Chapter XII: Preface: v (11)

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Here is an interesting little puzzle suggested to me by Mr. W. T. Whyte. Mark off on a sheet of paper a rectangular space 5 inches by 3 inches, and then find the greatest number of halfpennies that can be placed within the enclosure under the following conditions. A halfpenny is exactly an inch in diameter. Place your first halfpenny where you like, then place your second coin at exactly the distance of an inch from the first, the third an inch distance from the second, and so on. No halfpenny may touch another halfpenny or cross the boundary. Our illustration will make the matter perfectly clear. No. 2 coin is an inch from No. 1; No. 3 an inch from No. 2; No. 4 an inch from No. 3; but after No. 10 is placed we can go no further in this attempt. Yet several more halfpennies might have been got in. How many can the reader place?

430.--FIND THE MAN'S WIFE.

One summer day in 1903 I was loitering on the Brighton front, watching the people strolling about on the beach, when the friend who was with me suddenly drew my attention to an individual who was standing alone, and said, "Can you point out that man's wife? They are stopping at the same hotel as I am, and the lady is one of those in view." After a few minutes' observation, I was successful in indicating the lady correctly. My friend was curious to know by what method of reasoning I had arrived at the result. This was my answer:--

"We may at once exclude that Sister of Mercy and the girl in the short frock; also the woman selling oranges. It cannot be the lady in widows' weeds. It is not the lady in the bath chair, because she is not staying at your hotel, for I happened to see her come out of a private house this morning assisted by her maid. The two ladies in red breakfasted at my hotel this morning, and as they were not wearing outdoor dress I conclude they are staying there. It therefore rests between the lady in blue and the one with the green parasol. But the left hand that holds the parasol is, you see, ungloved and bears no wedding-ring. Consequently I am driven to the conclusion that the lady in blue is the man's wife--and you say this is correct."

Now, as my friend was an artist, and as I thought an amusing puzzle might be devised on the lines of his question, I asked him to make me a drawing according to some directions that I gave him, and I have pleasure in presenting his production to my readers. It will be seen that the picture shows six men and six ladies: Nos. 1, 3, 5, 7, 9, and 11 are ladies, and Nos. 2, 4, 6, 8, 10, and 12 are men. These twelve individuals represent six married couples, all strangers to one another, who, in walking aimlessly about, have got mixed up. But we are only concerned with the man that is wearing a straw hat--Number 10. The puzzle is to find this man's wife. Examine the six ladies carefully, and see if you can determine which one of them it is.

I showed the picture at the time to a few friends, and they expressed very different opinions on the matter. One said, "I don't believe he would marry a girl like Number 7." Another said, "I am sure a nice girl like Number 3 would not marry such a fellow!" Another said, "It must be Number 1, because she has got as far away as possible from the brute!" It was suggested, again, that it must be Number 11, because "he seems to be looking towards her;" but a cynic retorted, "For that very reason, if he is really looking at her, I should say that she is not his wife!"

I now leave the question in the hands of my readers. Which is really Number 10's wife?

The illustration is of necessity considerably reduced from the large scale on which it originally appeared in _The Weekly Dispatch_ (24th May 1903), but it is hoped that the details will be sufficiently clear to allow the reader to derive entertainment from its examination. In any case the solution given will enable him to follow the points with interest.

SOLUTIONS.

1.--A POST-OFFICE PERPLEXITY.

The young lady supplied 5 twopenny stamps, 30 penny stamps, and 8 twopence-halfpenny stamps, which delivery exactly fulfils the conditions and represents a cost of five shillings.

2.--YOUTHFUL PRECOCITY.

The price of the banana must have been one penny farthing. Thus, 960 bananas would cost £5, and 480 sixpences would buy 2,304 bananas.

3.--AT A CATTLE MARKET.

Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must have taken 21. There were thus 39 animals altogether.

4.--THE BEANFEAST PUZZLE.

The cobblers spent 35s., the tailors spent also 35s., the hatters spent 42s., and the glovers spent 21s. Thus, they spent altogether £6,13s., while it will be found that the five cobblers spent as much as four tailors, twelve tailors as much as nine hatters, and six hatters as much as eight glovers.

5.--A QUEER COINCIDENCE.

Puzzles of this class are generally solved in the old books by the tedious process of "working backwards." But a simple general solution is as follows: If there are n players, the amount held by every player at the end will be m(2^n), the last winner must have held m(n + 1) at the start, the next m(2n + 1), the next m(4n + 1), the next m(8n + 1), and so on to the first player, who must have held m(2^{n - 1}n + 1).

Thus, in this case, n = 7, and the amount held by every player at the end was 2^7 farthings. Therefore m = 1, and G started with 8 farthings, F with 15, E with 29, D with 57, C with 113, B with 225, and A with 449 farthings.

6.--A CHARITABLE BEQUEST.

There are seven different ways in which the money may be distributed: 5 women and 19 men, 10 women and 16 men, 15 women and 13 men, 20 women and 10 men, 25 women and 7 men, 30 women and 4 men, and 35 women and 1 man. But the last case must not be counted, because the condition was that there should be "men," and a single man is not men. Therefore the answer is six years.

7.--THE WIDOW'S LEGACY.

The widow's share of the legacy must be £205, 2s. 6d. and 10/13 of a penny.

8.--INDISCRIMINATE CHARITY

The gentleman must have had 3s. 6d. in his pocket when he set out for home.

9.--THE TWO AEROPLANES.

