG CABINETSay the number of drawers is n. Then there will be 2n - 1 strips one wayand 2n - 3 strips the other, resulting in 4n2 - 4n cells and 4n - 4 strips.Thus, in the twelfth drawer we shall get 23 and 21 strips (44 together), and

528 cells. This applies to all drawers except the second, where we may haveany number of strips one way, and a single one the other. So I and 1 willhere serve (a single strip is not admissible, because "intersecting" was stipulated). There are thus only 262 strips in all, and 2,284 cells (not 264 and 2,288).

253. THE IRON CHAINThe inner width of a link, multiplied by the number of links, and added totwice the thickness of the iron, gives the exact length. Every link put on the

284 Answerschain loses a length equal to twice the thickness of the iron. The inner widthmust have been 2YJ in. This, multiplied by 9 and added to 1 makes 22 in.,and multiplied by 15 and added to 1 makes 36 in. The two pieces of chain,therefore, contained 9 and 15 links respectively.

254. LOCATING THE COINSIf his answer be "even," then the dime is in the right pocket and the nickelin the left; if it be "odd," then the dime is in the left pocket and the nickel inthe right.

255. THE THREE SUGAR BASINSThe number in each basin was originally 36 lumps, and after each cup hadreceived 2 (VI 8) every cup would then hold 6, and every basin IS-a difference of 12.

256. A RAIL PROBLEMThere must have been 51 divisions and 23 whole rails in every division.There were thus 1,173 whole rails, and 50 pairs of halves, making together

1,223 rails as stated.

257. MAKING A PENTAGONLet AB be the required 1 inch in tagon can be marked off-l inch inlength. Make BC perpendicular to AB length.and equal to half AB. Draw AC, Dwhich produce until CD equals CB. AThen join BD, and BD is the radiusof the circumscribing circle. If youdraw the circle the sides of the pen- A B

258. WITH COMPASSES ONLYIn order to mark off the four corners of a square, using the compasses only,first describe a circle, as in the diagram. Then, with the compasses open at thesame distance, and starting from anypoint, A, in the circumference, markoff the points B, C, D. Now, with thecenters A and D and the distanceAC, describe arcs at E, and the distance EO is the side of the squaresought. If, therefore, we mark off Fand G from A with this distance, thepoints A, F, D, G will be the fourcomers of a perfect square.

259. LINES AND SQUARESAnswers 285If you draw 15 lines in the manner A p, C J) E Fshown in the diagram, you will haveformed exactly 100 squares. There are

40 with sides of the length AB, 28 ofthe length AC, 18 of the length AD,

10 of the length AE, and 4 squareswith sides of the length AF, making

100 in all. It is possible with 15 straightlines to form 112 squares, but we wererestricted to 100. With 14 straightlines you cannot form more than

91 squares.The general formula is that with n straight lines we can form as many as(n - 3)(n - l)(n + 1) squares ifn be odd and (n - 2)n(n - 1) ifnbeeven

24 '24'If there are m straight lines at right angles to n straight lines, m being lessm(m - 1)(3n - m - 1) than n, then 6 = number of squares.

260. MR. GRINDLE'S GARDENThe rule is this. When the four sides are in arithmetical progression thegreatest area is equal to the square root of their continual product. Thesquare root of 7 X 8 X 9 X 10 is 70.99, or very nearly 71 square rods. This isthe correct answer.

286 Answers

261. THE GARDEN PATHThe area of the path is exactly 66¥! square yards, which is clearly seen ifyou imagine the little triangular piece cut off at the bottom and removed tothe top right-hand comer. Here is theproof. The area of the garden path is 5S"

55 X 40 = 2,200. And (53YJ X 40) +

66¥.J also equals 2,200. Finally, thesum of the squares of 53YJ and 40must equal the square of 66¥!, as it 4Pdoes.The general solution is as follows:Call breadth of rectangle B, length ofrectangle L, width of path C, andlength of path x.Then ± By(B2 - C2)(B2 + LZ) + C2LZ - BCL x= B2_C2In the case given above x = 66¥!, from which we find the length, 53 YJ.

262. THE GARDEN BEDBisect the three sides in A, B, andE. If you join AB and drop the perpendiculars AD and BC, then ABCD willbe the largest possible rectangle andexactly half the area of the triangle.The two other solutions, FEAG andKEBH, would also serve (all theserectangles being of the same area)except for the fact that they would Genclose the tree. This applies to anytriangle with acute angles, but in thecase of a right-angled triangle there are only two equal ways of proceeding.

263. A PROBLEM FOR SURVEYORSA rectilinear figure of any number of sides can be reduced to a triangle ofAnswers 287equal area, and as AGF happens to be a right-angle the thing is quite easy inthis way. Continue the line GA. Now lay a parallel ruler from A to C, run itup to B and mark the point 1. Then lay the ruler from 1 to D and run it downto C, marking point 2. Then lay it from 2 to E, run it up to D and mark point 3.Then lay it from 3 to F, run it up to E and mark point 4. If you now draw theline 4 to F the triangle G4F is equal in area to the irregular field. As ourscale map shows GF to be 7 inches (rods), and we find the length G4 in thiscase to be exactly 6 inches (rods), we know that the area of the field is halfof 7 times 6, or 21 square rods. The simple and valuable rule I have shownshould be known by everybody-but is not.

264. A FENCE PROBLEMThe diagram gives all the measurements. Generally a solution involves 60a biquadratic equation, but as I saidthe answer was in "exact feet," thesquare of 91 is found to be the sumof two squares in only one way-the

60

60squares of 84 and 35. Insert these numbers as shown and the rest is easy andproves itself. The required distance is 35 feet.

265. THE FOUR CHECKERSDraw a line from A to D. Then draw CE perpendicular to AD, and equalin length to AD. Then E will be the center of another square. Draw a line from

288 AnswersE to B and extend it on both sides. Also draw a line FG through C andparallel to EB and the lines through A and D perpendicular to EB and FG.Now, as H is the center of a corner square, we can mark off the length HE allround the square and we find the board is 10 X 10.F &If the size of the men were not given we might subdivide into more squares,but the men would be too large for the squares. As the distance between thecenters of squares is the same as the width of the squares, we can now complete the board with ease, as shown in the diagram inset.

266. A MILITARY PUZZLEThe illustration shows the supremely easy way of solving thispuzzle. The central star is the officer,and the dots are the men.Answers 289

267. THE HIDDEN STARThe illustration shows the symmetrical star in its exact position in the silkpatchwork cloth. All the other pieces contributed are omitted for the sake ofclearness, and it can be at once located on reference to the illustration givenwith the puzzle. It is surprising how bewildering it is to find the star until youhave been once shown it or have found it, and then it will appear prettyobvious.

268. A GARDEN PUZZLEThe trapezium will be inscribable in a circle. Half the sum of the sides is 29.From this deduct the sides in turn, and we get 9, 13, 17, 19, which, multipliedtogether, make 37,791. The square root of this, 194.4 square rods, will be thearea.

269. A TRIANGLE PUZZLEIf you extend the table on the following page you may get as many rationaltriangles, with consecutive numbers, as you like.

290 AnswersP Q Height Area

2 4 3 6

8 14 12 84

30 52 45 1,170

112 194 168 16,296

418 724 627 226,974

1,560 2,702 2,340 3,161,340Here three times the square of P, added to 4, will make the square of Q.Every value of P is four times the last number less the previous one, and Q isfound in the same way after the first step. The height is half as much again asP, and the area is the height multiplied by half of Q. The middle number ofour three sides will always be found as Q. The last line will give us the first casewhere the area is divisible by 20. The triangle is 2,701, 2,702, 2,703, withheight 2,340.

270. THE DONJON KEEP WINDOWThe illustration will show how the eight lights "whose sides are allsquare window may be divided into equal." Every side of the pane is ofthe same length.It was understood ( though notactually stated) that the lights shouldbe all of the same area, but the fourirregular lights are each one-quarterlarger than the square lights. Andneither the shape nor the number ofsides of the lights are equal. Yet thesolution strictly complies with theconditions as given. If you shut out allthese tricks and quibbles in a puzzleyou spoil it by overloading the conditions. It is better (except in the caseof competitions) to leave certainthings to be understood.

271. THE SQUARE WINDOWThe diagram shows the originalwindow, a yard square. Mter he hadblocked out the four triangles indicated by the dotted lines, he still hada square window, as seen, measuringa yard in height and a yard in breadth.Answers 291l"J.

272. DIVIDING THE BOARDThe distance from end B at which the cut must be made is 5 y'IO - 10 =

5.811+.

273. A RUNNING PUZZLE~ ____ ~~ ____ -,c.

274. THREE TABLECLOTHSThe three tablecloths, each 4 ft. by

4 ft., will cover a table 5 ft. 1 in. by 5ft. 1 in. if laid in the manner hereshown. ABeD is the table top, and

1, 2, and 3 are the three square cloths.Portions of 2 and 3, of course, fallover the edge of the table.Each side of the field is 440 yards;BAE is a right-angled triangle, AEbeing 330 yards and BE 550 yards.Now, if Brown could run 550 yardswhile Adams ran 360 (330 + 30),then Brown can run the remaining

110 yards while Adams runs 72 yards.But 30 + 72 = 102 yards leavesAdams just 8 yards behind. Brownwon by 8 yards.Ai---___ ,;.": .... B , . , : , . 1

292 Answers

275. AN ARTIST'S PUZZLEThe canvas must be 10 in. in width and 20 in. in height; the picture itself

6 in. wide and 12 in. high. The margin will then be as required.

276. IN A GARDENThe garden bed must have been 14 ft. long and 10 ft. in width.

277. COUNTING THE TRIANGLESThere are various ways of making the count, and the answer is 3S.

278. A HURDLES PUZZLEThe old answer is that you can arrange them as in A, and then, by addingone more hurdle at each end, as in B, you double the area. No particular formA 24-

24 11.B

43 12c..D

1156 ~~[~} 8EI'?>was stated. But even if you admit that the original pen was 24 X 1 the answerfails, for if you arrange the SO as in Figure C, the area is increased from 24 squarehurdles to IS6, with accommodation for 6S0 sheep with no extra hurdles. OrAnswers 293you can double the area, as in D, with 28 hurdles only. If all the hurdles mustbe used you might construct it as in Figure E.

279. THE ROSE GARDENMake AD a quarter of the distanceAB, and measure DE and AF each aquarter of Be. Now, if we make Gthe same distance from E that D isfrom F, then AG is the correct widthof the path. If the garden is, forexample, 12 ft. by 5 ft., the path willbe 1 ft. wide, yet it cannot always beAG D Egiven in exact figures, though correctin measurement.

280. CORRECfING A BLUNDERThe correctness of the diagram can be easily proved, since the squares of 15and 20 equal the square of 25; the squares of 15 and 36 equal the square of

39; and the squares of 15 and 8 equal the square of 17. Also, 20 + 8 = 28. Ifa right-angled triangle had been allowed, the one on the left, 15,25,20, woulditself give the solution, since the height on the base 25 would be 12, and themedian line 12~.Perhaps our readers would like to try their hand at constructing the generalsolution to triangles of this class.[Victor Meally has found a second solution to this problem: an obtuse triangle with base 66, sides of 41 and 85, and altitude of 40. The line bisectingthe base is 58. In this case the altitude is exterior to the triangle, meeting theextended base line to form a right triangle with a base of 9, and sides of 40 and

41.-M. G.]

294 Answers

281. THE RUSSIAN MOTORCYCLISTSThe two distances given were 15 miles and 6 miles. Now, all you need dois to divide 15 by 6 and add 2, which gives us 41h. Now divide 15 by 4\6, andthe result (3YJ miles) is the required distance between the two points. Thispretty little rule applies to all such cases where the road forms a right-angledtriangle. A simple solution by algebra will show why that constant 2 is added.We can prove the answer in this way. The three sides of the triangle are 15miles, 9YJ miles (6 plus 3YJ miles) and 171) miles (to make it 21 miles eachway). Multiply by 3 to get rid of the fractions, and we have 45, 28, and 53.Now, if the square of 45 (2,025) added to the square of 28 (784) equal thesquare of 53 (2,809) then it is correct-and it will be found that they do so.

282. THOSE RUSSIAN CYCLISTS AGAINThe diagram gives all the correct distances. All the General had to do wasto square Pipipoff's 60 miles (3,600) and divide by twice the sum of that 60and Sliponsky's 12 miles-that is, by 144. Doing it in his head, he, of course,saw that this is the same as dividing 300 by 12, which at once gave him thecorrect answer, 25 miles, as the distance from A to B. I need not show how allthe other distances are now easily obtained, if we want them.

283. THE PRICE OF A GARDENThe measurements given are absurd, and will not form a triangle. To do sothe two shorter sides must together be greater than the third. The Professorgave it to his pupils just to test their alertness.Answers 295

284. CHOOSING A SITEThis was another little jest. He may build wherever he pleases, for if perpendiculars are drawn to the sides of an equilateral triangle from any pointin the triangle, their united length will be equal to the altitude of the triangle.

285. THE COUNTER CROSSThere are 19 different squares to beindicated. Of these, nine will be of thesize shown by the four A's in the diagram, four of the size shown by theB's, four of the size shown by the C's,and two of the size shown by the D's.If you now remove the six coinsmarked "E," not one of these squarescan be formed from the counters thatremain.[Actually, there are 21 squares. Canthe reader find the two that Dudeneymissed? The answer to the secondpart of the puzzle continues, however, to be correct.-M. G.]

286. THE TRIANGULAR PLANTATION. B~The number of ways in which 3trees may be selected from 21 isTI X 20 X 12 or 1 330' 1 2 3' , ,and a triangle may be formed withanyone of these selections that doesnot happen to be three in a straightline. Let us enumerate these cases ofthree in a line. Three trees may beselected from the dotted line, AB, in

20 ways; from the next line of 5 treesparallel with it, in 10 ways; from the

296 Answersnext line of4, in 4 ways; and from the next parallelof3 trees, in I way-making,in aU, 35 ways in that direction. Similarly BC and the lines parallel with it willgive 35 ways, and AC and the lines parallel with it, 35 ways. Then AD and thetwo lines paraUei with it will give 3 ways, and similarly BF and CE, with theirparallels, will give 3 ways each. Hence 3 trees in a straight line may be selectedin 35 + 35 + 35 + 3 + 3 + 3 = 114 different ways. Therefore 1,330 - 114 =

1,216 must be the required number of ways of selecting three trees that wiUform the points of a triangle.

11rI. THE CIRCLE AND DISCSIn our diagram the dotted lines represent the circumference of the red circleand an inscribed pentagon. The center of both is C. Find D, a point equidistant from A, B, and C, and with radius AD draw the circle ABC. Five discsof this size will cover the circle if placed with their centers at D, E, F, G, andH. If the diameter of the large circle is 6 inches, the diameter of the discs is alittle less than 4 inches, or 4 inches "to the nearest half-inch." It requires alittle care and practice correctly to place the five discs without shifting, unlessyou make some secret markings that would not be noticed by others.Answers 297If readers require a closer approximation or further information as to themanner of solving this puzzle, I cannot do better than refer them to a paper,on "Solutions of Numerical Functional Equations, illustrated by an accountof a Popular Puzzle and its Solution," by Mr. Eric H. Neville, in the Proceedings of London Mathematical Society, Series II, Vol. 14, Part 4. I will just addthat covering is possible if the ratio of the two diameters exceeds .6094185, andimpossible if the ratio is less than .6094180. In my case above, where all fivediscs touch the center, the ratio is .6180340.

288. THE THREE FENCESTo divide a circular field into fourequal parts by three fences of equallength, first divide the diameter ofcircle in four parts and then describesemicircles on each side of the line inthe manner shown in the diagram.The curved lines will be the requiredfences.

289. SQUARING THE CIRCLEIf you make a rectangle with one side equal to the diameter, and the otherthree times the diameter, then the diagonal will be something near correct. InM

298 Answersfact, it would be 1 to l/y'IO, or 1 to 3.1622+. The method we recommend isthe following:In the diagram, AB is the diameter. Bisect the semicircle in D. Now, withthe radius AC mark off the points E and F from A and B, and draw the linesDE and DF. The distance DG, added to the distance GH, gives a quarter ofthe length of the circumference (IK), correct within a five-thousandth part.IKLM is the length of complete straight line.There is another way, correct to a seventeen-thousandth part, but it is alittle more difficult. [See W. W. Rouse Ball, Mathematical Recreations andEssays, revised 11th edition, Macmillan, 1960, p. 348.-M. G.)

290. THE CIRCLING CARSince the outside wheels go twice as fast as the inside ones, the circle theydescribe is twice the length of the inner circle. Therefore, one circle is twicethe diameter of the other, and, since the wheels are 5 ft. apart, the diameterof the larger circle is 20 ft. Multiply 20 ft. by 3.1416 (the familiar approximatevalue for "pi") to get 62.832 ft. as the length of the circumference of thelarger circle.

291. SHARING A GRINDSTONEThe first man should use the stone until he has reduced the radius by 1.754in. The second man will then reduce it by an additional 2.246 in., leaving thelast man 4 in. and the aperture. This is a very close approximation.

292. THE WHEELS OF THE CARThe circumference of the front wheel and the rear wheel respectively musthave been 15 ft. and 18 ft. Thus IS ft. goes 24 times in 360 ft., and 18 ft. 20times-a difference of four revolutions. But if we reduced the circumferenceby 3 ft., then 12 goes 30 times, and IS goes 24 times-a difference of 6revolutions.Answers 299

293. A WHEEL FALLACYThe inner circle has half the diameter of the whole wheel, and thereforehas half the circumference. If it merely ran along the imaginary line CD itwould require two revolutions: after the first, the point D would be at E. Butthe point B would be at F, instead of at G, which is absurd. The fact is theinner circle makes only one revolution, but in passing from one position tothe other it progresses partly by its own revolution and partly by carriage onthe wheel. The point A gets to B entirely by its own revolution, but if youimagine a point at the very center of the wheel (a point has no dimensionsand therefore no circumference), it goes the same distance entirely by whatI have called carriage. The curve described by the passage of the point A toB is a common cycloid, but the point C in going to D describes a trochoid.We have seen that if a bicycle wheel makes one complete revolution,so that the point A touches the ground again at B, the distance AB is the exact length of the circumference, though we cannot, if we are given the lengthof the diameter, state it in exact figures. Now that point A travels in the direction of the curved line shown in our illustration. This curve is called, as Ihave said, a "common cycloid." Now, if the diameter of the wheel is 28 inches,we can give the exact length of that curve. This is remarkable-that we cannot give exactly the length from A to B in a straight line, but can stateexactly the length of the curve. What is that length? I will give the answer atonce. The length of the cycloid is exactly four times that of the diameter.

300 AnswersTherefore, four times 28 gives us 112 inches as its length. And the area of thespace enclosed by the curve and the straight line AB is exactly three times thearea of the circle. Therefore, the enclosed space on either side of the circle isequal in area to the circle.

294. A FAMOUS PARADOXOf course, every part of the wheel revolves round the axle at a uniformspeed, and, therefore, in the case of a fixed wheel, such as a grindstone, theanswer is in the negative. But in the case of a bicycle wheel in motion along aroad it is an undoubted fact that what is the upper part for the time being always moves faster through space than the lower part. If it did not do so, noprogress would be made and the cyclist would have to remain as stationaryas the grindstone.Look at our diagram and you will see the wheel in four different positionsthat occur during one complete revolution from A1 to A4 • I have earlier explained the peculiar curve, called a common cycloid, that is described by apoint on the edge of the tire. The curve is shown here for two points at A1and B1. Note that in a half-revolution A1 goes to A3 and B1 to B3, equal distances. But neither point moves throughout at a uniform speed. This is atonce seen if we examine the quarter-revolution, where A1 has only moved asfar as A2, while B1 has gone all the way to B2• We thus see that a point onthe rim moves slowest through space when at the bottom, and fastest whennear the top.And here is a simple practical way of demonstrating it to your unbelievingfriends without the aid of my diagrams. Draw a straight line on a sheetof paper and lay down a penny with the base of Lincoln's head on the line.Now make the penny run along the line a very short distance to the right andthen to the left. That the base hardly leaves the original point on the line,Answers 301while Lincoln's head travels a considerable distance, ought to be at once obvious to everybody. It should be quite convincing that the part of the wheelthat is for the time being at the top moves faster through space than the partat the bottom.

295. ANOTHER WHEEL PARADOXI have already shown that, if you mark a spot on the circumference of abicycle wheel, that spot, when the wheel is progressing, will describe in spacea curve known as a common cycloid. If, however, you mark the edge of theflange of a locomotive or railroad-car wheel, the spot will describe a trochoidcurve, terminating in nodes or loops, as shown in the diagram. I have showna wheel, with flanges below the railway line, in three positions-the start, ahalf-revolution, and a complete revolution. The spot marked Al, has gone toA2 and A3. As the wheel is supposed to move from left to right, trace withyour pencil the curve in that direction. You will then find that at the lowerpart of the loop you are actually going from right to left.The fact is that "at any given moment" certain points at the bottom of theloop must be moving in the opposite direction to the train. As there is an infinite number of such points on the flange's circumference, there must be aninfinite number of these loops being described while the train is in motion.In fact, certain points on the flanges are always moving in a direction oppositeto that in which the train is going.

296. A MECHANICAL PARADOXThe machine shown in our illustration on page 302 consists of two pieces ofthin wood, B, C, made into a frame by being joined at the corners. This frame,by means of the handle, n, may be turned round an axle, a, which pierces theframe and is fixed in a stationary board or table, A, and carries within the frame

302 AnswersnA~ ____________________________ ~an immovable wheel. This first wheel, D, when the frame revolves, turns asecond and thick wheel, E, which, like the remaining three wheels, F, G, and H,moves freely on its axis. The thin wheels, F, G, and H, are driven by thewheel E in such a manner that when the frame revolves H turns the sameway as E does, G turns the contrary way, and F remains stationary. The secret lies in the fact that though the wheels may be all of the same diameter,and D, E, and F may (D and F must) have an equal number of teeth, yet Gmust have at least one tooth fewer, and H at least one tooth more, than D.Readers will find a full account of this paradox and its inventor in a littlebook, Remarkable Men, published by the Society for the Promotion of Christian Knowledge.

297. THE FOUR HOUSEHOLDERSThe simplest, though not the only solution, is that shown in our illustration.

298. THE FIVE FENCESThe illustration explains itself.Answers 303

299. THE FARMER'S SONSThe illustration shows the simplesolution to this puzzle. The land isdivided into eight equal parts, eachcontaining three trees.

300. A VOIDING THE MINESEThe illustration shows the passage through the mines in two straight courses.

304 Answers

301. SIX STRAIGHT FENCESThe six straight fences are so drawnthat every one of the twenty trees is ina separate enclosure. We stated thattwenty-two trees might be so enclosedin the square by six straight fences iftheir positions were more accommodating. We will here state that in sucha case every line must cross everyother line without any two crossingscoinciding. As there are in our puzzleonly twenty trees, this is not necessary, and it will be seen that four ofthe fences cross only four othersinstead of five.

12

302. DISSECTING THE MOONThe illustration shows that the fivecuts can be so cunningly made as toproduce as many as twenty-one pieces.Calling the number of cuts n, thenin the case of a circle the maximumnumber of pieces will be n2 i n + I,but in the case of the crescent it willb n2 + 3n 1 e 2 +.

303. DRAWING A STRAIGHT LINETake pieces of thick cardboard (they need not have straight edges!) and jointhem with shoemakers' eyelets, as in the illustration. The two long pieces areof equal length from center of eyelet to eyelet, and the four pieces at the bot-tom forming a diamond are all ofequal length. Nails or pins at A and Bfasten the instrument to the table, Bbeing so fixed that the distance fromA to B is the same as from B to C.Then the pencil at D (if all is accurately made and adjusted) will drawthe straight line shown.[For other linkages that do the job,see A. B. Kempe, How to Draw aStraight Line (1877), reprinted byChelsea, 1953.--M. G.]

304. DRAWING AN ELLIPSEAnswers 305DDraw the two lines CD and EF at right angles (CD being equal to the required length, 12 inches, and EF to the required breadth, 8 inches), intersectEC~~~--------+---~~--~--~DFing midway. Find the points A and B, so that AF and FB each equals half thelength CD, that is 6 inches, and place your pins at A and B, making the lengthof your loop of string equal to ABF A. Say the distance CA = x. Then, when

306 Answersthe pencil is at F the length of string is 12 + (12 - 2x) = 24 - 2x, andwhen the pencil is at C the length of string is 2(12 - x) = 24 - 2x also,proving the correctness of the solution.

305. THE BRICKLAYER'S TASKA glance at the illustration will show that if you could cut off the portionof wall marked 1 and place it in the position indicated by 2, you would have.-----~1Z~~~; __ .·.·~ A B Ca piece of straight wall, BC, enclosed by the dotted lines, exactly similar to thewall AB. Therefore, both men were wrong, and the price should be the samefor the portion of wall that went over the hill as for the part on the level. Ofcourse, the reader will see at a glance that this will only apply within a certainlimitation. But an actual drawing of the wall was given.

306. MEASURING THE RIVERMeasure any convenient distancealong the bank from A to C, say 40yards. Then measure any distanceperpendicularly to D, say 12 yards.Now sight along DB and find thepoint E. You can then measure thedistance from A to E, which will herebe 24 yards, and from E to C, whichwill be 16 yards. Now AB: DC =AE: EC, from which it is evident thatAB, the width of the river, must be

18 yards.Answers 307

307. PAT AND HIS PIGThe pig will run and be caught at 667) yards, and Pat will run 133YJ yards.The curve of Pat's line is one of those curves the length of which may beexactly measured. It is 2an2 I' where the pig's speed is assumed to be 1, andn -Pat runs n times as fast, and a is the initial distance between Pat and the pig.

308. THE LADDERThe distance from the top of the ladder to the ground was ~ of the lengthof the ladder. Multiply the distance from the wall-4 yards-by the denominator of this fraction-5-and you get 20. Now deduct the square of thenumerator from the square of the denominator of~, and you have 9, whichis the square of 3. Finally, divide 20 by 3, and there is the answer 67) yards.

309. A MA VPOLE PUZZLEThe height of the pole above ground must have been 50 ft. In the first caseit was broken at a distance of 29 ft. from the top, and in the second case

34 ft. from the top.

310. THE BELL ROPEThe bell rope must have been 32 ft. I ~ in. in length from ceiling to floor.

311. THE DISPATCH RIDER IN FLANDERSOf course, a straight line from A to C would not be the quickest route. Itwould be quicker to ride from A to E and then direct to C. The quickest possible route of all is that shown in the diagram on the following page by thedotted line from A to G (exactly I mile from E) and then direct to C.It is necessary that the sine of the angle FGC shall be double the sine ofAGH. In the first case the sine is 6 divided by the square root of 62 + 32,which is 6 divided by the square root of 45, or the same as 2 divided by the

308 Answerssquare root of 5. In the second case the sine is I divided by the square rootof 12 + 22, which is I divided by the square root of 5. Thus the first is exactlydouble the second.D ~~~ I MILE. F ________ ~6~M~'~"~S~ ________ ~C.- ,.,.... ..' ~t.<l" Til 'VE.. /G .. : oJ .- l ,t'l , SAND,A H

312. THE SIX SUBMARINESIt will be seen from the illustrationthat this puzzle is absurdly easywhen you know how to do it! And yetI have not the slightest doubt thatmany readers found it a hard nut tocrack. It will be seen that every matchundoubtedly touches every othermatch.[The number of matches can beincreased to seven and still meet thepuzzle's conditions. See my ScientificAmerican Book of Mathematical Puzzles & Diversions (Simon & Schuster,

1959), p. lI5.-M. G.]

313. ECONOMY IN STRINGThe total length of string that passes along the length, breadth, or depthmust in every case be the same to allow of the maximum dimensions-that is,

4 feet. When the reader is told this, or has found it for himself (and I thinkthe point will be found interesting), the rest is exceedingly easy. For the stringAnswers 309passes 2 times along length, 4 times along breadth, and 6 times along depth.Therefore 4 feet divided by 2, 4, and 6 will give us 2 feet, 1 foot, and "'3 footrespectively for the length, breadth, and depth of the largest possible parcel.The following general solution is by Mr. Alexander Fraser. Let the stringpass a times along length x, b times along breadth y, and c times along depthz, and let length of string be m.Then ax + by + cz = m. Find maximum value of xyz.First find maximum area of xy._ _ n - by _ n b 2 dxy _ n 2b_ Put ax + by _ n, x _ --, xy _ - y - - y, dy _ - - - y _ 0, a a a a a. n b n· .y = 2b' or ~ = 2'· .. ax also = 1, and ax = by.Similarly, ax = by = cz = ~.. m m m m3· . x = 3a'y = 3b' z =)c' and xyz = 27abc'In the case of the puzzle, a = 2, b = 4, c = 6, m = 12 .. '. x = 2,y = I, z = 13.xyz = IY.!.

314. THE STONE PEDESTALThe cube of a square number is always a square. Thus:-The cube of 1 is I, the square of l.The cube of 4 is 64, the square of 8.The cube of 9 is 729, the square of 27.The cube of 16 is 4,096, the square of 64.And so on.We were told to look at the illustration. If there were one block in pedestaland one in base, the base would be entirely covered, which it was not. If 64in pedestal and base, the side of the former would measure 4 feet, and the sideof square 8 feet. A glance will show that this is wrong. But 729 blocks in eachcase is quite in agreement with the illustration, for the width of the pedestal

310 Answers(9 feet) would be one-third of the width of the square (27 feet). In all thesuccessive higher cases the square will be increasingly too large for the pedestalto be in agreement with the illustration.

315. A CUBE PARADOXIt is a curious fact that a cube canbe passed through another cube ofsmaller dimensions. Suppose a cube tobe raised so that its diagonal AB isperpendicular to the plane on whichit rests, as in Figure I. Then the resulting projection will be a regularl~'O Bhexagon, as shown. In Figure 2 thesquare hole is cut for the passage ofa cube of the same dimensions. Butit will be seen that there is room forcutting a hole that would pass a cubeof even larger dimensions. Therefore,the one through which I cut a holewas not, as the reader may havehastily supposed, the larger one, butthe smaller! Consequently, the largercube would obviously remain theheavier. This could not happen if thesmaller were passed through thelarger.

316. THE CARDBOARD BOXThere are eleven different shapes in all, if turning over is allowed, and theyare as shown. If the outside of the box is blue and the inside white, and everypossible shape has to be laid out with white uppermost, then there are twentydifferent ways, for all except Nos. I and 5 can be reversed to be different.Answers 311

317. THE AUSTRIAN PRETZELThe pretzel may be divided into asmany as ten pieces by one straightcut of the knife in the direction indicated in the illustration. ~. .. 6

318. CUTTING THE CHEESEMark the mid-points in BC, CH, HE, EF, FG, and GB. Then insert theknife at the top and follow the direction indicated by the dotted plane. ThenrW"/:; , .... G. ••.•" ", .... r •

1 2.the two surfaces will each be a perfect hexagon, and the piece on the right will,in perspective, resemble Figure 2.

319. THE FLY'S JOURNEYA clever fly would select the routeshown by the line in the illustration,which will take him 2.236 minutes. Hewill not go in the direction indicatedby the dotted line that will probablyhave suggested itself to the reader.This is longer, and would take moretime.

320. THE TANK PUZZLE(I) The water rises 1.8 inches, and (2) rises an additional 2.2 inches.

312 Answers

321. THE NOUGAT PUZZLEFirst cut off the piece marked A from the end, I in. thick. The remaindercan then be cut in the manner shown, into twenty-four pieces of the requiredsize, 5 X 3 X 2 ~ in. All but four of the pieces are visible-two under Bandtwo under C.J6 i ...~ .... C ...-.':;.~ ......... : ........ :.: .... >~: ......... :: ........ ~ ...... : ... -. ....... - ... . , . • " I i .... ··+ ·····i·· .... i··· .. · : ....... ' , , . . . :

322. AN EASTER EGG PROBLEMThe volumes of similar solids are as the cubes of corresponding lengths. Thesimplest answer, to be exact as required, is that the three small eggs were I ~in., 2 in., and 2 ~ in. respectively in length. The cubes of these numbers are

2*, 8, and mAl, the sum of which is exactly 27-the cube of 3. The nexteasiest answer is 2% in., 2 in., and Y.J in. But there is an infinite numberof answers.

323. THE PEDESTAL PUZZLEThe man made a box 3 X I X I ft. inside, and into this he placed thepedestal. Then he filled the box with fine dry sand, shaking it down and levelling the top. Then he took out the pedestal, and the sand was shaken downand levelled, when the surface was found to be exactly 2 ft. from the top ofthe box. It was, therefore, obvious that the pedestal, when completed, contained 2 cubic ft. of wood, and that I cubic ft. had been removed.

324. THE SQUIRREL'S CLIMBThe squirrel climbs 5 ft. in ascending 4 ft. of the pole. Therefore he travels

20 ft. in a 16-ft. climb.Answers 313

325. THE FLY AND THE HONEYThe drop of honey is represented by H, the fly by F. The fly clearly has togo over the edge to the other side. Now, imagine we are dealing with a cylinderof cardboard. If we cut it we can lay it out flat. If we then extend the line ofthe side I inch to B, the line FB will cut the edge at A, which will be the pointat which the fly must go over. The shortest distance is thus the hypotenuse ofa right-angled triangle, whose height is 4 and base 3. This we know is 5, so thatthe fly has to go exactly 5 inches.B

3'·

326. PACKING CIGARETTESSay the diameter of a cigarette is 2 units and that 8 rows of 20 each, as inFigure A (that is, 160 cigarettes) exactly fit the box. The inside length of the box

314 Answersis therefore 40 and the depth 16. Now, if we place 20 in the bottom row, and,instead of placing 20 in the next row, we drop 19 into the position shown inFigure B, we save .268 (i.e., 2 - V3) in height. This second row, and every additional row of 20 and 19 alternately, will increase the height by 1.732.Therefore, we shall have 9 rows reaching to a height of 2 + 8 X 1.732 or

15.856, which is less than our depth of 16. We shall thus increase the numberof cigarettes by 20 (through the additional row), and reduce it by 4 (l in eachrow of 19), making a net increase of 16 cigarettes.

327. A NEW CUTTING-OUT PUZZLEMake the cuts as shown in the illustration and fit the pieces into the placesenclosed by the dotted lines.[The cuts can be shifted to make a more symmetrical solution. See HarryLindgren, Geometric Dissections, (D. Van Nostrand, 1964), p. 40, Figure 9.2.-M.G.]Answers 315

328. THE SQUARE TABLE TOPThe illustration shows the simplest, and I think the prettiest, solution insix pieces. Move piece A up a step on B and you have the original piece

12 X 12. Move C up a step on D and the two pieces will join E and form thesquare 15 X 15. The piece 16 X 16 is not cut.

16f E. l5

26

9 10

329. THE SQUARES OF VENEERThe sides of the two squares must be 24 in. and 7 in., respectively. Make thecuts as in the first diagram and the pieces A, B, and C will form a perfectsquare as in the second diagram. The square D is cut out intact.

18 r 6 18S :B CbIS

8 » ~

8 17 A 8 24b

17 A s D 1~lY C 1

6

18 7 18 6

316 Answers

330. DISSECTING THE LETTER EThe first illustration shows how to cut the letter into five pieces that will fittogether to form a perfect square, without turning over any pieces.If pieces may be turned over, the dissection can be accomplished with fourpieces, as shown in the second illustration.