The man must have paid £500 and £750 for the two machines, making together £1,250; but as he sold them for only £1,200, he lost £50 by the transaction.

10.--BUYING PRESENTS.

Jorkins had originally £19, 18s. in his pocket, and spent £9, 19s.

11.--THE CYCLISTS' FEAST.

There were ten cyclists at the feast. They should have paid 8s. each; but, owing to the departure of two persons, the remaining eight would pay 10s. each.

12.--A QUEER THING IN MONEY.

The answer is as follows: £44,444, 4s. 4d. = 28, and, reduced to pence, 10,666,612=28.

It is a curious little coincidence that in the answer 10,666,612 the four central figures indicate the only other answer, £66, 6s. 6d.

13.--A NEW MONEY PUZZLE.

The smallest sum of money, in pounds, shillings, pence, and farthings, containing all the nine digits once, and once only, is £2,567, 18s. 9¾d.

14.--SQUARE MONEY.

The answer is 1½d. and 3d. Added together they make 4½d., and 1½d. multiplied by 3 is also 4½d.

15.--POCKET MONEY.

The largest possible sum is 15s. 9d., composed of a crown and a half-crown (or three half-crowns), four florins, and a threepenny piece.

16.--THE MILLIONAIRE'S PERPLEXITY.

The answer to this quite easy puzzle may, of course, be readily obtained by trial, deducting the largest power of 7 that is contained in one million dollars, then the next largest power from the remainder, and so on. But the little problem is intended to illustrate a simple direct method. The answer is given at once by converting 1,000,000 to the septenary scale, and it is on this subject of scales of notation that I propose to write a few words for the benefit of those who have never sufficiently considered the matter.

Our manner of figuring is a sort of perfected arithmetical shorthand, a system devised to enable us to manipulate numbers as rapidly and correctly as possible by means of symbols. If we write the number 2,341 to represent two thousand three hundred and forty-one dollars, we wish to imply 1 dollar, added to four times 10 dollars, added to three times 100 dollars, added to two times 1,000 dollars. From the number in the units place on the right, every figure to the left is understood to represent a multiple of the particular power of 10 that its position indicates, while a cipher (0) must be inserted where necessary in order to prevent confusion, for if instead of 207 we wrote 27 it would be obviously misleading. We thus only require ten figures, because directly a number exceeds 9 we put a second figure to the left, directly it exceeds 99 we put a third figure to the left, and so on. It will be seen that this is a purely arbitrary method. It is working in the denary (or ten) scale of notation, a system undoubtedly derived from the fact that our forefathers who devised it had ten fingers upon which they were accustomed to count, like our children of to-day. It is unnecessary for us ordinarily to state that we are using the denary scale, because this is always understood in the common affairs of life.

But if a man said that he had 6,553 dollars in the septenary (or seven) scale of notation, you will find that this is precisely the same amount as 2,341 in our ordinary denary scale. Instead of using powers of ten, he uses powers of 7, so that he never needs any figure higher than 6, and 6,553 really stands for 3, added to five times 7, added to five times 49, added to six times 343 (in the ordinary notation), or 2,341. To reverse the operation, and convert 2,341 from the denary to the septenary scale, we divide it by 7, and get 334 and remainder 3; divide 334 by 7, and get 47 and remainder 5; and so keep on dividing by 7 as long as there is anything to divide. The remainders, read backwards, 6, 5, 5, 3, give us the answer, 6,553.

Now, as I have said, our puzzle may be solved at once by merely converting 1,000,000 dollars to the septenary scale. Keep on dividing this number by 7 until there is nothing more left to divide, and the remainders will be found to be 11333311 which is 1,000,000 expressed in the septenary scale. Therefore, 1 gift of 1 dollar, 1 gift of 7 dollars, 3 gifts of 49 dollars, 3 gifts of 343 dollars, 3 gifts of 2,401 dollars, 3 gifts of 16,807 dollars, 1 gift of 117,649 dollars, and one substantial gift of 823,543 dollars, satisfactorily solves our problem. And it is the only possible solution. It is thus seen that no "trials" are necessary; by converting to the septenary scale of notation we go direct to the answer.

17.--THE PUZZLING MONEY BOXES.

The correct answer to this puzzle is as follows: John put into his money-box two double florins (8s.), William a half-sovereign and a florin (12s.), Charles a crown (5s.), and Thomas a sovereign (20s.). There are six coins in all, of a total value of 45s. If John had 2s. more, William 2s. less, Charles twice as much, and Thomas half as much as they really possessed, they would each have had exactly 10s.

18.--THE MARKET WOMEN.

The price received was in every case 105 farthings. Therefore the greatest number of women is eight, as the goods could only be sold at the following rates: 105 lbs. at 1 farthing, 35 at 3, 21 at 5, 15 at 7, 7 at 15, 5 at 21, 3 at 35, and 1 lb. at 105 farthings.

19.--THE NEW YEAR'S EVE SUPPERS.

The company present on the occasion must have consisted of seven pairs, ten single men, and one single lady. Thus, there were twenty-five persons in all, and at the prices stated they would pay exactly £5 together.

20.--BEEF AND SAUSAGES.

The lady bought 48 lbs. of beef at 2s., and the same quantity of sausages at 1s. 6d., thus spending £8, 8s. Had she bought 42 lbs. of beef and 56 lbs. of sausages she would have spent £4, 4s. on each, and have obtained 98 lbs. instead of 96 lbs.--a gain in weight of 2 lbs.

21.--A DEAL IN APPLES.

I was first offered sixteen apples for my shilling, which would be at the rate of ninepence a dozen. The two extra apples gave me eighteen for a shilling, which is at the rate of eightpence a dozen, or one penny a dozen less than the first price asked.

22.--A DEAL IN EGGS.

The man must have bought ten eggs at fivepence, ten eggs at one penny, and eighty eggs at a halfpenny. He would then have one hundred eggs at a cost of eight shillings and fourpence, and the same number of eggs of two of the qualities.

23.--THE CHRISTMAS-BOXES.

The distribution took place "some years ago," when the fourpenny-piece was in circulation. Nineteen persons must each have received nineteen pence. There are five different ways in which this sum may have been paid in silver coins. We need only use two of these ways. Thus if fourteen men each received four four-penny-pieces and one threepenny-piece, and five men each received five threepenny-pieces and one fourpenny-piece, each man would receive nineteen pence, and there would be exactly one hundred coins of a total value of £1, 10s. 1d.

24.--A SHOPPING PERPLEXITY.

The first purchase amounted to 1s. 5¾d., the second to 1s. 11½d., and together they make 3s. 5¼d. Not one of these three amounts can be paid in fewer than six current coins of the realm.

25.--CHINESE MONEY.

As a ching-chang is worth twopence and four-fifteenths of a ching-chang, the remaining eleven-fifteenths of a ching-chang must be worth twopence. Therefore eleven ching-changs are worth exactly thirty pence, or half a crown. Now, the exchange must be made with seven round-holed coins and one square-holed coin. Thus it will be seen that 7 round-holed coins are worth seven-elevenths of 15 ching-changs, and 1 square-holed coin is worth one-eleventh of 16 ching-changs--that is, 77 rounds equal 105 ching-changs and 11 squares equal 16 ching-changs. Therefore 77 rounds added to 11 squares equal 121 ching-changs; or 7 rounds and 1 square equal 11 ching-changs, or its equivalent, half a crown. This is more simple in practice than it looks here.

26.--THE JUNIOR CLERKS' PUZZLE.

Although Snoggs's _reason_ for wishing to take his rise at £2, 10s. half-yearly did not concern our puzzle, the _fact_ that he was duping his employer into paying him more than was intended did concern it. Many readers will be surprised to find that, although Moggs only received £350 in five years, the artful Snoggs actually obtained £362, 10s. in the same time. The rest is simplicity itself. It is evident that if Moggs saved £87, 10s. and Snoggs £181, 5s., the latter would be saving twice as great a proportion of his salary as the former (namely, one-half as against one-quarter), and the two sums added together make £268, 15s.

27.--GIVING CHANGE.

The way to help the American tradesman out of his dilemma is this. Describing the coins by the number of cents that they represent, the tradesman puts on the counter 50 and 25; the buyer puts down 100, 3, and 2; the stranger adds his 10, 10, 5, 2, and 1. Now, considering that the cost of the purchase amounted to 34 cents, it is clear that out of this pooled money the tradesman has to receive 109, the buyer 71, and the stranger his 28 cents. Therefore it is obvious at a glance that the 100-piece must go to the tradesman, and it then follows that the 50-piece must go to the buyer, and then the 25-piece can only go to the stranger. Another glance will now make it clear that the two 10-cent pieces must go to the buyer, because the tradesman now only wants 9 and the stranger 3. Then it becomes obvious that the buyer must take the 1 cent, that the stranger must take the 3 cents, and the tradesman the 5, 2, and 2. To sum up, the tradesman takes 100, 5, 2, and 2; the buyer, 50, 10, 10, and 1; the stranger, 25 and 3. It will be seen that not one of the three persons retains any one of his own coins.

28.--DEFECTIVE OBSERVATION.

Of course the date on a penny is on the same side as Britannia--the "tail" side. Six pennies may be laid around another penny, all flat on the table, so that every one of them touches the central one. The number of threepenny-pieces that may be laid on the surface of a half-crown, so that no piece lies on another or overlaps the edge of the half-crown, is one. A second threepenny-piece will overlap the edge of the larger coin. Few people guess fewer than three, and many persons give an absurdly high number.

29.--THE BROKEN COINS.

If the three broken coins when perfect were worth 253 pence, and are now in their broken condition worth 240 pence, it should be obvious that 13/253 of the original value has been lost. And as the same fraction of each coin has been broken away, each coin has lost 13/253 of its original bulk.

30.--TWO QUESTIONS IN PROBABILITIES.

In tossing with the five pennies all at the same time, it is obvious that there are 32 different ways in which the coins may fall, because the first coin may fall in either of two ways, then the second coin may also fall in either of two ways, and so on. Therefore five 2's multiplied together make 32. Now, how are these 32 ways made up? Here they are:--

(a) 5 heads 1 way
(b) 5 tails 1 way
(c) 4 heads and 1 tail 5 ways
(d) 4 tails and 1 head 5 ways
(e) 3 heads and 2 tails 10 ways
(f) 3 tails and 2 heads 10 ways

Now, it will be seen that the only favourable cases are a, b, c, and d--12 cases. The remaining 20 cases are unfavourable, because they do not give at least four heads or four tails. Therefore the chances are only 12 to 20 in your favour, or (which is the same thing) 3 to 5. Put another way, you have only 3 chances out of 8.