331. HEXAGON TO SQUARECut your hexagon in half and place the two parts together to form the figure ABCD. Continue the line DC to E, making CE equal to the heightCF. Then, with the point of your compasses at G, describe the semicircleDHE, and draw the line CH perpendicular to DE. Now CH is the mean pro-Answers 317portional between DC and CE, and therefore the side of the required square.From C describe the arc HJ, and with the point of your compasses at Kdescribe the semicircle DJe. Draw CJ and DJ. Make JL equal to JC, andcomplete the square. The rest requires no explanation.This solution was first published by me in the Weekly Dispatch, in August,

1901.[Victor Meally informs me that Dudeney's solution had earlier been discovered by Paul Busschop. It appeared in Nouvelle Correspondance Mathematique, Brussels, 1875, Vol. II, p. 83.-M. G.]

332. SQUARING A STARI give the very neat solution by Mr. E. B. Escott, of Oak Park, Illinois. Thefive pieces of the star form a perfect square. Find side of equal square (a meanproportional between AB and BC) and make BD equal to such side. Dropperpendicular from A on BD at E and AE will equal BD. The rest is obvious.

318 Answers[Edward B. Escott, a mathematics teacher and insurance company actuary,was a number-theory expert who contributed to many mathematical journalsuntil his death in 1946. See also Puzzle No. 348. For other five-piece solutionsof the star to square, see Lindgren, Geometric Dissections, p. 18, Figures 3.3and 3.4.-M. G.)

333. THE MUTILATED CROSSThe illustration shows clearly how to cut the mutilated cross into four piecesto form a square. Just continue each side of the square until you strikea corner, and there you are!I(sB1~

334. THE VICTORIA CROSSThe illustration will show how to cut the cross into seven pieces to form asquare.This solution was sent to me by Mr. A. E. Hill.Answers 319[I have been unable to identify Mr. Hill, who discovered this truly remarkable dissection, and do not even know if he lived in England or somewhereelse.-M. G.]

335. SQUARING THE SWASTIKAThe illustration shows how the swastika should be cut into four parts andplaced together to form a square. The direction of the nearly horizontal cut isobvious, the other is at right angles to it.[See Lindgren, Geometric Dissections, p. 43, Figures 9.15 and 9.16, for moresymmetrical four-part solutions.-M. G.]

336. THE MALTESE CROSSCut the star in four pieces across the center, and place them in the fourcorners of the frame. Then you have a perfect Maltese Cross in white, asindicated.

320 Answers

337. THE PIKA TES' FLAGThe illustration will show that theflag need only be cut in two piecesalong the zigzag line. If the lower pieceis then moved up one step we shallget a flag with the required ten stripes.

338. THE CARPENTER'S PUZZLEIn order that it may be cut in two pieces, on the step principle, to forma square, take any rectangular board whose sides are the squares of two consecutive whole numbers. Thus, in the following table, the square of 1 (1) andthe square of 2 (4); or 2 (4) and 3 (9); or 3 (9) and 4 (l6);-and soon. The table may be extended to any length desired.Sides No. of Steps Side of Square

1 X 4 1 2

4X9 2 6

9 X 16 3 12

16 X 25 4 20

25 X 36 5 30In Figure I is shown the simple case of a board 1 X 4, in Figure II the caseof 4 X 9, and Figure III shows the case of 16 X 25. It will be seen thatthe number of steps increases regularly as we advance, but with the table theyare easily found. Thus, for the case 16 X 25, as the side of the square will be

20, the steps will be 20 - 16 = 4 in height and 25 - 20 = 5 in breadth.As the sides are square numbers, and two square numbers multipliedtogether always make another square, the area will always be a perfectsquare. But we must not conclude from this that a board, say, 9 X 25 wouldAnswers 321work just because its area is a square with side IS. Figure IV shows thebest that can be done in that case, but there are three pieces instead oftwo as required. This is because 9 is not a multiple of the added height, 6, nor

25 a multiple of the reduced length, 10. Consequently, the steps cannotbe formed.II

25 -------------lB i DZ' --i!--: II I

9 5 Bz.sOf course, any multiple of the sides will work. Thus, a board 8 X 18is solved exactly like 4 X 9, in two steps, by just doubling all the measurements. Similarly, a board 4 X 6\4 will work, for it is the same ratio as 16 X 25,the steps being I in height and I \4 in breadth. In the former case we shouldreduce it like a fraction, and in the second case multiply it by 4 to get rid ofthe fraction. Then we should see that 4 X 9 and 16 X 25 were, in each case,squares of consecutive numbers and know that a solution is possible.

322 Answers

339. THE CRESCENT AND THE STARThough we cannot square a circle, certain portions of a circle may besquared, as Hippocrates first discovered. If we draw the circle in the diagramand then, with the point of the compasses at E, draw the arc BA, the area ofAcB.' .,-., •the lune or crescent is exactly the same as the area of the triangle ABE. As weknow the line AB to be 12 in., the area of the triangle (and therefore of thecrescent) is obviously 36 sq. in. Also, as the triangle D is known to contain

3 sq. in., the star, which is built up of twelve such triangles, contains 36 sq. in.Therefore the areas of the crescent and the star were exactly the same.

340. THE PATCHWORK QUILTExcept for my warning the reader might have supposed that the dark zigzag line from A to B would solve the puzzle. But it will not, because the piecesare not of the same size and shape. It would be all right if we could go alongthe dotted line D instead of C, but that would mean cutting a piece. We mustcut out all the shaded portion in one piece, which will exactly match the other.Answers 323One portion of the patchwork is drawn in just to guide the eye when comparing with the original.

341. THE IMPROVISED CHECKERBOARDThe illustration shows how to cut into two pieces, A and B, that will fit together and form the square board ....-.--~ IA

18 B

324 Answers

342. TESSELLATED PAVEMENTSThe illustration shows how the square space may be covered with twentynine square tiles by laying down seventeen whole and cutting each of the remaining twelve tiles in two parts. Two parts having a similar number form awhole tile.

343. SQUARE OF SQUARESThere is, we believe, only one solution to this puzzle, here shown. The fewest pieces must be 11, the portions must be of the sizes given, the three largestpieces must be arranged as shown, and the remaining group of eight squaresmay be "reflected" but cannot be differently arranged.[For a discussion of the general problem, still unsolved, of dividing a squarelattice of any size, along lattice lines, into the minimum number of smallersquares, see J. H. Conway, "Mrs. Perkins's Quilt," Proceedings of the Cambridge Philosophical Society, VoL 60, 1964, pp. 363-368; G. B. Trustrum'spaper of the same title, in the same journal, VoL 61, 1965, pages 7-11; and myScientific American column for September 1966. The corresponding problemon a triangular lattice has not, to my knowledge, yet been investigated.Although this puzzle also appears in Dudeney's earlier book, Amusementsin Mathematics (1917), under the title "Mrs. Perkins's Quilt" (Problem 173,Answers 325'I S

6

2 11S'i

10

9

1

3on p. 47), I have allowed it to remain in this volume because of currentinterest in the problem. Sam Loyd gave the same problem in the first issue ofOur Puzzle Magazine, which he edited in 1907, the pages of which later became the pages of his posthumously published Cyclopedia of P1fzzles (1914).This is the earliest appearance of the problem I have been able to trace. IfLoyd took it from Dudeney, then Dudeney must have published it in a magazine or newspaper before 1907.-M. G.]

344. STARS AND CROSSESThe illustration shows how the square may be cut into four pieces, each ofthe same size and shape, so that each part shaJI contain a star and a cross.

326 Answers

345. GREEK CROSS PUZZLEPlace the four pieces together in the manner shown, and the symmetricalGreek cross will be found in the center.

346. SQUARE AND CROSSIfwe cut the smaller Greek cross in the manner shown in Figure I, the fourpieces A, B, C, and D will fit together and form a perfect square, as shown inFigure 2.

2.

1 AA BBD

347. THREE GREEK CROSSES FROM ONECut off the upper and lower arms of your large cross and place them in thepositions A and B, so as to form the rectangle in Figure I. Now cut the largerAnswers 327Apiece, as shown, into three pieces so that the five will form the rectangle inFigure II. This figure may be said to be built up of fifteen equal squares, fiveof which will be required for each new cross. Cutting is then not difficult, and

2,5,8,9 clearly form one cross; 13,6, 10, 7, and 11 will form the second cross,as in Figure III; and 1, 3,4, 12 will form the third cross, as in Figure IV. Thesmaller arms are one-third of the area of the larger arms. It is shown on page

232 of The Canterbury Puzzles how to find the side of the smaller squares.The rest is now easy.[Lindgren, in Geometric Dissections, pp. 55-56, shows how the number ofpieces can be reduced to twelve.-M. G.]

348. MAKING A SQUAREThe diagram will make it clear howthe figure should be cut into fourpieces of the same size and shapethat will fit together and form a perfect square.

328 Answers

349. TABLE TOP AND STOOLSFigure A shows the circle divided into four equal pieces, forming the GreatMonad, and in Figure B we have two pieces reassembled to form one of thetwo stools, the other stool being similarly constructed from 3 and 4. Unfortunately the hand holes are across instead of lengthways, but no conditionis broken.

350. TRIANGLE AND SQUARECut one triangle in half, and place the pieces together as in Figure 1. Nowcut in the direction of the dotted lines, making ab and cd each equal to theside of the required square. Then fit together the six pieces as in Figure 2,

1 2.Answers 329sliding the pieces F and C upwards to the left, and bringing down the littlepiece D from one corner to the other.[Lindgren has found a way to accomplish this result in as few as five pieces.See his Geometric Dissections, p. 9.-M. G.]

351. CHANGING THE SUITThe diagram shows how the spade may be cut into three parts that will fittogether and form a heart.

1

352. PROBLEM OF THE EXTRA CELLThe fallacy lies in the fact that the oblique edges of the pieces do not coincide in direction. If you carefully layout the pieces so that the outer edges"""'" ~ -....;;:~~;::::- .........~~ ~;....

330 Answersform a true rectangle, then there is a long diamond-shaped space in themiddle uncovered, as in the diagram. This space is exactly equal in area toone of the little square cells. Therefore we must deduct one from 65 to get 64as the actual area covered. The size of the diamond-shaped piece has beenexaggerated to make it quite clear to the eye of the reader.[For a discussion of many new and closely related paradoxes of this type,see the two chapters on "Geometrical Vanishes" in my Mathematics, Magic,and Mystery (Dover, 1956, pp. 114-155).-M. G.]

353. PROBLEM OF THE MISSING CELLThe diagram shows how the four pieces may be put together in a differentway, so that, on first sight, it may appear that we have lost a cell, there now....... ...... ...... ...... ~ ..... r--... ...... ....being only sixty-three of these. The explanation, as in the case of the preceding fallacy, lies in the fact that the lines formed by the slanting cuts do notcoincide in direction. In the case of the fallacy shown in the puzzle here, ifthe pieces are so replaced that the outside edges form a true rectangle, thereAnswers 331will be a long diamond-shaped space not covered, exactly equal in area to thesupposed extra cell. In this case the pieces, if truly laid, will overlap, and thearea of the overlapping is exactly equal to the supposed missing cell. This isthe simple explanation in a nutshell.

354. A HORSESHOE PUZZLEFirst make cut AB. Then so placethe three pieces together that with oneclip of the scissors you can make thecut CD together with EF and GH.A-+-+----+-+-- B

355. SQUARE TABLE TOPThe eight pieces of veneer may be fitted together, as in the illustration, toform a perfect square, and the arrangement is symmetrical and pleasing.

3

8 5

6

1.

332 Answers

356. TWO SQUARES IN ONEPlace the two squares together, sothat AB and CD are straight lines.Then find the center of the largersquare, and draw through it the lineEF parallel to AD. If you now makeGH (also through the center) perpendicular to EF, you can cut out thefour pieces and form the lower square,as shown.[This dissection was discoveredabout 1830 by Henry Perigal, a Britishstockbroker and amateur mathematician, and first published by him in

1873. It is one of the best of manyways to demonstrate the Pythagoreantheorem by cutting. See the chapteron "Paper Cutting" in my New Mathematical Diversions from Scientific~~-------+---,BAmerican (Simon & Schuster, 1966).-M.G.]

357. CUTTING THE VENEERThe illustration will show clearly how the veneer may be cut. Squares Aand B are cut out entire, as in Figure I, and the four pieces C, D, E, F will fittogether, as in Figure 2, to form a third square.

1C: A! :E. ~-!---1 .... : ......... - ... . 2_E.... : .... : ....... . - . :B: ...... : ......... ; .... : ......... ~ C F . . . .... : ....FDAnswers 333[Victor Meally found many ways to solve this problem with as few as fivepieces. Can the reader discover a five-piece pattern on which the total lengthof the cuts is as low as 16 units?-M. G.]

358. IMPROVISED CHESSBOARDCut out the piece A, and inset it again after having given it a quarter tumin a clockwise direction, and the chessboard will be formed.

359. THE PATCHWORK CUSHIONThe illustration shows how the twenty pieces will form a perfect square.

334 Answers

360. THE DAMAGED RUGIf we cut as in Figure I, the two pieces will fit together, as in Figure 2, andform a square. The steps are 2 ft. wide and 1 ft. in height.

361. FOLDING A HEXAGONFold through the midpoints of theopposite sides and get the lines AOBand COD. Also fold EH and FG, bisecting AO and OB. Turn over AK sothat K lies on the line EH, at thepoint E, and then fold AE and EOG.Similarly find H and fold AH andHOF. Now fold BF, BG, EF andHG, and EFBGHAE is the regularhexagon required.

362. FOLDING A PENTAGONFold AB on itself and find the mid-point E. Fold through EC. Lay EB onEC and fold so as to get EF and FG. Make CH equal to CG. Find K,Answers 335the mid-point on BH, and make CL equal to BK. BC is said to be divided inmedial section, and we have found KL, the side of the pentagon. Now (seeA 1Esecond diagram) lay KM and LN equal to KL, so that M and N may lie onBA and CD respectively. Fold PQ and lay MO and NO equal to KM andLN. Then KMONL is the pentagon required. For this solution I am indebtedto a little book, Geometrical Exercises in Paper Folding, by T. Sundara Row(Madras, 1893). *

363. FOLDING AN OCTAGONBy folding the edge CD over ABwe can crease the middle points Eand G. In a similar way we can findthe points F and H, and then creasethe square EHGF. Now fold CH onEH and EC on EH, and the pointwhere the creases cross will be I. Proceed in the same way at the otherthree corners, and the regular octagon, as shown, will be marked out bythe creases and may be at once cutout.l* Currently available in a Dover paperback repnnt -M. G.)

336 Answers

364. SQUARE AND TRIANGLE•••••... HFold the square in half and makethe crease FE. Fold the side AB sothat the point B lies on FE, and youwill get the points G and H fromwhich you can fold HGJ. While B ison G, fold AB back on AH, and youwill have the line AK. You can nowfold the triangle AJK, which is thelargest possible equilateral triangleobtainable. ·~~----~----~R

365. STRIP TO PENTAGONBy folding A over, find C, so that BC equals AB. Then fold as in Figure I,across the point A, and this will give you the point D. Now fold as in Figure 2,making the edge of the ribbon lie along AB, and you will have the point E.

8 CContinue the fold as in Figure 3, and so on, until all the ribbon lies on thepentagon. This, as we have said, is simple, but it is interesting and instructive.Answers 337

366. A CREASE PROBLEMBisect AB in C and draw the line CO, parallel to BR. Then bisect AC in Dand draw the semicircle DB, cutting the line CO in E. Now the line DEFgives the direction of the shortest possible crease under the conditions.G. HFA

367. FOLDING POSTAGE STAMPSNumber the stamps as in the diagram shown previously-that is, I 2 3 4 inthe first row and 5 6 7 8 in the second row. To get the order I 5 6 4 8 7 3 2(with No. I face upwards, only visible), start this way, with all faces downwards:

567 8

123 4Fold 7 over 6. Lay 4 flat on 8 and tuck them both in between 7 and 6 sothat these four are in the order 7 846. Now bring 5 and 1 under 6, andit is done.The order I 3 7 5 6 8 4 2 is more difficult and might well have been overlooked, if one had not been convinced that, according to law, it must be possible. First fold so that 5 6 7 8 only are visible with their faces uppermost.Then fold 5 on 6. Now, between I and 5 you have to tuck in 7 and 8, so that

7 lies on the top of 5, and 8 bends round under 6. Then the order willbe as required.

338 Answers

368. COUNTER SOLITAIREPlay in the following manner and all the counters except one will beremoved in seven moves, and the final leap will be made by number 1, as required: 2-10,4-12,6-5,3-6,7-15 (8-16,8-7,8-14, 8-3), (1-9, 1-2, 1-11,

1-8, 1-13, 1-4).

369. A NEW LEAP-FROG PUZZLEPlay 9 over 13, 14,6,4,3,1,2,7,15, 17, 16,11. Play 12 over 8. Play 10 over 5and 12. Play 9 over 10.

370. TRANSFERRING THE COUNTERSMake a pile of five counters (1 to 5) on B in 9 moves. Make a pile of four(6 to 9) on C in 7 moves. Make a pile of three (10 to 12) on D in 5 moves.Make a pile of two (13 and 14) on E in 3 moves. Place one (15) on F in 1move. Replace 13 and 14 on F in 3, 10 to 12 on F in 5, 6 to 9 in 7, and I to 5in 9 moves. Forty-nine moves in all.

371. MAGIC FIFTEEN PUZZLEMove the counters in the following order: 12, 8,4, 3, 2, 6, 10, 9, 13, 15, 14,

12, 8, 4, 7, 10, 9, 14, 12, 8, 4, 7, 10, 9,6, 2, 3, 10, 9, 6, 5, 1, 2, 3, 6, 5, 3, 2, 1,

13, 14, 3, 2, 1, 13, 14, 3, 12, 15, 3-fifty moves in all.[If the 14 and 15 counters are in correct serial order at the start, a magicsquare can be achieved in 37 moves: IS, 14, 10, 6, 7, 3, 2, 7, 6, 11, 3,2, 7, 6,

11, 10, 14,3,2, 11, 10,9,5, 1,6, 10,9,5, 1,6, 10,9, 5,2, 12, 15, 3.-M. G.]

372. TRANSFERRING THE COUNTERSThe two additional counters should be placed one on the fourth square inthe second row from the top and the other in the second square in the fourthrow. The puzzle is then quite possible, and so easy that it is quite unnecessaryto give all the moves.Answers 339

373. ODDS AND EVENSThe fewest possible moves are 24. Playas follows. (It is only necessary togive, by the letters, the circles from which and to which the counters aremoved. Only a single counter can be moved at a time.) E to A, E to B, E toC, E to D, B to D, E to B, C to B, A to B, E to C, E to A, B to A, C to E, B~~A~~B~~C~RC~~B~~E~~E~RC~RD~aD to B, E to B-24 moves.

374. RAILWAY SHUNTINGMake a rough sketch like our diagram and use five counters marked X, L,R, A, and B. The engines are Land R, and the two cars on the right A andB. The three cars on the left are never separated, so we call them X. The sidetrack is marked S. Now, playas follows: R to left, R to S, XL to right, R toleft, XLA to left, L takes A to S, L to left, XL to right, R to A, RA to left,®©

5XLB to left, L takes B to S, L to left, LX right away, RA to B, RABright away. Fourteen moves, because the first and third moves (R to left andXL to right) do not involve a change of direction. It cannot be done in fewermoves.

375. ADJUSTING THE COUNTERSMake the exchanges of pairs as follows: (1-7, 7-20, 20-16, 16-11, 11-2,

2-24), (3-10, 10-23,23-14, 14-18, 18-5), (14-19, 19-9,9-22), (6-12, 12-15,

15-13,13-25), (17-21). The counters are now all correctly arranged in 19 exchanges. The numbers within a pair of brackets represent a complete cycle,

340 Answersall being put in their proper places. Write out the numbers in their originalorder, and beneath them in their required order, thus:

7 24 10 19 3 12 20 8 22, etc.I 2 3 4 5 6 7 8 9, etc.The construction of the cycles is obvious, for I in the bottom row is exchanged with the 7 above it, then this 7 with the 20 above it, and so on untilthe cycle completes itself, when we come to 24 under 1.

376. NINE MEN IN A TRENCHLet the men move in the following order: 2-1,3-2,4-3,5-11,6-4,7-5,8-6,

9-7,1-13,9-10,8-9,1-12,7-13,6-8,5-7,1-11,4-12,3-6, 2-5, 1-1,2-2,3-3,

4-4,5-5,6-6, 7-7, 8-8, 9-9, and the sergeant is in his place in 28 moves.The first number in a move is that of a man, and the second number thatof his new position, the places being numbered I to 10 in the row, and the recesses II to 13 above.

377. BLACK AND WHITEIn the first case move the pairs in the following order: 67 before the I, then

34, 7 I, and 4 8 to the vacant spaces, leaving the order 64 827 I 5 3.In the second case move 3 4 and replace them as 4 3 before the I. Then remove and reverse 6 7 (as 7 6), 6 5 (as 56),3 I (as I 3), and 6 8 (as 8 6), leavingthe order 4 8 6 2 7 I 3 5-five moves in this case.

378. THE ANGELICA PUZZLEThough we start with the A's in correct positions, the puzzle can onlybe solved by making them change places. Represent the A in the bottom rowwith a capital letter, and the A in the top comer with a small letter. Then hereis a solution in 36 moves: A N LEG AN G C I A N G C I AN GeL E aANG ILCIL aECaL I (36 moves).[Dudeney's solution is not minimal. Can the reader find a way to solve itin 30 moves?-M. G.]Answers 341

379. THE FLANDERS WHEELMove the counters in the following order: AND A F L N 0 A F 0 N LDRS 0 L N A FRS E R S L N A L-30 moves in all.[The solution can be reduced to 28 moves, the lowest possible. Readers mayhave noticed that the problem is isomorphic with a square, eight-countersliding puzzle like the preceding one. For a general discussion of eight-countersliding puzzles of this type, see my Scientific American columns for March andJune 1965.-M. G.]

380. CATCHING THE PRISONERSIt is impossible for W I (warder) to catch P 2 (prisoner), or for W 2 tocatch P I. In the example we gave it was therefore hopeless, for each warder

1 2 3 4 S 6 7 B

9 .10 11 J2 n . 1+ 15 16@1 Ja ~9 20 '2.1 @~ 2;i (fJ ~5 :u -q 'l8 '2, lc 31 .;J~

33 31- 35 36 ~ 16 39 40would not be chasing "his prisoner," but the other fellow's prisoner. It is a caseof what we call in chess "gaining the opposition." Between W I and P 2 thereis only one square (an odd number), but between W I and P 1 (as also between W 2 and P 2) there are four squares (an even number). In the second

342 Answerscase the warders have the opposition, and can win. We will give a specimengame. The warders' moves are above the line, the prisoners' below:

19-2022-14 20-21 14-13 21-2213-12 22-23 12-20 23-3120-19

17-1824-23' 18-26 23-31 '26-27 31-32' 27-26 32-40' 40-3226-34'

31-32 Capture, 19-27 27-26 26-25 Capture.

34-33 33-25'There is no possible escape for the prisoners if each warder pursues hisproper man.

381. GRASSHOPPERS' QUADRILLEIf we regard only the central column containing three white and three blackpieces, these can be made to change places in fifteen moves. Number the sevensquares downwards I to 7. Now play 3 to 4,5 to 3, 6 to 5, 4 to 6, 2 to 4, I to

2,3 to 1,5 to 3, 7 to 5, 6 to 7, 4 to 6, 2 to 4,3 to 2,5 to 3, 4 to 5. Six of thesemoves are simple moves, and nine are leaps.There are seven horizontal rows of three white and three black pieces, if weexclude that central column. Each of these rows may be similarly interchangedin fifteen moves, and because there is some opportunity of doing this inevery case while we are manipulating the column-that is to say, there isalways, at some time or other, a vacant space in the center of every row-itshould be obvious that all the pieces may be interchanged in 8 X 15 = 120moves.

382. THE FOUR PENNIESFirst place the four pennies together as in the first diagram; then removenumber I to the new position shown in the second diagram; and finally,Answers 343carefully withdraw number 4 downwards and replace it above against numbers 2 and 3. Then they will be in the position shown in the third diagram,and the fifth penny may be added so that it will exactly touch all four.A glance at the last diagram will show how difficult it is to judge by the eyealone the correct distance from number I to number 3. One is almost certainto place them too near together.

383. THE SIX PENNIESFirst arrange the pennies as in Figure A. Then carefully shift 6 and getposition B. Next place 5 against 2 and 3 to get the position C. Number 3 cannow be placed in the position indicated by the dotted circle.

344 Answers

384. AN IRREGULAR MAGIC SQUAREIf for the 2 and 15 you substitute 7 bers can be arranged to form a magicand 10, repeated, the square can be square if they can be written in thisformed as shown. Any sixteen num- way, so that all the horizontal differ1 10l~ 10

8 3

12 11

9

5

16

4

14-

6'7'1ences are alike and all the verticaldifferences also alike. The differenceshere are 3 and 2:I 4 7 10

3 6 9 12

5 8 11 14

7 10 13 16

385. A MAGIC SQUARE DELUSIONIf you make nine squares preciselysimilar to this one and then place themtogether to form a larger square, thenyou can pick out a square of 25 cellsin any position and it will always be amagic square, so it is obvious you canarrange for any number you like to bein the central cell. It is, in fact, whatis called a Nasik square (so named bythe late Mr. Frost after the place in

9

3

12

2.1

20

11

25

19S

2-

18 5 :22

1 14<i 16

1 2~ 10IS 11 4

24 6 I?>India where he resided), and it is only can be treated in the manner deperfect squares of this character that scribed.

386. DIFFERENCE SQUARESThe three examples I give are, I believe, the only cases possible. The difference throughout is 5.

2 1 4 S 1 4 2 1 6?, 5 7 3 S 1 3 5 '1

6 9 S 6 9 2. -it 9 SAnswers 345

387. SWASTIKA MAGIC SQUAREIt will be seen that the square is perfectly "magic" and that all the primenumbers are placed within the swastika.

1( S j.~.~ .. ~~ ... ~. 14 j 12 25 8 16

11 ~l4~"'i'~20j 3

10 18 1 14jZ2 ~ .. ~ ..... _. ..... .

2, 6 19 i 2 )5

388. IS IT VERY EASY?All that is necessary is to push up the second figure in every cell and so formpowers of 2, as in the first square. Then the numbers become those in thesecond square, where all the eight rows give the same product-4,096. Ofcourse, every arithmetician knows that 20 equals 1.

21 2° 2s 128 1 32,

22 2~ 2 6 i- 16 64-

23 211 21 S 256 Z

389. MAGIC SQUARE TRICKThough the figures in each cell must be different in every case, it is not required that the numbers shall be different. In our smaller square the rows ofcells add to 15 in ten different directions, for in addition to the rows, columns,and two long diagonals, two of the short diagonals also add to 15. This is thegreatest number of directions possible. Now, all we have to do is to expresseach number with a different figure by repeating it with arithmetical signs. The

346 Answersyt 5 2t0 5 10

1£ 5 21,(SXg)-iS" ,.,.c, " ~ 1 + 1.+-1+1t-i -+--r !( '+'-t-(.,i-t-S,3-3 S (1'1. t)4-1t1t'IIi-X4-)+ Ij.t Ij.t~ qtqi'<1tqtq '2* 1.. +"J.. 1"'2. ..,. -1L q I+T~larger square shows how this may be done. All the conditions of the puzzle arethus complied with, and the maximum ten directions obtained.[The squares using 4,7 and 8 are needlessly complex. Victor Meally suggests:

4 + 4 __ 4_ = 7V2

4+4

7+7+7+7_10 7 -

8 - 8! 8 = 7lh..-M.O.]

390. A FOUR-FIGUREMAGIC SQUAREThe solution explains itself. Thecolumns, rows, and two diagonals alladd up alike to 6,726, and nine ofeach of the figures 1,2,3,4 have beenemployed.

2243

3141

1342Answers 347

131:1 3142

2242 1343

3143 2241

391. PROGRESSIVE SQUARESFill in the following numbers in the order given, beginning at the cell in thetop right-hand corner and proceeding downwards and "round the square": 13,

81,78,6,75,8, IS, 16,77,70, 19, 79, 21, 9, 23, 2, 69, 66, 67, 74, 7, 76, 4, 1,5,

80,59,73,61,3,63, 12. It is obvious that opposite numbers in the border mustsum to 82 in all cases, but a correct adjustment of them is not very easy. Ofcourse there are other solutions.

18 22 1

24- 3 r

5 9 1?>

6 15 19

12. 16 2.5

10

11

17

23

4

14

20

21

2

8

392. CONDITIONALMAGIC SQUAREThe example here shown is a solution, with the odd numbers and evennumbers placed in the positionsdesired.

393. THE FIVE-POINTED STARReferring to Figure I, on page 348, we will call A, B, C, D, E the "pentagon,"andF, G, H, J, K the "points." Write in the numbers 1,2,3,4,5 in the pentagon

348 Answersin the order shown in Figure II, where you go round in a clockwise direction,starting with I and jumping over a disc to the place for 2,jumping over anotherfor 3, and so on. Now to complete the star for the constant summation of 24,as required, use this simple rule. To find H subtract the sum of Band C fromhalf the constant plus E. That is, subtract 6 from 15. We thus get 9 as the required number for H. Now you are able to write in successively IO at F (tomake 24), 6 at J, 12 at G, and 8 at K. There is your solution.You can write any five numbers you like in the pentagon, in any order, andwith any constant summation that you wish, and you will always get, by therule shown, the only possible solution for that pentagon and constant. Butthat solution may require the use of repeated numbers and even negativenumbers. Suppose, for example, I make the pentagon I, 3, II, 7, 4, and theconstant 26, as in Figure III, then I shall find the 3 is repeated, and the repeated

4 is negative and must be deducted instead of added. You will also find thatif we had written our pentagon numbers in Figure II in any other orderwe should always get repeated numbers.Answers 349Let us confine our attention to solutions with ten different positive wholenumbers. Then 24 is the smallest possible constant. A solution for any higherconstant can be derived from it. Thus, if we want 26, add I at each of thepoints; if we want 28, add 2 at every point or I at every place in both pointsand pentagon. Odd constants are impossible unless we use fractions. Everysolution can be "turned inside out." Thus, Figure IV is simply a differentarrangement of Figure II. Also the four numbers in G, K, D, ] may always bechanged, if repetitions do not occur. For example, in Figure II substitute 13,

7, 6, 5 for 12, 8, 5, 6 respectively. Finally, in any solution the constant will betwo-fifths of the sum of all the ten numbers. So, if we are given a particularset of numbers we at once know the constimt, and for any constant we candetermine the sum of the numbers to be used.

394. THE SIX-POINTED STARI have insufficient space to explain fully the solution to this interestingproblem, but I will give the reader the main points.(I) In every solution the sum of the numbers in the triangle ABC (Figure I)

350 Answersmust equal the sum of the triangle DEF. This sum may be anything from 12to 27 inclusive, except 14 and 25, which are impossible. We need only obtainsolutions for 12, 13, 15, 16, 17, 18, and 19, because from these all the complementaries, 27, 26, 24, 23, 22, 21, and 20, may be derived by substituting forevery number in the star its difference from 13.(2) Every arrangement is composed of three independent diamonds,AGHF, DKBL, and EMCI, each of which must always sum to 26.(3) The sum of the numbers in opposite external triangles will always beequal. Thus AIK equals LMF.( 4) If the difference between 26 and the triangle sum ABC be added to anynumber at a point, say A, it will give the sum of the two numbers in the relative positions of Land M. Thus (in Figure II) 10 + 13 = II + 12, and

6+13=8+11.(5) There are six pairs summing to 13; they are 12 + I, II + 2, 10 + 3,

9 + 4, 8 + 5, 7 + 6, and one pair or two pairs may occur among the numbersat the points, but never three. The relative positions of these pairs determinethe type of solution. In the regular type, as in Figure II, A and F and also Gand H, as indicated by the dotted lines, always sum to 13, though I subdividethis class. Figures III and IV are examples of the two irregular types. Thereare 37 solutions in all (or 74, if we count the complementaries described in myfirst paragraph), of which 32 are regular and 5 irregular.Of the 37 solutions, 6 have their points summing to 26. They are as follows:

10 6 2 3 I 4 7 9 5 12 II 8

9 7 4 3 2 6 II 5 10 12 8

5 4 6 8 2 I 9 12 3 II 7 10

5 2 7 8 3 II 10 4 12 6 9

10 3 I 4 2 6 9 8 7 12 II 5

8 5 3 I 2 7 10 4 II 9 12 6The first is our Figure II, and the last but one our Figure III, so a referenceto those diagrams will show how to write the numbers in the star. The readershould write them all out in star form and remember that the 6 are increasedto 12 if you also write out their complementaries. The first four are of theregular type and the last two of the irregular. If the reader should be temptedto find all the 37 (or 74) solutions to the puzzle it will help him to know that,where the six points sum to 24, 26, 30, 32, 34, 36, 38, the respective numberof solutions is 3, 6, 2, 4,7,6,9, making 37 in all.Answers 351

395. THE SEVEN-POINTED STARPlace 5 at the top point, as indicatedin diagram. Then let the four numbersin the horizontal line (7, 11,9,3) besuch that the two outside numbersshall sum to 10 and the inner numbersto 20, and that the difference betweenthe two outer numbers shall be twicethe difference between the two innernumbers. Then their complementariesto 15 are placed in the relative positions shown by the dotted lines. Theremaining four numbers (13, 2, 14, I)are easily adjusted. From this fundamental arrangement we can get threeothers. (I) Change the 13 with the Iand the 14 with the 2. (2 and 3) Substitute for every number in the twoarrangements already found its difference from IS. Thus, 10 for 5, 8 for 7,

4 for II, and so on. Now, the readershould be able to construct a secondgroup of four solutions for himself,by following the rules.The general solution is too lengthyto be given here in full, but there are,in all, 56 different arrangements,counting complementaries. I dividethem into three classes. Class I includes all cases like the above example, where the pairs in the positionsof7-8, 13-2,3-12, 14-1 all sum to IS,and there are 20 such cases. Class IIincludes cases where the pairs in thepositions of 7-2, 8-13, 3-1,12-14 allsum to IS. There are, again, 20 suchcases. Class III includes cases wherethe pairs in the positions of 7-8, 13-2,

3-1,12-14 all sum to IS. There are 16such cases. Thus we get 56 in all.[Dudeney erred in his enumeration of solutions for both the sevenpointed and the six-pointed stars.When I published his results in acolumn on magic stars (ScientificAmerican, December 1965), tworeaders-E. J. Ulrich, Enid, Oklahoma, and A. Domergue, Parisindependently confirmed that thereare 80 patterns for the six-pointedstar, or 6 more than listed by Dudeney.That the seven-pointed star has 72distinct solutions (as against Dudeney's figure of 56) was first reportedto me by Mrs. Peter W. Montgomery,North Saint Paul, Minn. This wasconfirmed by the independent work ofUlrich and Domergue. In 1966 AlanMoldon, Toronto, using a computerat the University of Waterloo, alsoobtained 72 solutions for the sevenpointed star, so there seems littledoubt that this is correct.-M. G.]

352 Answers

396. TWO EIGHT-POINTED STARSThe illustration is the requiredsolution. Every line of four numbersadds up 34. If you now find any solution to one of the stars, you can immediately transfer it to the other bynoting the relative positions in thecase given.I have not succeeded in enumerating the stars of this order. The taskis, I think, a particularly difficult one.Perhaps readers may like to attemptthe solution.[Domergue (see preceding note),in his 1963 analysis of the six-, seven-,and eight-pointed stars, found 112different solutions for the eightpointed star. He estimated that thenine-point star has more than 2,000distinct patterns.-M. G.]

397. FORT GARRISONSThe illustration shows one way ofarranging the men so that the numbers in every straight line of four fortsadd up to one hundred.