The amount that should be paid for a draw from the bag that contains three sovereigns and one shilling is 15s. 3d. Many persons will say that, as one's chances of drawing a sovereign were 3 out of 4, one should pay three-fourths of a pound, or 15s., overlooking the fact that one must draw at least a shilling--there being no blanks.

31.--DOMESTIC ECONOMY.

Without the hint that I gave, my readers would probably have been unanimous in deciding that Mr. Perkins's income must have been £1,710. But this is quite wrong. Mrs. Perkins says, "We have spent a third of his yearly income in rent," etc., etc.--that is, in two years they have spent an amount in rent, etc., equal to one-third of his yearly income. Note that she does _not_ say that they have spent _each year_ this sum, whatever it is, but that _during the two years_ that amount has been spent. The only possible answer, according to the exact reading of her words, is, therefore, that his income was £180 per annum. Thus the amount spent in two years, during which his income has amounted to £360, will be £60 in rent, etc., £90 in domestic expenses, £20 in other ways, leaving the balance of £190 in the bank as stated.

32.--THE EXCURSION TICKET PUZZLE.

Nineteen shillings and ninepence may be paid in 458,908,622 different ways.

I do not propose to give my method of solution. Any such explanation would occupy an amount of space out of proportion to its interest or value. If I could give within reasonable limits a general solution for all money payments, I would strain a point to find room; but such a solution would be extremely complex and cumbersome, and I do not consider it worth the labour of working out.

Just to give an idea of what such a solution would involve, I will merely say that I find that, dealing only with those sums of money that are multiples of threepence, if we only use bronze coins any sum can be paid in (n + 1)² ways where n always represents the number of pence. If threepenny-pieces are admitted, there are

2n³ + 15n² + 33n
--------------------- + 1 ways.
18

If sixpences are also used there are

n^{4} + 22n³ + 159n² + 414n + 216
---------------------------------
216

ways, when the sum is a multiple of sixpence, and the constant, 216, changes to 324 when the money is not such a multiple. And so the formulas increase in complexity in an accelerating ratio as we go on to the other coins.

I will, however, add an interesting little table of the possible ways of changing our current coins which I believe has never been given in a book before. Change may be given for a

Farthing in 0 way.
Halfpenny in 1 way.
Penny in 3 ways.
Threepenny-piece in 16 ways.
Sixpence in 66 ways.
Shilling in 402 ways.
Florin in 3,818 ways.
Half-crown in 8,709 ways.
Double florin in 60,239 ways.
Crown in 166,651 ways.
Half-sovereign in 6,261,622 ways.
Sovereign in 500,291,833 ways.

It is a little surprising to find that a sovereign may be changed in over five hundred million different ways. But I have no doubt as to the correctness of my figures.

33.--A PUZZLE IN REVERSALS.

(i) £13. (2) £23, 19s. 11d. The words "the number of pounds exceeds that of the pence" exclude such sums of money as £2, 16s. 2d. and all sums under £1.

34.--THE GROCER AND DRAPER.

The grocer was delayed half a minute and the draper eight minutes and a half (seventeen times as long as the grocer), making together nine minutes. Now, the grocer took twenty-four minutes to weigh out the sugar, and, with the half-minute delay, spent 24 min. 30 sec. over the task; but the draper had only to make _forty-seven_ cuts to divide the roll of cloth, containing forty-eight yards, into yard pieces! This took him 15 min. 40 sec., and when we add the eight minutes and a half delay we get 24 min. 10 sec., from which it is clear that the draper won the race by twenty seconds. The majority of solvers make forty-eight cuts to divide the roll into forty-eight pieces!

35.--JUDKINS'S CATTLE.

As there were five droves with an equal number of animals in each drove, the number must be divisible by 5; and as every one of the eight dealers bought the same number of animals, the number must be divisible by 8. Therefore the number must be a multiple of 40. The highest possible multiple of 40 that will work will be found to be 120, and this number could be made up in one of two ways--1 ox, 23 pigs, and 96 sheep, or 3 oxen, 8 pigs, and 109 sheep. But the first is excluded by the statement that the animals consisted of "oxen, pigs, and sheep," because a single ox is not oxen. Therefore the second grouping is the correct answer.

36.--BUYING APPLES.

As there were the same number of boys as girls, it is clear that the number of children must be even, and, apart from a careful and exact reading of the question, there would be three different answers. There might be two, six, or fourteen children. In the first of these cases there are ten different ways in which the apples could be bought. But we were told there was an equal number of "boys and girls," and one boy and one girl are not boys and girls, so this case has to be excluded. In the case of fourteen children, the only possible distribution is that each child receives one halfpenny apple. But we were told that each child was to receive an equal distribution of "apples," and one apple is not apples, so this case has also to be excluded. We are therefore driven back on our third case, which exactly fits in with all the conditions. Three boys and three girls each receive 1 halfpenny apple and 2 third-penny apples. The value of these 3 apples is one penny and one-sixth, which multiplied by six makes sevenpence. Consequently, the correct answer is that there were six children--three girls and three boys.

37.--BUYING CHESTNUTS.

In solving this little puzzle we are concerned with the exact interpretation of the words used by the buyer and seller. I will give the question again, this time adding a few words to make the matter more clear. The added words are printed in italics.

"A man went into a shop to buy chestnuts. He said he wanted a pennyworth, and was given five chestnuts. 'It is not enough; I ought to have a sixth _of a chestnut more_,' he remarked. 'But if I give you one chestnut more,' the shopman replied, 'you will have _five-sixths_ too many.' Now, strange to say, they were both right. How many chestnuts should the buyer receive for half a crown?"

The answer is that the price was 155 chestnuts for half a crown. Divide this number by 30, and we find that the buyer was entitled to 5+1/6 chestnuts in exchange for his penny. He was, therefore, right when he said, after receiving five only, that he still wanted a sixth. And the salesman was also correct in saying that if he gave one chestnut more (that is, six chestnuts in all) he would be giving five-sixths of a chestnut in excess.

38.--THE BICYCLE THIEF.

People give all sorts of absurd answers to this question, and yet it is perfectly simple if one just considers that the salesman cannot possibly have lost more than the cyclist actually stole. The latter rode away with a bicycle which cost the salesman eleven pounds, and the ten pounds "change;" he thus made off with twenty-one pounds, in exchange for a worthless bit of paper. This is the exact amount of the salesman's loss, and the other operations of changing the cheque and borrowing from a friend do not affect the question in the slightest. The loss of prospective profit on the sale of the bicycle is, of course, not direct loss of money out of pocket.

39.--THE COSTERMONGER'S PUZZLE.

Bill must have paid 8s. per hundred for his oranges--that is, 125 for 10s. At 8s. 4d. per hundred, he would only have received 120 oranges for 10s. This exactly agrees with Bill's statement.

40.--MAMMA'S AGE.

The age of Mamma must have been 29 years 2 months; that of Papa, 35 years; and that of the child, Tommy, 5 years 10 months. Added together, these make seventy years. The father is six times the age of the son, and, after 23 years 4 months have elapsed, their united ages will amount to 140 years, and Tommy will be just half the age of his father.

41.--THEIR AGES.

The gentleman's age must have been 54 years and that of his wife 45 years.

42.--THE FAMILY AGES.

The ages were as follows: Billie, 3½ years; Gertrude, 1¾ year; Henrietta, 5¼ years; Charlie, 10½; years; and Janet, 21 years.

43.--MRS. TIMPKINS'S AGE.

The age of the younger at marriage is always the same as the number of years that expire before the elder becomes twice her age, if he was three times as old at marriage. In our case it was eighteen years afterwards; therefore Mrs. Timpkins was eighteen years of age on the wedding-day, and her husband fifty-four.

44.--A CENSUS PUZZLE.

Miss Ada Jorkins must have been twenty-four and her little brother Johnnie three years of age, with thirteen brothers and sisters between. There was a trap for the solver in the words "seven times older than little Johnnie." Of course, "seven times older" is equal to eight times as old. It is surprising how many people hastily assume that it is the same as "seven times as old." Some of the best writers have committed this blunder. Probably many of my readers thought that the ages 24½ and 3½ were correct.

45.--MOTHER AND DAUGHTER.

In four and a half years, when the daughter will be sixteen years and a half and the mother forty-nine and a half years of age.

46.--MARY AND MARMADUKE.

Marmaduke's age must have been twenty-nine years and two-fifths, and Mary's nineteen years and three-fifths. When Marmaduke was aged nineteen and three-fifths, Mary was only nine and four-fifths; so Marmaduke was at that time twice her age.

47.--ROVER'S AGE.

Rover's present age is ten years and Mildred's thirty years. Five years ago their respective ages were five and twenty-five. Remember that we said "four times older than the dog," which is the same as "five times as old." (See answer to No. 44.)

48.--CONCERNING TOMMY'S AGE.

Tommy Smart's age must have been nine years and three-fifths. Ann's age was sixteen and four-fifths, the mother's thirty-eight and two-fifths, and the father's fifty and two-fifths.

49.--NEXT-DOOR NEIGHBOURS.

Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 13; Mr. Simkin 42; Mrs. Simkin 40; Sophy 10; and Sammy 8.

50.--THE BAG OF NUTS.

It will be found that when Herbert takes twelve, Robert and Christopher will take nine and fourteen respectively, and that they will have together taken thirty-five nuts. As 35 is contained in 770 twenty-two times, we have merely to multiply 12, 9, and 14 by 22 to discover that Herbert's share was 264, Robert's 198, and Christopher's 308. Then, as the total of their ages is 17½ years or half the sum of 12, 9, and 14, their respective ages must be 6, 4½, and 7 years.

51.--HOW OLD WAS MARY?

The age of Mary to that of Ann must be as 5 to 3. And as the sum of their ages was 44, Mary was 27½ and Ann 16½. One is exactly 11 years older than the other. I will now insert in brackets in the original statement the various ages specified: "Mary is (27½) twice as old as Ann was (13¾) when Mary was half as old (24¾) as Ann will be (49½) when Ann is three times as old (49½) as Mary was (16½) when Mary was (16½) three times as old as Ann (5½)." Now, check this backwards. When Mary was three times as old as Ann, Mary was 16½ and Ann 5½ (11 years younger). Then we get 49½ for the age Ann will be when she is three times as old as Mary was then. When Mary was half this she was 24¾. And at that time Ann must have been 13¾ (11 years younger). Therefore Mary is now twice as old--27½, and Ann 11 years younger--16½.

52.--QUEER RELATIONSHIPS.

If a man marries a woman, who dies, and he then marries his deceased wife's sister and himself dies, it may be correctly said that he had (previously) married the sister of his widow.

The youth was not the nephew of Jane Brown, because he happened to be her son. Her surname was the same as that of her brother, because she had married a man of the same name as herself.

53.--HEARD ON THE TUBE RAILWAY.

The gentleman was the second lady's uncle.