398. THE CARD PENTAGONDeal the cards 1, 2, 3, 4, 5 in the manner indicated by the dotted lines(that is, drop one at every alternate angle in a clockwise direction roundthe pentagon), and then deal the 6, 7,

8, 9, 10 in the opposite direction, asshown, taking care to start with the 6on the correct side of the 5. The pipson every side add to 14. If you deal the

6, 7, 8, 9, 10 in the first manner and the

1,2,3,4,5 in the second manner, youwill get another solution, adding up to

19. Now work with the two sets ofnumbers 1,3,5,7,9 and 2,4,6,8,10in the same way and you will get twomore solutions, adding, respectively,to 16 and 17.There are six different solutions inall. The last two are peculiar. Write in,in the same order, 1,4, 7, 10, 13 andAnswers 353

6,9, 12, IS, 18; also write in 8, II, 14,

17,20 and 3, 6, 9,12, IS. Then deduct

10 from every number greater than 10.

399. A HEPTAGON PUZZLEThe diagram shows the solution.Starting at the highest point, write inthe numbers I to 7 in a clockwisedirection at alternate points. Then,starting just above the 7, write 8 to 14successively in the opposite direction,taking every vacant circle in tum. Ifinstead you write in 1,3,5,7,9, II, 13,and then 2, 4, 6,8, 10, 12, 14, you willget a solution with the sides addingto 22 instead of 19. If you substitutefor every number in these solutions itsdifference from 15 you will get the spectively to 26 and 23 (the differencecomplementary solutions, adding re- of 19 and 22 from 45).

354 Answers

400. ROSES, SHAMROCKS,AND THISTLESIt is clear that the sum obtained inthe different ways must be 26. Hereis one of many arrangements thatsolve the puzzle.

401. THE MAGIC HEXAGONOur illustration shows the only correct answer.

402. THE WHEEL PUZZLEAll you have to do is to place 10 in the center and write in their properorder round the circle 1,2,3,4,5,6,7,8,9,19,18, 17, 16, IS, 14, 13, 12, II.Answers 355

403. AT THE BROOKA B

15 16 15 160 16* 15 5* *15 0 0 11

15 1* 0 5 0 15 15 110 I 5 0 15 15 *10 160 5 16 *14 16 10 0

16 15 6* 14 0 0 10

15 2* 0 6 0 14 15 100 2 6 0 15 14 *9 16

2 0 6 16 *13 16 9 0

2 16 15 7* 13 0 0 9

15 3* 0 7 0 13 15 90 3 7 0 15 13 *8 16

3 0 7 16 *12 16

3 16 15 8* 12 0

15 4* 0 8 0 120 4 8 0 15 12

4 0 8 16 *11 16

4 16 11 0Every line shows a transaction. Thus, in column A, we first fill the 16measure; then fill the 15 from the 16, leaving I, if we want it; then empty the

15; then transfer the I from 16 to 15; and so on. The asterisks show how tomeasure successively I, 2, 3, 4, etc. Or we can start, as in column B, by firstfilling the 15 and so measure in turn, 14, 13, 12, II, etc. If we continue A weget B read upwards, or vice versa. It will thus be seen that to measure from Iup to 7 inclusive in the fewest transactions we must use the method A, but toget from 8 to 14 we must use method B. To measure 8 in the A direction willtake 30 transactions, but in the B manner only 28, which is the correct answer.It is a surprising fact that with any two measures that are prime to each other(that have no common divisor, like 15 and 16) we can measure any wholenumber from I up to the largest measure. With measures 4 and 6 (each divisible by 2) we can only measure 2, 4, and 6. With 3 and 9 we could only measure

3, 6, and 9. In our tables the quantities measured come in regular numerical

356 Answersures 9 and 16, under A we should get the order 7, 14,5, 12,3, etc., a cyclicaldifference of 7 (since 16 - 9 = 7). After adding 7 to 14 we must deduct 16to get 5, and after adding 7 to 12 we must deduct 16 to get 3, and so on.[For a clever method of solving liquid pouring puzzles of this type bygraphing the problem on isometric paper see my Scientific American column,September 1963, and the fuller discussion of the method in T. H. O'Beirne,Puzzles and Paradoxes (Oxford University Press, 1965), Chapter 4.-M. G.]

404. A PROHIBITION POSERFirst fill and empty the 7-quart measure 14 times and you will have thrownaway 98 and leave 22 quarts in the barrel in 28 transactions. (Filling andemptying are 2 transactions.) Then, fill 7-qt.; fill 5-qt. from 7-qt., leaving 2 in

7-qt.; empty 5-qt.; transfer 2 from 7-qt. to 5-qt.; fill 7-qt.; fill up 5-qt. from

7-qt., leaving 4 in 7-qt.; empty 5-qt.; transfer 4 to 5-qt.; fill 7-qt.; fill up 5-qt.from 7-qt., leaving 6 in 7-qt.; empty 5-qt.; fill 5-qt. from 7-qt., leaving 1in 7-qt.; empty 5-qt., leaving 1 in 7-qt.; draw off remaining l-qt. from barrelinto 5-qt., and the thing is done in 14 more transactions, making, with the 28above, 42 transactions. Or you can start by wasting 104 and leaving 16 inbarrel. These 16 can be dealt with in 10 transactions, and the 104 require 32in the emptying (12 times 7 and 4 times 5 is the quickest way).

405. PROHIBITION AGAINFill 7-qt.; fill 5-qt.; empty 108 quarts from barrel; empty 5-qt. into barrel;fill 5-qt. from 7-qt.; empty 5-qt. into barrel; pour 2 quarts from 7-qt. into

5-qt.; fill 7-qt. from barrel; fill up 5-qt. from 7-qt.; empty 5-qt. into barrel;pour 4 quarts from 7-qt. into 5-qt.; fill 7-qt. from barrel; fill up 5-qt. from

7-qt.; throwaway contents of 5-qt.; fill 5-qt. from barrel; throwaway 5quarts from 5-qt.; empty 1 quart from barrel into 5-qt. The feat is thus performed in 17 transactions-the fewest possible.

406. THE KEG OF WINEThe capacity of the jug must have been a little less than 3 gallons. To bemore exact, it was 2.93 gallons.Answers 357

407. WATER MEASUREMENTTwo pints may be measured in fourteen transactions as below, where thevessels above the line are empty and every other row shows a transaction.

7 II

7 0o 7

7 7

3 II

3 0o 3

7 3o 10

7 10

6 II

6 0o 6

7 6

2 IIThe contents of the vessels, after each transaction, will make everythingclear.

408. MIXING THE WINEThe mixture will contain )24 wine, and 1)24 water.

409. THE STOLEN BALSAMOne of several solutions is as follows:oz. oz. oz. oz.The vessels can hold .... 24 13 II 5Their contents at the start 24 0 0 0Make their contents ..... 0 8 II 5Then .................. 16 8 0 0Then .................. 16 0 8 0Then .................. 3 13 8 0Then .................. 3 8 8 5Finally ................ 8 8 8 0

358 Answers[Victor Meally has found a better solution that requires only five transactions:

8 0 11 5

8 11 0 5

8 I3 3 0

8 8 3 5

8 8 8 0-M.G.]

410. DELIVERING THE MILKThe simplest way of showing the solution is as follows: At the top we havefour vessels, in the second line their contents at the start, and in every subsequent line the contents after a transaction:

80-pt. 80-pt. 5-pt. 4-pt.can can jug Jug

80 80 0 0

75 80 5 0

75 80 I 4

79 80 1 0

79 80 0

74 80 5

74 80 2 4

78 80 2 0

78 76 2 4

80 76 2 2Thus we first fill the 5-pint jug from one of the cans, then fill the 4-pint jugfrom the 5-pint, then empty the 4-pint back into the can, and so on. It canbe followed quite easily this way. Note the ingenuity of the last two transactions-filling the 4-pint jug from the second can and then filling up the firstcan to the brim.

411. THE WAY TO TIPPERARYThe thick line in the illustration shows a route from London to TipperaryAnswers 359in eighteen moves. It is absolutely necessary to include the stage marked"Irish Sea" in order to perform the journey in an even number of stages.

412. MARKING A TENNIS COURTThe ten points lettered in the illustration are all "odd nodes," that is pointsfrom which you can go in an odd number of directions-three. Therefore weknow that five lines (one-half of 10) will be required to draw the figure. TheA'r---.8",:r---------,.C-.-.-... --· - D.!:E~:---------------~F· • •G ~ •••••••• • ~~ _____________ ~:.:...7 ____ __4IKdotted lines will be the four shortest possible between nodes. Note that youcannot here use a node twice or it would be an improvement to make

360 AnswersEH and CF dotted lines instead of CD and GH. Having fixed our fourshortest lines, the remainder may all be drawn in one continuous line fromA to K, as shown. When you get to D you must run up to C and back to D,from G go to H and back, and so on. Or you can wait until you get to C andgo to D and back, etc. The dotted lines will thus be gone over twice and themethod shown gives us the minimum distance that must be thus repeated.

413. WATER, GAS, AND ELECTRICITYThis puzzle can only be solved by a trick. If one householder will allow onepipe for a neighbor to pass through his house there is no difficulty, and theconditions did not prohibit this not very unreasonable arrangement. Look atFigure I, and you will see that the water pipe for supplying house C passesthrough house A, but no pipe anywhere crosses another pipe. I am, however,often asked to prove that there is no solution without any trick, and I will nowgive such a proof for the first time in a book.Assume that only two houses, A and B, are to be supplied. The relativepositions of the various buildings clearly make no difference whatever. I givetwo positions for the two houses in Figures 2 and 3. Wherever you buildthose houses the effect will be the same-one of the supply stations will becut off. In the examples it will be seen that if you build a third house on the

1Answers 361outside (say in the position indicated by one of the black crosses) the gas cannever reach you without crossing a pipe. Whereas if you put the house inside one of the enclosures (as indicated by the stars), then you must be cutoff either from the water or the electricity-one or the other. But the housemust be either inside or outside. Therefore a position is impossible in whichit can be supplied from all three stations without one pipe crossing another.I hope that is thoroughly convincing. Build your two houses wherever youlike and you will find that those conditions that I have described will alwaysobtain.[In modem graph theory this is called the "utilities problem," and thenonplanar graph that supplies the desired connections is known as a Thomsengraph. The graph cannot be drawn on the plane without one line crossing another; can it be drawn, without such a crossing, on the surface of adoughnut?-M. G. J

414. CROSSING THE LINESLet us suppose that we cross the lines by bridges, represented in Figure Iby the little parallels. Now, in Figure 2, I transform the diagram, reducing the

5./\. A~ i-BJE

362 Answersspaces A, B, C, D, E to mere points, and representing the bridges thatconnect these spaces by lines or roads. This transformation does not affectthe conditions, for there are 16 bridges or roads in one case, and 16 roads orlines in the other, and they connect with A, B, C, D, E in precisely the sameway. It will be seen that 9 bridges or roads connect with the outside. Obviously we are free to join these up in pairs in any way we choose, provided theroads do not cross one another. The simplest way is shown in Figure 3, whereon coming out from A, B, C, or E, we immediately return to the same pointby the adjacent bridge, leaving one point, X, necessarily in the open. InFigure 2 there are 4 odd nodes, A, B, D, and X (if we decide on the exits andentrances, as in Figure 3), so, as I have already explained, we require 2strokes (half of 4) to go over all the roads, proving a perfect solution to beimpossible.Now, let us cancel the line AB. Then A and B become even nodes, but wemust begin and end at the odd nodes, D and X. Follow the line in Figure 3,and you will see that this can be done, omitting the line from A to B. Thisroute the reader will easily transform into Figure 4 if he says to himself, "Gofrom X to D, from D to E, from E to the outside and return into E," and soon. The route can be varied by linking up those outside bridges differently,by making X an outside bridge to A or B, instead of D, and by takingthe cancelled line either at AB, AD, BD, XA, XB, or XD. In Figure 5 I makeX lead to B. We still omit AB, but we must start and end at D and X. Transformed in Figure 6, this will be seen to be the precise example that I gave instating the problem. The reader can now write out as many routes as he likesfor himself, but he will always find it necessary to omit one line or crossing.It is thus seen how easily sometimes a little cunning, like that of the transformation shown, will settle a perplexing question of this kind.

415. THE NINE BRIDGESTransform the map as follows. Reduce the four islands, A, B, C, and D, tomere points and extend the bridges into lines, as in Figure I, and the conditions are unchanged. If you link A and B for outside communication, andalso C and D, the conditions are as in Figure 2; if you link A and D,Answers 363and also Band C, you get Figure 3; if you link A and C, and also Band D,you get Figure 4. In each case Band D are "odd nodes" (points from whichyou can proceed in an odd number of ways, three), so in every route youAmust start and finish at B or D, to go over every line once, and once only.Therefore, Tompkins must live at B or D: we will say B, and place Johnsonat D. There are 44 routes by scheme 2, 44 by scheme 3, and 44 by scheme 4,making 132 in all, not counting reverse routes as different. Taking Figure 2,and calling the outside curved lines 0, if you start BOAB, BOAC, BAOB, orBAC, there are 6 ways of continuing in each case. If you start BOAD, BAD,BCOD, BCA, or BCD, there are always 4 ways of continuing. In the case ofFigure 3, BOCA, BOCB, BCA, or BCOB give 6 ways. BOCD, BAOD, BAC,BAD, or BCD give 4 ways each. Similarly, in the case of Figure 4.

364 Answers

416. SINKING THE FISHING-BOATSThe diagram shows how the warship sinks all the forty-nine boats in twelvestraight courses, ending at the point from which she sets out. Follow everyline to its end before changing your direction.[M. S. Klamkin, in American Mathematical Monthly, February 1955, p. 124,proved that a continuous path of as few as 2n - 2 straight line segmentscould be drawn through all the dots in a square array of n dots on the side,provided n is greater than 2. The case of n = 3 is a well-known puzzle whichmost people fail to solve in four moves because they do not think of extending line segments beyond the square's borders. The 5 X 5 is the smallestsquare that can be solved in 2n - 2 segments without going outside theborders.S. W. Golomb has shown that a path of 2n - 2 segments is also sufficientfor a closed path (one that ends, as in the above problem, at the startingpoint) on all squares with sides greater than 3. The 7 X 7 is the smallest oddsided square with a closed path of 2n - 2 segments that lies entirely withinits border. (The smallest even-sided square on which such a path can bedrawn is the 6 X 6.) The solution given here by Dudeney also appearsAnswers 365in Sam Loyd's Cyclopedia of Puzzles as the solution to his Presidential Puzzle on page 293. Loyd says he first gave the puzzle in 1908, and describes thesolution as "a wonderfully difficult trick." Whether Loyd copied the puzzlefrom Dudeney, or vice versa, has not yet been determined.Note that this solution for the 7 X 7 array also provides a solution to theproblem of making a closed "queen's tour" on a 7 X 7 chessboard in 2n - 2moves. Closed queen's tours in 2n - 2 moves are also possible on all boardswith sides greater than 7. The closed 14-move queen's tour on the standard

8 X 8 board was first solved by Sam Loyd, who considered it one of his bestpuzzles. (See Alain C. White, Sam Loyd and His Chess Problems, 1913,reprinted by Dover in 1962, pp. 42-43.)A proof that 2n - 2 segments are also necessary for any square array wasgiven by John L. Selfridge in American Mathematical Monthly, Vol. 62, 1955,p. 443.-M. G.]

417. GOING TO CHURCHStarting from the house, H, there is only one way of getting to each of thepoints in a northerly direction, and also going direct east, so I write in the

366 Answersfigure I, as shown. Now take the second column, and you will find that thereare three ways of going to the second point from the bottom, five ways to thenext above, seven to the next, and so on, continually adding two. The sameapplies to the second row from the bottom. Write in these numbers. Then thecentral point of all can be reached in thirteen ways, because we can enter iteither from the point below that can be reached in five ways, from the pointto the left that can also be reached in five ways, or from the diagonal point belowthat can be reached in three ways, making together thirteen. So all we haveto do is to write in turn at every point the sum of those three numbers fromwhich it can be immediately reached. We thus find that the total numberof different routes from H to C is 321.

418. THE SUBMARINE NETThe illustration will show the best way of cutting the net. It will be seen thateight cuts are made from A to B, dividing the net into two parts.Answers 367

419. THE TWENTY-TWO BRIDGESIt will be found that every department has an even number (2,4, or 6) ofbridges leading from it, except C and L, which can each be approached bythree bridges-an odd number. Therefore to go over every bridge once, andonly once, it is necessary to begin and end at C and L, which are the two departments in which the houses stand. Thus, starting from C, we may take thefollowing route: C, G, F, C, B, A, D, H, E, I, H, J, K, L, M, G, I, F, B, E, F,I, L.

420. FOOTPRINTS IN THE SNOWThe illustration explains itself.,., -------- ire····· _-----_ .. -. . '~. .. ",- -"

421. A MONMOUTH TOMBSTONEThe number of different ways in which "HERE LIES JOHN RENIE" canbe read is 45,760; or, if diagonal readings are allowed, 91,520, because on

368 Answersreaching one of the corner I's we have the option of ending in the extremecorner or going backwards to an E diagonally. There is not space to give thesolution in detail. The only other information on the stone is "who diedMay 31, 1832, aged 32 years."

422. THE FLY'S TOURThe line here shows the fly's route under the con<1Illl ... __

423. INSPECTING THE ROADSThe shortest possible route is as follows: ABCHCDEIEFGBHDIHGIFAG.Thus he has gone 211 miles, and passed along the two short roads CH andEI twice.Answers 369

424. RAILWAY ROUTESThere are 2,501 ways of going from B to D, as follows:I-line journey I route 2 variations 2

2-line journey I route 9 variations 9

3-line journey 2 routes 12 variations 24

4-line journey 5 routes 18 variations 90

5-line journey 4 routes 72 variations 288

6-line journey 14 routes 36 variations 504

8-line journey 22 routes 72 variations 1,584

2,501We have only to consider the routes from B to D. The I-line route is directto D. The 2-line route is CD. The two 3-line routes are CBD and DCD. Thefive 4-line routes are DB CD, DCBD, CBCD, CDCD, and CDBD. Eachof these routes is subject to a number of variations in the actual lines used, andfor a journey of a given number of lines there is always the same number ofvariations, whatever the actual route. A 7-line journey is not possible.