54.--A FAMILY PARTY.

The party consisted of two little girls and a boy, their father and mother, and their father's father and mother.

55.--A MIXED PEDIGREE.

]

The letter m stands for "married." It will be seen that John Snoggs can say to Joseph Bloggs, "You are my _father's brother-in-law_, because my father married your sister Kate; you are my _brother's father-in-law_, because my brother Alfred married your daughter Mary; and you are my _father-in-law's brother_, because my wife Jane was your brother Henry's daughter."

56.--WILSON'S POSER.

If there are two men, each of whom marries the mother of the other, and there is a son of each marriage, then each of such sons will be at the same time uncle and nephew of the other. There are other ways in which the relationship may be brought about, but this is the simplest.

57.--WHAT WAS THE TIME?

The time must have been 9.36 p.m. A quarter of the time since noon is 2 hr. 24 min., and a half of the time till noon next day is 7 hr. 12 min. These added together make 9 hr. 36 min.

58.--A TIME PUZZLE.

Twenty-six minutes.

59.--A PUZZLING WATCH.

If the 65 minutes be counted on the face of the same watch, then the problem would be impossible: for the hands must coincide every 65+5/11 minutes as shown by its face, and it matters not whether it runs fast or slow; but if it is measured by true time, it gains 5/11 of a minute in 65 minutes, or 60/143 of a minute per hour.

60.--THE WAPSHAW'S WHARF MYSTERY.

There are eleven different times in twelve hours when the hour and minute hands of a clock are exactly one above the other. If we divide 12 hours by 11 we get 1 hr. 5 min. 27+3/11 sec., and this is the time after twelve o'clock when they are first together, and also the time that elapses between one occasion of the hands being together and the next. They are together for the second time at 2 hr. 10 min. 54+6/11 sec. (twice the above time); next at 3 hr. 16 min. 21+9/11 sec.; next at 4 hr. 21 min. 49+1/11 sec. This last is the only occasion on which the two hands are together with the second hand "just past the forty-ninth second." This, then, is the time at which the watch must have stopped. Guy Boothby, in the opening sentence of his _Across the World for a Wife_, says, "It was a cold, dreary winter's afternoon, and by the time the hands of the clock on my mantelpiece joined forces and stood at twenty minutes past four, my chambers were well-nigh as dark as midnight." It is evident that the author here made a slip, for, as we have seen above, he is 1 min. 49+1/11 sec. out in his reckoning.

61.--CHANGING PLACES.

There are thirty-six pairs of times when the hands exactly change places between three p.m. and midnight. The number of pairs of times from any hour (n) to midnight is the sum of 12 - (n + 1) natural numbers. In the case of the puzzle n = 3; therefore 12 - (3 + 1) = 8 and 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, the required answer.

The first pair of times is 3 hr. 21+57/143 min. and 4 hr. 16+112/143 min., and the last pair is 10 hr. 59+83/143 min. and 11 hr. 54+138/143 min. I will not give all the remainder of the thirty-six pairs of times, but supply a formula by which any of the sixty-six pairs that occur from midday to midnight may be at once found:--

720b + 60a 720a + 60b min.
a hr ---------- min. and b hr. ---------------
143 143

For the letter a may be substituted any hour from 0, 1, 2, 3 up to 10 (where nought stands for 12 o'clock midday); and b may represent any hour, later than a, up to 11.

By the aid of this formula there is no difficulty in discovering the answer to the second question: a = 8 and b = 11 will give the pair 8 hr. 58+106/143 min. and 11 hr. 44+128/143 min., the latter being the time when the minute hand is nearest of all to the point IX--in fact, it is only 15/143 of a minute distant.

Readers may find it instructive to make a table of all the sixty-six pairs of times when the hands of a clock change places. An easy way is as follows: Make a column for the first times and a second column for the second times of the pairs. By making a = 0 and b = 1 in the above expressions we find the first case, and enter hr. 5+5/143 min. at the head of the first column, and 1 hr. 0+60/143 min. at the head of the second column. Now, by successively adding 5+5/143 min. in the first, and 1 hr. 0+60/143 min. in the second column, we get all the _eleven_ pairs in which the first time is a certain number of minutes after nought, or mid-day. Then there is a "jump" in the times, but you can find the next pair by making a = 1 and b = 2, and then by successively adding these two times as before you will get all the _ten_ pairs after 1 o'clock. Then there is another "jump," and you will be able to get by addition all the _nine_ pairs after 2 o'clock. And so on to the end. I will leave readers to investigate for themselves the nature and cause of the "jumps." In this way we get under the successive hours, 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agrees with the formula in the first paragraph of this article.

Some time ago the principal of a Civil Service Training College, who conducts a "Civil Service Column" in one of the periodicals, had the query addressed to him, "How soon after XII o'clock will a clock with both hands of the same length be ambiguous?" His first answer was, "Some time past one o'clock," but he varied the answer from issue to issue. At length some of his readers convinced him that the answer is, "At 5+5/143 min. past XII;" and this he finally gave as correct, together with the reason for it that at that time _the time indicated is the same whichever hand you may assume as hour hand!_

62.--THE CLUB CLOCK.

The positions of the hands shown in the illustration could only indicate that the clock stopped at 44 min. 51+1143/1427 sec. after eleven o'clock. The second hand would next be "exactly midway between the other two hands" at 45 min. 52+496/1427 sec. after eleven o'clock. If we had been dealing with the points on the circle to which the three hands are directed, the answer would be 45 min. 22+106/1427 sec. after eleven; but the question applied to the hands, and the second hand would not be between the others at that time, but outside them.