425. A CAR TOURThe number of different routes is 264. It is quite a difficult puzzle, and consideration of space does not admit of my showing the best method of makingthe count.

426. MRS. SIMPER'S HOLIDAY TOURThere are sixty different routes by which Mrs. Simper, starting from H,might visit every one of the towns once, and once only, by the roads shown,and return to H, counting reversals of routes as different. But if the lady is toavoid going through the tunnels between Nand 0, and Sand R, it will befound that she is restricted to eight different routes.If the reader is sufficiently interested he may wish to discover these eightroutes for himself. If he does so, he will find that the route that complies withthe conditions, avoids the two tunnels, and delays her visit to D as long aspossible, is as follows: HISTLKBCMNUQRGFPODEAH. This is, therefore,undoubtedly her best route.

370 Answers

427. SIXTEEN STRAIGHT RUNSThe illustration shows how the traveller could have driven his car 76 milesin sixteen straight runs and left only three towns unvisited. This is not an easypuzzle, and the solution is only to be found after considerable trial.•~••[Victor Meally improved this answer by finding a 76-mile, l6-segment paththat leaves only one town unvisited. It is believed to be the best possible solution. The reader may enjoy searching for it.-M. G.]

428. PLANNING TOURSIn our illustration, in which the roads not used are omitted for the sake ofclearness, every man's route is shown. It will be seen that no two drivers evergo along the same road, and that no man ever crosses the track of another.Although no hard-and-fast rules can be laid down for the solution of puzzlesin this particular class, a little careful thought will generally overcome ourdifficulties. For example, we showed in the puzzle that if A goes direct to Ain a straight line, we shall hopelessly cut off C, D, and E. It soon becomesevident that A must go to the left of the upper D, and having done so, mustAnswers 371then get to the right of C. Also the route from D to D then becomes evident,as does that from B to B, and the rest is easy.

429. A MADAM PROBLEMEvery reading must begin with an M, and as there are only four M's therecan only be four starting-points. It will be found that there are 20 differentways of reading MADAM, always starting from the same M, therefore thecorrect answer is that there are 80 ways in all.

430. THE ENCIRCLED TRIANGLESThis puzzle may be solved in the astonishingly small number of fourteen

372 Answersstrokes, starting from A and ending at Z. Places in the figure are purposelynot joined up, in order to make the route perfectly clear.

431. THE SIAMESE SERPENTThe drawing cannot be executedunder the conditions in fewer thanthirteen lines. We have therefore tofind the longest of these thirteen lines.In the illustration we start at A andend at B, or the reverse. The dottedlines represent the lines omitted. Itrequires a little thought. Thus, theunbroken line from D to C is longerthan the dotted line, therefore we takethe former. Again, we can get in alittle more of the drawing by takingthe tongue rather than the mouth, butthe part of the tongue that ends in astraight line has to be omitted.

432. A BUNCH OF GRAPESThere are various routes possible,and our illustration shows one ofthem. But it is absolutely necessarythat you begin at A and end at B, orthe reverse. At any other point in thedrawing a departure can be made intwo or four ways (even numbers),but at A and B there are three waysof going (an odd number), thereforethe rule is that you must begin andend at A and B.Answers 373

433. A HOPSCOTCH PUZZLEThe hopscotch puzzle may be drawn in one continuous line without takingthe pencil off the paper, or going over the same line twice. But it is necessaryto begin at the point A and end at B, or begin at B and end at A. It cannototherwise be done.A

434. A WILY PUZZLEIn the diagram it will be seen that the prisoner's course is undoubtedly allright until we get to B. Ifwe had been the prisoner, when we got to that pointwe should have placed one foot at C, in the neighboring cell, and have said,"As one foot has been in cell C we have undoubtedly entered it, and yet whenwe withdraw that foot into B we do not enter B a second time, for the simplereason that we have never left it since we first went in!"~++~+:+~+l+~ :.+ ....... ;+:+:+!+:+:",::+,; +1 +! +: +:+!b ' •• c········: : ........ : :~/+++ .................................. +++~ :

374 Answers

435. A TREE PLANTING PUZZLEThe illustration shows the graceful manner of planting the trees so as to getnine rows with four trees in every row.

436. THE TWENTY PENNIESArrange the sixteen pennies in the form of a square 4 by 4. Then place onepenny on top of the first one in the first row; one on the third in the secondrow; one on the fourth in the third row; and one on the second in thefourth row.

437. TRANSPLANTING THE TREESThe illustration explains itself. Only six trees have been transplanted, andthey now form twenty rows with four trees in every row.Answers 375

438. A PEG PUZZLE~... . . . . . ',".· .. 3/. . · · .. '~;"'... . . . . " . • ~ \, • ''&f. • • • \ I"": ''Q.o/ ".• • .. ~, • :.t • •• .~;,". \. • • .••• -l:- • , . , "-.,,i:;:. /j,t ....... ~.-....... * .'::SFThe diagram shows how to place the pegs. The three removed from the holesbearing a cross are replaced in the top left-hand corner. The ten pegs nowform five rows with four pegs in every row. If you reflect the diagram in amirror you will get the only other solution.

439. FIVE LINES OF FOURThe illustration shows the solution to this puzzle. The ten counters form fiverows with four counters in every row ..... ~.... •..... :.- " .

376 Answers

440. DEPLOYING BATTLESHIPS••••The illustration shows that the ships form five lines with four in every line,and the white phantom ships indicate the positions from which four of themhave been removed.

441. CONSTELLATION PUZZLEHere is a symmetrical solution in which the 21 stars form II straight lines,with 5 stars in every line.'. ,.,.~. I . . ,. .'Answers 377

442. THE FOUR-COLOR MAP THEOREMWith two or more contiguous countries, two colors at least are obviouslynecessary (Figure I). If three countries are contiguous each with each, threecolors are necessary (Figure 2). With four countries, three will be required ifthe fourth (Y) is contiguous with each of two that are already contiguous(Figure 3). (For, as in Figure 4, G may be contiguous with two countries notcontiguous with one another, when only two colors are needed.) And fourcolors will be necessary if the fourth country is contiguous with each of threealready contiguous each with each (Figure 5).With five contiguous countries, three will be required if one country is contiguous with two already contiguous (Figure 6). And four will be necessaryif the fifth country is contiguous with each of three countries that are contiguous each with each (Figure 7). Yet five colors would be needed if a fifth

378 Answerscountry were contiguous with each of four that are contiguous each with each.If such a map is possible, the theorem breaks down.First, let us consider four countries contiguous each with each. We will usea simple transformation and suppose every two contiguous countries to beconnected by a bridge across the boundary line. The bridge may be as long aswe like, and the countries may be reduced to mere points, without affectingthe conditions. In Figures 8 and 9 I show four countries (points) connectedby bridges (lines), each with each. The relative positions of these pointsis quite immaterial, and it will be found in every possible case that onecountry (point) must be unapproachable from the outside.The proof of this is easy. If three points are connected each with each bystraight lines these points must either form a triangle or lie in a straight line.First suppose they form a triangle, YRG, as in Figure 16. Then a fourth contiguous country, B, must lie either within or without the triangle. If within, itis obviously enclosed. Place it outside and make it contiguous with Y and G,as shown: then B cannot be made contiguous with R without enclosing eitherY or G. Make B contiguous with Y and R: then B cannot be made contiguouswith G without enclosing either Y or R. Make B contiguous with Rand G:then B cannot be made contiguous with Y without enclosing either R or G.Take the second case, where RYG lie in a straight line, as in Figure 17. IfB lies within the figure it is enclosed. Place B outside and make it contiguouswith Rand G, as shown: then B cannot be made contiguous with Y withoutenclosing either R or G. Make B contiguous with Rand Y: then B cannot bemade contiguous with G without enclosing either R or Y. Make B contiguouswith Y and G: then B cannot be made contiguous with R without enclosingeither Y or G.We have thus taken every possible case and found that if three countries arecontiguous each with each a fourth country cannot be made contiguous withall three without enclosing one country.Figure 10 is Figure 8 before the transformation, and Figure II is the sameas Figure 9, and it will be seen at once that you cannot possibly reach R fromthe outside. Therefore four countries cannot possibly be drawn so that a fifthmay be contiguous with every one of them, and consequently the fifth countrymay certainly repeat the color R. And if you cannot draw five countries, it isquite obvious that you cannot draw any greater number contiguous eachwith each.Answers 379It is now clear that, at each successive addition of a new country, allthe countries previously drawn must be contiguous each with each to preventthe employment of a repeated color. We can draw four countries underthis condition, only one country must always be enclosed. Now, we canmake the fifth country contiguous with only one country (as in Figure 12),with two countries (as in Figure 13), or with three countries (as in Figure 14).In the first case the new country can be Y or B or R, in the second caseB or R, and in the third case R only. We take the last case (Figure 14)and "bring out," or repeat, R. But in doing so we have been compelledto enclose G. In drawing a sixth country the best we can do (in trying to upset the theorem) is to "bring out" G (as in Figure 15), and the result is thatwe must enclose R. And so on into infinity. We can never avoid enclosing acolor at every successive step, and thus making the color available for the nextstep. If you cannot artificially construct a map that will require a fifth color,such a map cannot possibly occur. Therefore a fifth color can never benecessary, and the truth of the theorem is proved.[Dudeney correctly shows that no more than four regions can be drawn sothat each borders all the others, but his proof fails to show that four colorsare sufficient for all maps. It is true that if any four regions of a map are considered in isolation, no fifth color is necessary for any fifth region. What isrequired, however, is a proof that on a map with a large number of regions,these various sets of five will never conflict with each other in such a way thatfive colors are demanded.The difficulty is best seen by actually constructing a complicated map,using the step by step procedure proposed by Dudeney. If each new region isdrawn so as to border three others, its color is determined automatically andfour-color maps can indeed be extended to infinity. But if many new regionsare added that touch only one, two, or no other previously drawn regions, thechoice of colors for these regions becomes arbitrary. As the map grows in sizeand complexity, one suddenly discovers that it is possible to construct a newregion that will require a fifth color. By backtracking and altering previouscolors, it appears that it is always possible to rectify the mistake and take careof the new region without going to a fifth color. But is it always possible? Thisis the conjecture that remains unproved. For a discussion of the problem, anda list of recent references, see the chapter on the four-color map theorem inmy New Mathematical Diversions from Scientific American.-M. G.]

380 Answers

443. A SW ASTIKLAND MAP-~A - - C r--B IThree different colors are necessary. The bottom right-hand corner of themap is here reproduced. The Lord High Keeper of the Maps had introducedthat little line dividing A and B by mistake, and this was his undoing. A, B,and C must be different colors. Except for this slip, two colors would havebeen sufficient.

444. COLORING THE MAPTwo! The map requires four colors. If the boy had three pigments (red,blue, and yellow) in his box, he could have obtained green, orange, or purpleby mixing any two. But he cannot obtain four colors from fewer than three;consequently, there must have been only two ("not enough colors by one")in his box. "Color" refers to red, orange, yellow, green, blue and purple.Variant shades, such as bluish-green and yellowish-green, are not to beconsidered.

445. PICTURE PRESENTATIONMultiply together as many 2's as there are pictures and deduct l. Thus 2raised to the tenth power is 1,024, and deducting I we get 1,023 as the correctanswer. Suppose there had been only three pictures. Then one can be selectedAnswers 381in 3 ways, two in 3 ways, and three in I way, making together 7 ways, whichis the same as the cube of 2 less l.

446. A GENERAL ELECTIONThe answer is 39,147,416 different ways. Add 3 to the number of members(making 618) and deduct I from the number of parties (making 3). Then theanswer will be the number of ways in which 3 things may be selected from 618.That is:

618 X 617 X 616 _ 39147416 I X 2 X 3 - , , ways.The general solution is as follows. Let p = parties and m = members. Thenumber of ways the parliament can be elected is equal to the number of combinations in which p - I objects may be selected from m + p - I objects.

447. THE MAGISTERIAL BENCHApart from any conditions, ten men can be arranged in line in IO! ways =

3,628,800. Now how many of these cases are barred? Regard two of a nationality in brackets as one item.(I) Then (EE) (SS) (WW) FISA can be permuted in 7! X 23 ways =

40,320. Remember the two E's can change places within their bracket whereverplaced, and so with the S's and the W's. Hence the 23.(2) But we may get (EE) (SS) WWFISA, where the W's are not bracketed,but free. This gives 8! X 22 cases, but we must deduct result (I) or these willbe included a second time. Result, 120,960.(3) Deal similarly with the two S's unbracketed. Result, 120,960.(4) Deal again, with the E's unbracketed. Result, 120,960.(5) But we may have (EE) SSWWFISA, where both Sand Ware unbracketed. This gives 9! X 2 cases, but we must deduct results (I), (2), and(3) for reasons that will now be obvious. Result, 443,520.(6) When only S is bracketed, deducting (I), (2), and (4). Result, 443,520.(7) When only W is bracketed, deducting (I), (3), and (4). Result,

443,520.

382 AnswersAdd these seven results together and you get 1,733,760, which deducted fromthe number first given above leaves 1,895,040 as the number of ways in whichthe ten men may sit.

448. CROSSING THE FERRYThe puzzle can be solved in as few as nine crossings, as follows: (I) Mr. andMrs. Webster cross. (2) Mrs. Webster returns. (3) Mother and daughter-inlaw cross. (4) Mr. Webster returns. (5) Father-in-law and son cross.(6) Daughter-in-law returns. (7) Mr. Webster and daughter-in-law cross.(8) Mr. Webster returns. (9) Mr. and Mrs. Webster cross.

449. MISSIONARIES AND CANNIBALSCall the three missionaries M m m, and the three cannibals C c c, thecapitals denoting the missionary and the cannibal who can row the boat. ThenC c row across; C returns with the boat; C c row across; C returns; M m rowacross; M c return; M C row across; M c return; M m row across; C returns;C c row across; C returns; C c row across; and all have crossed the river withinthe conditions stated.[River crossing problems of this and the preceding type lend themselves tosolution by a simple graph technique. See Robert Fraley, Kenneth L. Cooke,and Peter Detrick, "Graphical Solution of Difficult Crossing Puzzles," inMathematics Magazine, Vol. 39, May 1966, pp. 151-157. See also the firstchapter, "One More River to Cross," in Thomas H. O'Beirne, Puzzlesand Paradoxes (Oxford University Press, 1965).-M. G.]

450. CROSSING THE RIVERThe two children row to the opposite shore. One gets out and the otherbrings the boat back. One soldier rows across; soldier gets out, and boyreturns with boat. Thus it takes four crossings to get one man across and theboat brought back. Hence it takes four times 358, or 1,432 journeys, to getthe officer and his 357 men across the river and the children left in jointpossession of their boat.Answers 383

451. A GOLF COMPETITION PUZZLEThe players may be paired and arranged as follows:ROUNDS

1 2- 3 4 5l"LINHS Be BF SF CE ADZ-UNJ(l; PA CD CA DF RE.

3'''LlNt<S DE: E.A DB AB CF

452. FOOTBALL RESULTSWe see at once from the table that England beat Ireland and drew withWales. As E. scored 2 goals to 0 in these games, they must have won 2-0 anddrawn 0-0. This disposes of E. and leaves three games, W. v. I., S. v. I., andS. v. w., to be determined. Now, S. had only I goal scored against them-byW. or I. I. scored only I goal, and that must have been against W. or S.Assume it was against S. In that case W. did not score against S. But W.scored 3 goals altogether; therefore these must have been scored against I. Wefind I. had 6 goals against them: 2 scored by E., as shown, 3 by W. (if we assume that I. scored v. S.), and the remaining goal was scored by S. But, as wehave just assumed I. scored I goal against S., the match would have beendrawn. It was won by S., and therefore I. could not have scored against S.Thus the goal against S. must have been scored by W. And as W. scored

3 goals, the other two must have been v. I., who must have scored their onlygoal against W. Thus S. beat W. by 2-1 and I. by 2-0, while W. won by

2-1 v. I.

453. THE DAMAGED MEASURELet the eight graduation marks divide the 33-inch measure into the following nine sections: 1,3, 1,9,2,7,2,6,2, and any length can be measured fromI inch up to 33 inches. Of course, the marks themselves will be at 1,4, 5, 14,

384 Answers

16,23,25, and 31 inches from one end. Another solution is I, I, I, 1,6,6,6,6,5.This puzzle may be solved in, at fewest, sixteen different ways. I havesought a rule for determining the fewest possible marks for any number ofinches, and for at once writing out a solution, but a general law governing allthe multiplicity of answers has still to be found.[Although no general rule has yet been found for the ruler problem, considerable progress has been made since it was proposed by Dudeney. Thereader is referred to John Leech's paper, "On the Representation of I, 2, ... ,n by Differences," in the Journal of the London Mathematical Society, Vol. 31,

1956, pp. 160-169. Leech discovered that eight marks are also sufficient formarking a 36-inch yardstick so that all integral values from I to 36 canbe measured. The reader may like to search for this pattern. Had Dudeneyknown it, he would have surely found it preferable to his broken yardstick of

33 inches.-M. G.]

454. THE SIX COTfAGESIf the distances between the cottages are as follows, in the order given, anydistance from one mile up to twenty-six inclusive may be found as from onecottage to another: I, 1,4,4,3, 14 miles round the circular road.[This problem is, of course, a circular version of the preceding one. As before, Dudeney could have increased the length of the "measuring stick" (inthis case, the road) without altering the other aspects of his puzzle. It turnsout that six cottages can be placed on a circular road of 31 miles so that allintegral distances from 1 to 30 are represented by a distance, on the circle,between a pair of houses. It is not hard to see that for n houses the maximumnumber of different ways to measure a length between houses is n(n - I).For n = 6, the formula gives 30, so in this case it is possible to place the 6houses on the 3 I-mile road so that no distance between any pair of housesduplicates any other distance. Similar optimum solutions can be found whenn = 1, 2, 3, 4, or 5. See Michael Goldberg's solution to Problem EI76 inAmerican Mathematical Monthly, September 1965, p. 786.-M. G.]

455. FOUR IN LINEThere are nine fundamentally different arrangements, as shown in the illustration, the first, A, being the arrangement given as an example. Of these, D,E, and I each give eight solutions, counting reversals and reflections as ex-Answers 385plained, and the others give only four solutions each. There are, therefore, inall, forty-eight different ways in which the four counters may be placed onthe board, so that every square shall be in line with at least one counter.HIt may interest the reader to know that on a chessboard, 8 X 8, five countersmay be placed in line under precisely the same conditions in four fundamentaldifferent ways, producing twenty different solutions in all. The arrangementsare given in the solution to Number 311, "The Five Dogs Puzzle," in the bookAmusements in Mathematics.

386 Answers

456. FLIES ON WINDOW PANESThe three lively flies have taken up new positions, as indicated by the arrows,and still no two flies are in a straight line.Q]i~~i" I~I~~ t ~ 1t{Jl4:)Ii{}

457. CITY LUNCHEONSIf the Pilkins staff had been II in number, and the Radson staff 12, theycould have sat down differently in 165 and 495 ways respectively, whichwould have solved the question. Only we were told that there happened to bethe same number of men in each staff. Therefore the answer is 15 in threesfor 455 days, and 15 in fours for 1,365 days.

458. THE NECKLACE PROBLEMThe number of different necklaces with eight beads under the conditionsis 30.A general solution for any number of beads is difficult, if not impossible.But with as few as eight beads the reader will have no difficulty in finding thecorrect answer by mere trial.[For a discussion of the general problem, with a recursive formula thatgives the number of different necklaces with n beads of m different colors, seeAnswers 387the chapter on "The Calculus of Finite Differences" in my book New Mathematical Diversionsfrom Scientific American.-M. G.]

459. AN EFFERVESCENT PUZZLEThe answer to the first case is 88,200 ways. There is an easy way of gettingthe answer, but it would require too much space to explain it in detail. In thesecond case the answer is reduced to 6,300 ways.

460. TESSELLATED TILESDiscard the first tile in each of the four horiwntal rows. Then the remaining sixteen may be arranged as shown in the illustration, in accordance withthe conditions.

461. THE THIRTY-SIX LETTER PUZZLEIf you have tried, as most people do, first to place all six of one letter, thenall six of another letter, and so on, you will find, after you have placed fourdifferent kinds of letters six times each, it is impossible to place more thantwo letters each of the last two letters, and you will get our arrangement

388 Answersnumber l. The secret is to place six of each of two letters, and five of each ofthe remaining four, when we get our second diagram, with only four blanks.A .B C D E F A B C D E F.D F E B A C D E A F B cE C D .B F C D AB D C E 13 D C EC. .B E D C A E B F DE D C b E F .D C A B

462. THE TEN BARRELSArrange the barrels in one of the following two ways, and the sides willadd up to 13 in every case-the smallest number possible:o

8 6

4 9 5

732o

7 8

593

642By changing the positions of the side numbers (without altering the numbers contained in any side) we get eight solutions in each case, not countingmere reversals and reflections as different.

463. LAMP SIGNALLINGWith 3 red, white, or green lamps we can make 15 variations each (45).With 1 red and 2 white we can make the same, 15, and each way will admitof 3 variations of color order, 45 in all. The same with 1 red and 2 green,

1 white and 2 red, 1 white and 2 green, 1 green and 2 white, 1 green and 2 red(270). With 1 red, 1 white, and 1 green we can get 6 by 15 variations (90).With 2 red or 2 white or 2 green we can get 7 patterns (21). With 1 red and

1 white, or 1 red and 1 green, or 1 white and 1 green, we can get 14variations each (42). With 1 lamp only we can get only 1 signal each (3).Add together the numbers in parentheses (45,270,90,21,42, and 3), and weget the answer, 471 ways.Answers 389

464. THE HANDCUFFED PRISONERSThe following is a solution. Every prisoner will be found to have been handcuffed to every other prisoner once, and only once.

1-2-3 2-6-8 6-1-7 1-4-8 7-2-9 4-3-1 4-5-6 5-9-1 9-4-2 2-5-7 3-6-4

5-8-2 7-8-9 3-7-4 8-3-5 6-9-3 8-1-5 9-7-6If the reader wants a hard puzzle to keep him engrossed during the wintermonths, let him try to arrange twenty-one prisoners so that they can all walkout, similarly handcuffed in triplets, on fifteen days without any two menbeing handcuffed together more than once.In case he should come to the opinion that the task is impossible, we willadd that we have written out a perfect solution. But it is a hard nut!

465. SEATING THE PARTYThe number of different ways in which the six occupants of the car can beseated under the conditions is 144.

466. QUEER GOLFThe two best distances are 100 yards (called "the approach"), and 125 yards("the drive"). Hole I can be reached in three approaches, hole 2 in two drives,hole 3 in two approaches, hole 4 in two approaches and one drive, hole 5 inthree drives and one backward approach, hole 6 in two drives and one approach, hole 7 in one drive and one approach, hole 8 in three drives, and hole 9in four approaches-26 strokes in all.

467. THE ARCHERY MATCHMrs. Finch scored 100 with four 17's and two 16's; Reggie Watson scoredllO with two 23's and four 16's; Miss Dora Talbot scored 120 with one 40 andfive 16's. Her score can be made up in various ways, except for the fact thatthe bull's-eye has to be got in somewhere, and this is the only place where itcan occur.

390 Answers

468. TARGET PRACfICEThe total score was 213, so each man scored 71, and this could be done inthe following manner: One man scored 50, 10, 5, 3, 2, and I; another scored

25, 20, 20, 3, 2, and I; and the third man 25, 20, 10, 10, 5, and I.

469. TOM TIDDLER'S GROUNDThe large majority of competitors who tried to solve "Tom Tiddler'sGround," when it first appeared in London's Daily News, succeeded in securing bags containing only $45.00.The correct answer is $47.00 contained in ten bags, all deposited on outsideplots, thus: 4, 5, 6 in the first row, 5 in the second, 4 in the third, 3 inthe fourth, 5 in the fifth, and 5, 6, 4 in the bottom row. If you includefive bags containing $6.00 each, you can secure only nine bags, and a valueof $46.00.

470. THE SEVEN CHILDRENThere are 5,040 ways of arranging the children, and 720 different ways ofplacing a girl at each end. Therefore, the chances are 720 in 5,040, or I in 7.Or, which is the same thing, the chances are I to 6 in favor, or 6 to I against,there being a girl at both ends.

471. TIC TAC TOENumber the board, as in Figure A. Mr. Nought (the first player) can openin one of three ways: he can play to the center, 5, or to a corner, I, 3, 7, or 9,or to a side, 2, 4, 6, 8. Let us take these openings in turn. If he leads with acenter, then Mr. Cross has the option of a corner or a side. If he takes a side,such as 2 in Figure A, then Nought plays I and 4 successively (or I and 7),and wins. Cross must take a corner, as in Figure B, and then Nought cannotdo better than draw. If Nought leads with a corner, say I, Cross has five different replies, as in Figures C, D, E, F, and G (for 4 here is the same as 2, 7the same as 3, and 8 the same as 6). If he plays as in Figure C, Nought winswith 5 and 4; if he plays as in D, Nought wins with 7 and 3; if as inE, Nought wins with 9 and 7; if as in F, Nought wins with 5 and 3. Cross isAnswers 391compelled to take the center, as in Figure G, to save the game, for this willresult in a draw. If Nought opens with a side, say 2, as in Figures H, J, K, L,and M, and Cross plays as in H, Nought wins with 5 and I; and if he playsas in J, Nought wins with I and 5. Cross must playas in K, L, or Mto secure a draw.

1 Il){ 1

3'I- So (, ., IS ~WInI have thus shown the play for Nought to win in seven cases where Crossmakes a bad first move, but I have not space to prove the draws in theremaining five positions B, G, K, L, and M. But the reader can easily tryeach of these cases for himself and be convinced that neither player can winwithout the bad play of his opponent. Of course, either player can throwaway the game. For example, if in Figure L Nought stupidly plays 3 on hissecond move, Cross can play 7 and 9 and win. Or if Nought plays 8, Crosscan play 5 and 7 and win.Now, if I were playing with an equally expert player I should know thatthe best I could possibly do (barring my opponent's blunders) would be tosecure a draw. As first player, Nought, I should know that I could safely lead

392 Answerswith any square on the board. As second player, Cross, I should take acomer if Nought led with a center and take the center if he led with anythingelse. This would avoid many complexities and should always draw. The factremains that it is a capital little game for children, and even for adults whohave never analyzed it, but two experts would be merely wasting their time inplaying it. To them it is not a game, but a mere puzzle that they havecompletely solved.

472. THE HORSESHOE GAMEJust as in tic tac toe, every game should be a draw. Neither player can winexcept by the bad play of his opponent.[For a complete analysis of the game, see Edouard Lucas, RecreationsMathematiques, Vol. III, pp. 128-131.-M. G.]

473. TURNING THE DIEThe best call for the first player is either "2" or "3," as in either case onlyone particular throw should defeat him. If he called" I," the throw of either

3 or 6 should defeat him. If he called "2," the throw of 5 only should defeathim. If he called "3," the throw of 4 only should defeat him. If he called "4,"the throw of either 3 or 4 should defeat him. If he called "5," a throwof either 2 or 3 should defeat him. And if he called "6," the throw of eitherI or 5 should defeat him. It is impossible to give here a complete analysis ofthe play, but I will just state that if at any time you score either 5, 6, 9, 10,

14, 15, 18, 19, or 23, with the die any side up, you ought to lose. If you score

7 or 16 with any side up you should win. The chance of winning with theother scores depends on the lie of the die.

474. THE THREE DICEMason's chance of winning was one in six. If Jackson had selected thenumbers 8 and 14 his chances would have been exactly the same.

475. THE 37 PUZZLE GAMEThe first player (A) can always win, but he must lead with 4. The winningscores to secure during the play are 4, 11, 17, 24, 30, 37. In the first gameAnswers 393below the second player (B) puts off defeat as long as possible. In the secondgame he prevents A scoring 17 or 30, but has to give him 24 and 37. In thethird game he prevents A scoring II or 24, but has to give him 17,30, and 37.Notice the important play of the 3 and the 5.A B A B A B

4 I (a) 4 I 4 I

3 I (b) 3 I 3 4(II) 2 I (11) 2 3 (d) (17) 5(17) 5 I (c) 5 I 3 4

3 2 (24) 4 3 (e) (30) 5 (j) I(24) I 2 5 I 3(30) 4 (37) 4 (37) 2

3 2(37) I(a) Or A will score II next move. (b) B could not prevent A scoring II or

17 next move. (c) Again, to prevent A immediately scoring 24. (d) Preventing A scoring 17, but giving him 24. (e) Preventing A scoring 30, but givinghim the 37. (j) Thus A can always score 24 (as in the last game) or 30 (asin this), either of which commands the winning 37.

476. THE 22 GAMEApart from the exhaustion of cards, the winning series is 7, 12, 17,22. If youcan score 17 and leave at least one 5-pair of both kinds (4-1, 3-2), you mustwin. If you can score 12 and leave two 5-pairs of both kinds, you must win.If you can score 7 and leave three 5-pairs of both kinds, you must win. Thus,if the first player plays a 3 or 4, you playa 4 or 3, as the case may be, and score

7. Nothing can now prevent the second player from scoring 12, 17, and 22. Thelead of 2 can also always be defeated if you reply with a 3 or a 2. Thus, 2-3,

2-3, 2-3, 2-3 (20), and, as there is no remaining 2, second player wins.Again, 2-3, 1-3, 3-2, 3-2 (19), and second player wins. Again, 2-3, 3-4 (12),or 2-3, 4-3 (12), also win for second player. The intricacies of the defence 2-2I leave to the reader. The best second play of first player is a I.The first player can always win if he plays I, and in no other way. Here arespecimen games: 1-1,4-1,4-1,4 (16) wins. 1-3, 1-2,4-1,4-1,4 (21) wins.

1-4,2 (7) wins. 1-2,4 (7) wins.

394 Answers

477. THE NINE SQUARES GAMEI should play MN. My opponent may play HL, and I play CD. (Ifhe hadplayed CD, I should have replied HL, leaving the same position.) The best hecan now do is DH (scoring one), but, as he has to play again, I win the remaining eight squares.

478. THE TEN CARDSThe first player can always win. He must turn down the third card from oneof the ends, leaving them thus: 00.0000000. Now, whatever the second playerdoes, the first can always leave either 000.000 or 00.00.0.0 or 0.00.000 (orderof groups does not matter). In the first case, whatever the second player doesin one triplet the other repeats in the other triplet, until he gets the last card.In the second case, the first player similarly repeats the play of his opponent,and wins. In the third case, whatever the second player does, the first canalways leave either 0.0, or 0.0.0.0, or 00.00, and again get an obvious win.[Victor Meally points out that the first player can also win by first turningthe second card from either end, or the fourth card from either end.-M. G.]

479. THE DOMINO SWASTIKAIt will be seen that by placing the four extra dominoes in the positionsshown a perfect swastika is formed within the frame.Answers 395

480. DOMINO FRACfIONSThe illustration shows how to arrange the dominoes so that each of the threerows of five sum to 10. Give every fraction the denominator 60. Then thenumerators of the fractions used must sum to 1,800, or 600 in each row,to produce the sum 10. The selection and adjustment require a little thoughtand cunning.

481. A NEW DOMINO PUZZLEThese four dominoes fulfill the conditions. It will be found that, takingcontiguous pips, we can make them sum to any number from I to 23 inclusive.[Victor Meally has improved Dudeney's solution. The following chain offour dominoes, 1-3,6-6,6-2, 3-2, gives sums from I through 29. Meally alsopoints out that three dominoes, 1-1,4-4,4-3, have sums from I through 17.-M.G.]

396 Answers

482. A DOMINO SQUARE• • • • • • ••• · .' • , • • • , • • ••••• • • • • · • • • • • • -• '-• '-' • -' - • • '. • • • • • • • • • • • • •"• • •• • • • , • • • • •• • - .-. • •• • • • • I ••• • ., • , • • • • •••• • • • • • • ••• 1 , • , • • •• • •The illustration explains itself. The eighteen dominoes are arranged so as toform the required square, and it will be found that in no column or row is anumber repeated. There are, of course, many other ways of doing it.

483. A DOMINO STARThe illustration shows a correct solution. The dominoes are placed togetheraccording to the ordinary rule, the pips in every ray sum to 21, and thecentral numbers are 1,2,3,4,5,6, and two blanks.

1,·- .-1.·'o.o •: ....o.· · . . :/ · .... .. : : : : .:.- o· · · · '.·:·:1:-:·: :1: : .••• ,Answers 397

484. DOMINO GROUPSThe illustration shows one way in which the dominoes may be laid out sothat, when the line is broken in four lengths of seven dominoes each, everylength shall contain forty-two pips.

485. LES QUADRILLESThe illustration shows a correct solution, the two blank squares being on theinside. If in the example shown before all the numbers had not happened tobe found somewhere on the edge, it would have been an easy matter, for we•• ••• • • •• ._.1 . , • • , ._. • • • • •• • • • - • • • • • • • • • • • • .!..!.. • • • •• • • • - - • -• • • - • • • • • • • • • •• • • • • • • • I , • • • • • • • • • •••• ••• - , ... • • ••• 1 ••• • • ••• • •• • • • • -Ie.• • • • • ·1- •-;- ._e .-- - - • :r-· ----;- • • • • • •• • • • • • • • • •• •• • • • •• -. • • • • 1 • • ' . • · .' .. •

398 Answersshould have had merely to exchange that rrussmg number with a blankwherever found. There would thus have been no puzzle. But in the circumstances it is impossible to avail oneself of such a simple maneuver.[For more domino problems of this type, known as quadrilles, see EdouardLucas, Recreations Mathematiques, Vol. 2, pp. 52-63, and Wade E. Philpott,"Quadrilles," in Recreational Mathematics Magazine, No. 14, JanuaryFebruary 1964, pp. 5-11.-M. G.]

486. DOMINO FRAMES· ,• •• ••• • •• •• • • • •• • r-------,---'--i• I . '···1·· •••••• •• •• • f--• •-• •• • f-;-; · . • •• •• •• • I I · . ~I· • • •• •• I ,- ... - .' - ••••• • • • • - • • • -• ~• • • - f-- -• • • • -• • •I • -. • :1-. I -•The three diagrams show a solution. The sum of all the pips is 132. Onethird of this is 44. First divide the dominoes into any three groups of 44 pipseach. Then, if we decide to try 12 for the sum of the sides, 4 times 12 being 4more than 44, we must arrange in every case that the four corners in a frameshall sum to 4. The rest is done by trial and exchanges from one groupto another of dominoes containing an equal number of pips.Answers 399

487. DOMINO HOLLOW SQUARESIt is shown in the illustration how the 28 dominoes may be arranged in theform of7 hollow squares, so that the pips in the four sides of every square addup alike. It is well to remember this little rule when forming your squares. Ifthe pips on your dominoes sum, say, to 7 (as in the first example), and youwish the sides to add up 3, then 4 X 3 - 7 gives us 5 as the sum of the fourcomers. This is absolutely necessary. Thus, in the last example, 4 X 16 =

64 - 43 tells us that the four comers must sum to 21, as it will be found they do.

488. DOMINO SEQUENCESIf we draw from the set the four dominoes 7-6, 5-4, 3-2, 1-0, the remaining dominoes may be put together in proper sequence. Any other combinationsof these particular numbers would do equally well; thus we might withdraw

7-0,6-1,5-2, and 4-3. Generally, for any set of dominoes ending in a doubleodd number, those withdrawn must contain together every number once fromblank up to two less than the highest number in the set.