63.--THE STOP-WATCH.

The time indicated on the watch was 5+5/11 min. past 9, when the second hand would be at 27+3/11 sec. The next time the hands would be similar distances apart would be 54+6/11 min. past 2, when the second hand would be at 32+8/11 sec. But you need only hold the watch (or our previous illustration of it) in front of a mirror, when you will see the second time reflected in it! Of course, when reflected, you will read XI as I, X as II, and so on.

64.--THE THREE CLOCKS.

As a mere arithmetical problem this question presents no difficulty. In order that the hands shall all point to twelve o'clock at the same time, it is necessary that B shall gain at least twelve hours and that C shall lose twelve hours. As B gains a minute in a day of twenty-four hours, and C loses a minute in precisely the same time, it is evident that one will have gained 720 minutes (just twelve hours) in 720 days, and the other will have lost 720 minutes in 720 days. Clock A keeping perfect time, all three clocks must indicate twelve o'clock simultaneously at noon on the 720th day from April 1, 1898. What day of the month will that be?

I published this little puzzle in 1898 to see how many people were aware of the fact that 1900 would not be a leap year. It was surprising how many were then ignorant on the point. Every year that can be divided by four without a remainder is bissextile or leap year, with the exception that one leap year is cut off in the century. 1800 was not a leap year, nor was 1900. On the other hand, however, to make the calendar more nearly agree with the sun's course, every fourth hundred year is still considered bissextile. Consequently, 2000, 2400, 2800, 3200, etc., will all be leap years. May my readers live to see them. We therefore find that 720 days from noon of April 1, 1898, brings us to noon of March 22, 1900.

65.--THE RAILWAY STATION CLOCK.

The time must have been 43+7/11 min. past two o'clock.

66.--THE VILLAGE SIMPLETON.

The day of the week on which the conversation took place was Sunday. For when the day after to-morrow (Tuesday) is "yesterday," "to-day" will be Wednesday; and when the day before yesterday (Friday) was "to-morrow," "to-day" was Thursday. There are two days between Thursday and Sunday, and between Sunday and Wednesday.

67.--AVERAGE SPEED.

The average speed is twelve miles an hour, not twelve and a half, as most people will hastily declare. Take any distance you like, say sixty miles. This would have taken six hours going and four hours returning. The double journey of 120 miles would thus take ten hours, and the average speed is clearly twelve miles an hour.

68.--THE TWO TRAINS.

One train was running just twice as fast as the other.

69.--THE THREE VILLAGES.

Calling the three villages by their initial letters, it is clear that the three roads form a triangle, A, B, C, with a perpendicular, measuring twelve miles, dropped from C to the base A, B. This divides our triangle into two right-angled triangles with a twelve-mile side in common. It is then found that the distance from A to C is 15 miles, from C to B 20 miles, and from A to B 25 (that is 9 and 16) miles. These figures are easily proved, for the square of 12 added to the square of 9 equals the square of 15, and the square of 12 added to the square of 16 equals the square of 20.

70.--DRAWING HER PENSION.

The distance must be 6¾ miles.

71.--SIR EDWYN DE TUDOR.

The distance must have been sixty miles. If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock--an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock--an hour too late. But if he went at 12 miles an hour, he would reach the castle of the wicked baron exactly at five o'clock--the time appointed.

72.--THE HYDROPLANE QUESTION.

The machine must have gone at the rate of seven-twenty-fourths of a mile per minute and the wind travelled five-twenty-fourths of a mile per minute. Thus, going, the wind would help, and the machine would do twelve-twenty-fourths, or half a mile a minute, and returning only two-twenty-fourths, or one-twelfth of a mile per minute, the wind being against it. The machine without any wind could therefore do the ten miles in thirty-four and two-sevenths minutes, since it could do seven miles in twenty-four minutes.

73.--DONKEY RIDING.

The complete mile was run in nine minutes. From the facts stated we cannot determine the time taken over the first and second quarter-miles separately, but together they, of course, took four and a half minutes. The last two quarters were run in two and a quarter minutes each.

74.--THE BASKET OF POTATOES.

Multiply together the number of potatoes, the number less one, and twice the number less one, then divide by 3. Thus 50, 49, and 99 multiplied together make 242,550, which, divided by 3, gives us 80,850 yards as the correct answer. The boy would thus have to travel 45 miles and fifteen-sixteenths--a nice little recreation after a day's work.

75.--THE PASSENGER'S FARE.

Mr. Tompkins should have paid fifteen shillings as his correct share of the motor-car fare. He only shared half the distance travelled for £3, and therefore should pay half of thirty shillings, or fifteen shillings.

76.--THE BARREL OF BEER.

Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which together sum to 29, whose digital root is 2. As the contents of the barrels sold must be a number divisible by 3, if one buyer purchased twice as much as the other, we must find a barrel with root 2, 5, or 8 to set on one side. There is only one barrel, that containing 20 gallons, that fulfils these conditions. So the man must have kept these 20 gallons of beer for his own use and sold one man 33 gallons (the 18-gallon and 15-gallon barrels) and sold the other man 66 gallons (the 16, 19, and 31 gallon barrels).

77.--DIGITS AND SQUARES.

The top row must be one of the four following numbers: 192, 219, 273, 327. The first was the example given.

78.--ODD AND EVEN DIGITS.