489. TWO DOMINO SQUARESThe illustration on page 400 shows how the twenty-eight dominoes may belaid out so as to form the two required squares with the pips in each of theeight sides summing to 22. With 22 as the constant sum, the comers must sum

400 Answers, • . .. • I I~ . . .. . .. ' . . . . .' .... . ••• ... . ·,1-·· .. • • , , . .......... •~ -· -;- • • · . • • · ~ .....- • .-·• · r.--.- · " -· · . · · • · . . r!--!- · . · · · .... · · .' • · • . · . I-;- - · -· -;- · . . , . ',1- ... . · . . '· • · . .. . . .... • · . · . •.• , ·.-1• ·1 • ':1' • , , . .. . .to 8, with 23 to 16, with 24 to 24, with 25 to 32, with 26 to 40. The constantsum cannot be less than 22 or more than 26.

490. DOMINO MULTIPLICATION... _ •.. .Here all the twenty-eight dominoes are used, and they form seven multiplication sums as required.

491. DOMINO RECTANGLE...• • • • • •• . . •• • • • •• • • 0 0 0· 0 • 0• · · • 0 I" • 0 " • • 0 • 0• 0- 0 • • • - • • • 0 0• " • • • 0 • ·• • • • • • • • · • • • • • • • • · •00••.-.. • .-. • • • • • • • • 0 • • • • 0• • • • • • • • • • • 0 • • • • 0 -• • • o.• • 0 · • • 0 • • • •-• 0• • · •• 0• -. 0 • • •••• •• • 0 •• •Answers 401The illustration shows how the twenty-eight dominoes may be arranged sothat the columns add up to 24 and the rows to 21.

492. THE DOMINO COLUMN• · •0." - o. . :- •• " . .. · · -. 0 · · . 0 " · " .. 0-·, .· · -. • · •• · -.. 0 0· .. - · .-. . • -1.- O • · . · .'. : : · . · . ., 0Place the second column under the first, and the third under the second (thecolumn is broken merely for convenience in printing), and the conditions willbe found to be fulfilled.

402 Answers

493. ARRANGING THE DOMINOESThere are just 126,720 different ways of playing those 15 dominoes, counting the two directions as different.

494. A NEW MA TeH PUZZLEThe smallest possible number is 36 matches. We can form triangle andsquare with I2 and 24 respectively, triangle and pentagon with 6 and 30,triangle and hexagon with 6 and 30, square and pentagon with 16 and 20,square and hexagon with 12 and 24, and pentagon and hexagon with 30 and

6. The pairs of numbers may be varied in all cases except the fourth and last.There cannot be fewer than 36. The triangle and hexagon require a numberdivisible by 3: the square and hexagon require an even number. Therefore thenumber must be divisible by 6, such as 12, 18,24,30,36, but this conditioncannot be fulfilled for a pentagon and hexagon with fewer than 36 matches.

495. HURDLES AND SHEEPIf the enclosure is to be rectangular,the nearer the rectangle approachesto the form of a square the greaterwill be the area. But the greatest areaof all will always be when the hurdlesare arranged in the form of a regularpolygon, inscribed in a circle, and ifthis can be done in more than one waythe greatest area will be when thereare as many sides as hurdles. Thus, thehexagon given earlier had a greaterarea than the triangle. The twelvesided figure or regular dodecagontherefore encloses the largest possiblearea for twelve hurdles-room forabout eleven sheep and one-fifth.Eleven hurdles would only accommodate a maximum of about nine andnine-twenty-fifths, so that twelvehurdles are necessary for ten sheep. IfAnswers 403you arrange the hurdles in the form of a square, as shown by the dotted lines,you only get room for nine sheep.

496. THE TWENTY MATCHESThe illustration shows how two enclosures may be formed with 13 and 7matches respectively, so that one area shall be exactly three times as large as~\ .. ·······::/:·--7L~0v :· . . .. _·_··::·· . .... . ';"'---'Q'" . , . " .. .~·~G/·the other, for one contains five of those little equilateral triangles, and theother fifteen.There are other solutions.

497. A MATCH PUZZLEThe illustration shows one of thefour distinctive ways of solving thispuzzle, with eleven, an odd numberof matches. If you first enclose anoutside row, as A, then you can enclose the square, B, in any position,and complete the solution with elevenmatches in all.

404 Answers

498. AN INGENIOUS MATCHPUZZLEIt will be seen that the second I inVII has been moved, so as to form thesign of square root. The square root ofI is, of course, I, so that the fractionalexpression itself represents I.IVI

499. FIFTY-SEVEN TO NOTHINGRemove the two cigarettes forming the letter L in the original arrangement,and replace them in the way shown in our illustration. We have the square rootL ...... ' .... n~ of I minus I (that is I less I), which clearly is O. In the second case we canremove the same two cigarettes and, by placing one against the V and the otheragainst the second I, form the word NIL, or nothing.

500. THE FIVE SQUARESPlace the twelve matches as in thediagram and five squares are enclosed. It is true that the one in thecenter (indicated by the arrow) isvery smaIl, but no conditions wereimposed as to dimensions.Answers 405SOl. A MATCH TRICKYou must secrete the match insidein the manner shown by the dottedlines in the illustration, so that thehead is just over the edge of the trayof the box. In closing the box youpress this extra match forward withthe thumbnail (which, if done carefully, will not be noticed), and it fallsinto its place. Of course, one of thematches first shown does not tumround, as that would be an impossibility, but nobody ever counts thematches.

562. THREE TIMES THE SIZEThe illustration shows how two enclosures may be formed with 13 and

7 matches respectively, so that onearea shall be exactly three times aslarge as the other. The dotted lineswill show that one figure contains twosquares and an equilateral triangle,and the other six squares and threesuch triangles. The 12 horizontal andvertical matches have not been moved.

563. A SIX-SIDED FIGUREThe illustration shows the simpleanswer. We did not ask for a planefigure, nor for the figure to be formedwith the 9 matches. We show (in perspective) a cube (a regular six-sidedfigure).

406 Answers

504. TWENTY-SIX MATCHESThe illustration shows how the 26 matches may be placed so that the squareis divided into two parts of exactly the same size and shape, one part contain·ing two stars, and the other two crosses.

506. EQUILATERAL TRIANGLESRemove the four matches indicatedby the dotted lines, and the remainderform four equal triangles.

505. THE THREE MATCHESArrange the three matches as shownin the illustration, and stand the boxon end in the center.

1... ...... .

508. HEXAGON TO DIAMONDSThe illustration will show by thedotted lines the original position ofthe two matches that have beenmoved.Answers 407

507. SQUARES WITH MATCHESIn the illustration the dotted linesindicate the six matches that havebeen removed. The thin lines showwhere they are replaced, and the thicklines indicate the six that have notbeen moved ..............a' ' ••

509. QUEER ARITHMETICArrange 10 matches thus-FIVE. Then take away the 7 matches formingFE (seven-tenths of the whole), and you leave IV, or four.

510. COUNTING THE MATCHESThere were 36 matches in the box with which he could form a triangle 17,

10, 9, the area of which was 36 sq. in. After 6 had been used, the remaining

30 formed a triangle 13, 12,5, with an area of30 sq. in.; and after using another

6, the 24 remaining would form a triangle 10, 8, 6, with an area of 24 sq. in.

511. A PUZZLE WITH CARDSArrange the pack in the following order face downwards with 9 of Clubs atthe top and 5 of Spades at bottom: 9 C, Jack D., 5 c., ace D., King R., King S.,

7 R., 2 D., 6 S., Queen D., 10 S., ace S., 3 C, 3 D., 8 C, King D., 8 R., 7 C,

408 Answers

4 D., 2 S., ace H., ace c., 7 S., 5 D., 9 H., 2 H., Jack S., 6 D., Queen c., 6 c.,

10 H., 3 S., 3 H., 7 D., 4 c., 2 c., 8 S., Jack H., 4 H., 8 D., Jack c., 4 S., QueenS., King C., 9 D., 5 H., 10 c., Queen H., 10 D., 9 S., 6 H., 5 S.[All such arrangements for spelling cards can be solved quickly by startingwith the last card to be spelled, then performing all the required operations inreverse order, finishing with the full pack, or packet, of cards.-M. G.]

512. CARD SHUFFLINGTo shuffle fourteen cards in the manner described, so that the cards shallreturn to their original order, requires fourteen shuffles, though with sixteencards we require only 5. We cannot go into the law of the thing, but the readerwill find it an interesting investigation.[For the mathematical theory of this shuffle, see W. W. Rouse Ball,Mathematical Recreations and Essays, revised eleventh edition, edited byH. S. M. Coxeter (Macmillan, 1960), pp. 310-311. The shuffle is sometimescalled "Monge's shuffle" after Gaspard Monge, a famous eighteenth centuryFrench mathematician who was the first to investigate it.-M. G.]

513. A CHAIN PUZZLETo open a link and join it again will cost 3¢. By opening one link at the endof each of the thirteen pieces the cost will be 39¢, so it would be cheaper thanthat to buy a new chain. If there happened to be a piece of twelve links, allthese twelve coUld be opened to join the remaining twelve pieces at a cost alsoof 36¢. If there had happened to be two pieces together, containing elevenlinks, all these could be opened to join the remaining eleven pieces at a costof 33¢.The best that can be done is to open three pieces containing together tenlinks to join the remaining ten pieces at a cost of 30¢. This is possible if webreak up the piece of four links and two pieces of three links. Thus, if we include the piece of three links that was shown in the middle row as one of thethree link pieces, we shall get altogether five large links and five small ones.If we had been able to find four pieces containing together nine linkswe should save another 3¢, but this is not possible, nor can we find five piecescontaining together eight links, and so on, therefore the correct answer is asstated, 30¢.Answers 409

514. A SQUARE WITH FOUR PENNIESThe illustration indicates how we may show a square with four pennies. Thesides of the square are the lines beneath Britannia.

515. SIMPLE ADDITIONAdd IV turned upside down below VI and you get XI.

516. A CALENDAR PUZZLEEvery year divisible by 4 without remainder is bissextile (leap year), exceptthat every year divisible by 100 without remainder is not leap year, unless itbe also divisible by 400 without remainder, when it is leap year. This is notgenerally understood. Thus 1800 was not leap year, nor was 1900; but 2000,

2400, 2800, etc., will all be leap years. The first day of the present century,January I, 1901, was Tuesday.The present century will contain 25 leap years, because 2000 is leap year,and therefore 36,525 (365 X 100 + 25) days, or 5217 weeks and 6 days; sothat January 1,2001, will be 6 days later than Tuesday-that is Monday. Thecentury beginning January I, 2001, will contain only 24 leap years, because

2100 is not leap year, and January 1,2101, will be 5 days later than Monday,

410 Answerslast mentioned, that is Saturday, because there are 5217 weeks and only

5 days. It will now be convenient to put the results into tabular form:January I, 1901-Tuesday.January I, 200I-Monday. 6 days later (2000, leap year)January I, 2IOI-Saturday. 5 days laterJanuary I, 220I-Thursday. 5 days laterJanuary I, 2301-Tuesday. 5 days laterJanuary 1, 2401-Monday. 6 days later (2400, leap year)It will thus be seen that the first days of successive centuries will be Tuesday,Monday, Saturday, and Thursday-perpetually recurring-so that the first dayof a century can never occur on a Sunday, Wednesday, or a Friday, as I havestated.

517. THE FLY'S TOURBefore you join the ends give one end of the ribbon a half-turn, so that thereis a twist in the ring. Then the fly can walk over all the squares without goingover the edge, for we have the curious paradox of a piece of paper with onlyone side and one edge![Dudeney has described what is now well known as the Moebius strip, oneof the great curiosities of topology.-M. G.]

518. A MUSICAL ENIGMAThe undoubtedly correct solutionto this enigma is BACH. If you turnthe cross round, you get successivelyB flat (treble clef), A (tenor clef), C(alto clef), and B natural (trebleclef). In German B flat is called "B"and B natural "H," making it readBACH.It reminds me of an organ fugue byB A C Hc. P. Emanuel Bach, based on thefamily name and beginning as in theillustration.Answers 411

519. SURPRISING RELATIONSHIPIf there were two men, each of whom marries the mother of the other, andthere is a son of each marriage, then each of such sons will at once be uncleand nephew to the other. This is the simplest answer.[Victor Meally supplies two more answers: (l) Each of two women marriesthe father of the other, (2) A man marries the mother of a woman, and thewoman marries the father of the man.-M. G.]

520. AN EPITAPH (A.D. 1538)If two widows had each a son, and each widow married the son of the otherand had a daughter by the marriage, all the relationships will be foundto result.

521. THE ENGINEER'S NAMEIt is clear that the guard cannot be Smith, for Mr. Smith is certainly theengineer's nearest neighbor, and his income is, therefore, exactly divisible by

3, which $10,000.00 is not. Also the stoker cannot be Smith, because Smithbeats him at billiards. Therefore the engineer must be Smith, and as we are concerned with him only, it is immaterial whether the guard is Jones and thestoker Robinson, or vice versa.[This is one of Dudeney's most popular puzzles. It became the prototypeof scores of later logic problems, sometimes called Smith-Jones-Robinsonpuzzles, in honor of Dudeney's original problem. James Joyce refers to theproblem ("Smith-Jones-Orbison") in the mathematics section of FinnegansWake (p. 302), and Dudeney himself is mentioned in footnote I on page 284.-M.G.]

522. STEPPING STONESNumber the stepping stones I to 8 in regular order. Then proceed as follows: I (bank), I, 2, 3, (2), 3, 4, 5, (4), 5, 6, 7, (6), 7, 8, bank, (8),bank. The steps in parentheses are taken in a backward direction. It will thus beseen that by returning to the bank after the first step, and then always going

412 Answersthree steps forward for one step backward, we perform the required feat innineteen steps.

523. AN AWKWARD TIMEThe Colonel's friend said that 12:50 is clearly an awkward time for a trainto start, because it is ten to one (10 to 1) if you catch it.

524. CRYPTIC ADDITIONIf you tum the page upside down you will find that one, nine, one, andeight added together correctly make nineteen.

525. THE TWO SNAKESWe cannot say how much of each snake must be swallowed before a vitalorgan is sufficiently affected to cause death. But we can say what will nothappen-that the snakes will go on swallowing one another until both disappear altogether! But where it will really end it is impossible to say.

526. TWO PARADOXESIf W and E were stationary points, and W, as at present, on your left whenadvancing towards N, then, after passing the Pole and turning round, Wwould be on your right, as stated. But Wand E are not fixed points, butdirections round the globe; so wherever you stand facing N, you will have theW direction on your left and the E direction on the right.In the reflection in a mirror you are not "turned round," for what appearsto be your right hand is your left, and what appears to be your left side is theright. The reflection sends back, so to speak, exactly what is opposite to it atevery point.[If the reader does not understand Dudeney's brief explanation of themirror paradox, he should consult the first chapter of my AmbidextrousUniverse (Basic Books, 1965), where it is considered at greater length.-M. G.]Answers 413

527. COIN AND HOLEStrange though it may at first appear, a half-dollar may be passedthrough that small hole (made bytracing around a penny) without tearing the paper. This is the largest cointhat can be used. First fold the paperacross the center of the hole and dropthe coin into the fold. Then, holdingthe paper at A and B, bring the handstogether upwards, and the coin maybe squeezed through the hole.

528. A LEAP YEAR PUZZLESince the new style was adopted in England in 1752, the first year with fiveWednesdays in February was 1764. Then 1792 and 1804. By adding 28, wethen get 1832, 1860, 1888. Then we make the jump to 1928, 1956, 1984, and

2012. The answer is, therefore, 1888 and 1956. Normally it occurs everytwenty-eighth year, except that 1800 and 1900 (which were not leap years)come in between, when the rule breaks down. As 2000 will be a leap year,twenty-eight years from 1984 is correctly 2012.

529. BLOWING OUT THE CANDLE·The trick is to lower the funnel until the dotted line, CA, on the top in thepuzzle illustration, is in line with the flame of the candle. Any attempt toblow the candle out with the flame opposite the center of the opening of thefunnel is hopeless.

530. RELEASING THE STICKIf you take the loop and pull through it as much of the coat above andbelow the buttonhole as is necessary, it will be possible to pass the lower endof the stick through the buttonhole and get it in the required position. To disentangle it you reverse the process, though it is rather more perplexing. If youput it on your friend's coat carefully while he is not looking, it will puzzlehim to remove it without cutting the string.

414 Answers

531. THE KEYS AND RINGFirst cut out the keys and ring inone piece, as here shown. Now cutonly half through the cardboard alongthe eight little dark lines, and similarlycut half through on the eight littledotted lines on the other side of thecardboard Then insert a penknife andsplit the card below the four littlesquares formed by these lines, andthe three pieces will come apart withthe keys loose on the ring and no join.

532. THE ENTANGLED SCISSORSSlacken the string throughout so as to bring the scissors J.U~ar the hand ofthe person holding the ends. Then work the loop, which is shown at the bottom of the figure, backwards along the double cord. You must be careful tomake the loop follow the double cord on its course, in and out, until the loopis free of the scissors. Then pass the loop right round the points of the scissors and follow the double cord backwards. The string will then, if you havebeen very careful, detach itself from the scissors. But it is important to avoidtwists and tangles, or you will get it in a hopeless muddle. But with a littlepractice it may be easily done.

533. INTELLIGENCE TESTSWhat is wrong with the dream story is the obvious fact that, as the dreamernever awoke from his dream, it is impossible that anything could be knownabout it. The story must, therefore, be a pure invention.The answer to the second question is: By immersing them in water and calculating from the rise of the water in the vessel.

534. AT THE MOUNTAIN TOPThe surface of water, or other liquid, is always spherical, and the greaterany sphere is the less is its convexity. The spherical surface of the water must,Answers 415therefore, be less above the brim of the vessel, and consequently it will holdless at the top of a mountain than at the bottom. This applies to any mountain whatever.

535. CUPID'S ARITHMETICAll the young mathematician had to do was to reverse the paper and holdit up to the light, or hold it in front of a mirror, when he would immediatelysee that his betrothed's curious jumble of figures will read: "Kiss me, dearest."

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