As we have to exclude complex and improper fractions and recurring decimals, the simplest solution is this: 79 + 5+1/3 and 84 + 2/6, both equal 84+1/3. Without any use of fractions it is obviously impossible.

79.--THE LOCKERS PUZZLE.

The smallest possible total is 356 = 107 + 249, and the largest sum possible is 981 = 235 + 746, or 657+324. The middle sum may be either 720 = 134 + 586, or 702 = 134 + 568, or 407 = 138 + 269. The total in this case must be made up of three of the figures 0, 2, 4, 7, but no sum other than the three given can possibly be obtained. We have therefore no choice in the case of the first locker, an alternative in the case of the third, and any one of three arrangements in the case of the middle locker. Here is one solution:--

107 134 235
249 586 746
--- --- ---
356 720 981

Of course, in each case figures in the first two lines may be exchanged vertically without altering the total, and as a result there are just 3,072 different ways in which the figures might be actually placed on the locker doors. I must content myself with showing one little principle involved in this puzzle. The sum of the digits in the total is always governed by the digit omitted. 9/9 - 7/10 - 5/11 -3/12 - 1/13 - 8/14 - 6/15 - 4/16 - 2/17 - 0/18. Whichever digit shown here in the upper line we omit, the sum of the digits in the total will be found beneath it. Thus in the case of locker A we omitted 8, and the figures in the total sum up to 14. If, therefore, we wanted to get 356, we may know at once to a certainty that it can only be obtained (if at all) by dropping the 8.

80.--THE THREE GROUPS.

There are nine solutions to this puzzle, as follows, and no more:--

12 × 483 = 5,796 27 × 198 = 5,346
42 × 138 = 5,796 39 × 186 = 7,254
18 × 297 = 5,346 48 × 159 = 7,632
28 × 157 = 4,396
4 × 1,738 = 6,952
4 × 1,963 = 7,852

The seventh answer is the one that is most likely to be overlooked by solvers of the puzzle.

81.--THE NINE COUNTERS.

In this case a certain amount of mere "trial" is unavoidable. But there are two kinds of "trials"--those that are purely haphazard, and those that are methodical. The true puzzle lover is never satisfied with mere haphazard trials. The reader will find that by just reversing the figures in 23 and 46 (making the multipliers 32 and 64) both products will be 5,056. This is an improvement, but it is not the correct answer. We can get as large a product as 5,568 if we multiply 174 by 32 and 96 by 58, but this solution is not to be found without the exercise of some judgment and patience.

82.--THE TEN COUNTERS.

As I pointed out, it is quite easy so to arrange the counters that they shall form a pair of simple multiplication sums, each of which will give the same product--in fact, this can be done by anybody in five minutes with a little patience. But it is quite another matter to find that pair which gives the largest product and that which gives the smallest product.

Now, in order to get the smallest product, it is necessary to select as multipliers the two smallest possible numbers. If, therefore, we place 1 and 2 as multipliers, all we have to do is to arrange the remaining eight counters in such a way that they shall form two numbers, one of which is just double the other; and in doing this we must, of course, try to make the smaller number as low as possible. Of course the lowest number we could get would be 3,045; but this will not work, neither will 3,405, 3,45O, etc., and it may be ascertained that 3,485 is the lowest possible. One of the required answers is 3,485 × 2 = 6,970, and 6,970 × 1 = 6,970.

The other part of the puzzle (finding the pair with the highest product) is, however, the real knotty point, for it is not at all easy to discover whether we should let the multiplier consist of one or of two figures, though it is clear that we must keep, so far as we can, the largest figures to the left in both multiplier and multiplicand. It will be seen that by the following arrangement so high a number as 58,560 may be obtained. Thus, 915 × 64 = 58,560, and 732 × 80 = 58,560.

83.--DIGITAL MULTIPLICATION.

The solution that gives the smallest possible sum of digits in the common product is 23 × 174 = 58 × 69 = 4,002, and the solution that gives the largest possible sum of digits, 9×654 =18×327=5,886. In the first case the digits sum to 6 and in the second case to 27. There is no way of obtaining the solution but by actual trial.

84.--THE PIERROT'S PUZZLE.

There are just six different solutions to this puzzle, as follows:--

8 multiplied by 473 equals 3784
9 " 351 " 3159
15 " 93 " 1395
21 " 87 " 1287
27 " 81 " 2187
35 " 41 " 1435

It will be seen that in every case the two multipliers contain exactly the same figures as the product.

85.--THE CAB NUMBERS.

The highest product is, I think, obtained by multiplying 8,745,231 by 96--namely, 839,542,176.

Dealing here with the problem generally, I have shown in the last puzzle that with three digits there are only two possible solutions, and with four digits only six different solutions.

These cases have all been given. With five digits there are just twenty-two solutions, as follows:--

3 × 4128 = 12384
3 × 4281 = 12843
3 × 7125 = 21375
3 × 7251 = 21753
2541 × 6 = 15246
651 × 24 = 15624
678 × 42 = 28476
246 × 51 = 12546
57 × 834 = 47538
75 × 231 = 17325
624 × 78 = 48672
435 × 87 = 37845
------
9 × 7461 = 67149
72 × 936 = 67392
------
2 × 8714 = 17428
2 × 8741 = 17482
65 × 281 = 18265
65 × 983 = 63985
------
4973 × 8 = 39784
6521 × 8 = 52168
14 × 926 = 12964
86 × 251 = 21586

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