Read Ebook: The Canterbury Puzzles and Other Curious Problems by Dudeney Henry Ernest

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Ebook has 528 lines and 32272 words, and 11 pages

Every child knows how to play this game. You make a square of nine cells, and each of the two players, playing alternately, puts his mark in a cell with the object of getting three in a line. Whichever player first gets three in a line wins with the exulting cry:--

"Tit, tat, toe, My last go; Three jolly butcher boys All in a row."

It is a very ancient game. But if the two players have a perfect knowledge of it, one of three things must always happen. The first player should win; the first player should lose; or the game should always be drawn. Which is correct?

A child may propose a problem that a sage cannot answer. A farmer propounded the following question: "That ten-acre meadow of mine will feed twelve bullocks for sixteen weeks or eighteen bullocks for eight weeks. How many bullocks could I feed on a forty-acre field for six weeks, the grass growing regularly all the time?"

It will be seen that the sting lies in the tail. That steady growth of the grass is such a reasonable point to be considered, and yet to some readers it will cause considerable perplexity. The grass is, of course, assumed to be of equal length and uniform thickness in every case when the cattle begin to eat. The difficulty is not so great as it appears, if you properly attack the question.

Mr. Stanton Mowbray was a very wealthy man, a reputed millionaire, residing in that beautiful old mansion that has figured so much in English history, Grangemoor Park. He was a bachelor, spent most of the year at home, and lived quietly enough.

According to the evidence given, on the day preceding the night of the crime he received by the second post a single letter, the contents of which evidently gave him a shock. At ten o'clock at night he dismissed the servants, saying that he had some important business matters to look into, and would be sitting up late. He would require no attendance. It was supposed that after all had gone to bed he had admitted some person to the house, for one of the servants was positive that she had heard loud conversation at a very late hour.

Next morning, at a quarter to seven o'clock, one of the man-servants, on entering the room, found Mr. Mowbray lying on the floor, shot through the head, and quite dead. Now we come to the curious circumstance of the case. It was clear that after the bullet had passed out of the dead man's head it had struck the tall clock in the room, right in the very centre of the face, and actually welded together the three hands; for the clock had a seconds hand that revolved round the same dial as the hour and minute hands. But although the three hands had become welded together exactly as they stood in relation to each other at the moment of impact, yet they were free to revolve round the swivel in one piece, and had been stupidly spun round several times by the servants before Mr. Wiley Slyman was called upon the spot. But they would not move separately.

Now, inquiries by the police in the neighbourhood led to the arrest in London of a stranger who was identified by several persons as having been seen in the district the day before the murder, but it was ascertained beyond doubt at what time on the fateful morning he went away by train. If the crime took place after his departure, his innocence was established. For this and other reasons it was of the first importance to fix the exact time of the pistol shot, the sound of which nobody in the house had heard. The clock face in the illustration shows exactly how the hands were found. Mr. Slyman was asked to give the police the benefit of his sagacity and experience, and directly he was shown the clock he smiled and said:

"The matter is supremely simple. You will notice that the three hands appear to be at equal distances from one another. The hour hand, for example, is exactly twenty minutes removed from the minute hand--that is, the third of the circumference of the dial. You attach a lot of importance to the fact that the servants have been revolving the welded hands, but their act is of no consequence whatever; for although they were welded instantaneously, as they are free on the swivel, they would swing round of themselves into equilibrium. Give me a few moments, and I can tell you beyond any doubt the exact time that the pistol was fired."

Mr. Wiley Slyman took from his pocket a notebook, and began to figure it out. In a few minutes he handed the police inspector a slip of paper, on which he had written the precise moment of the crime. The stranger was proved to be an old enemy of Mr. Mowbray's, was convicted on other evidence that was discovered; but before he paid the penalty for his wicked act, he admitted that Mr. Slyman's statement of the time was perfectly correct.

Can you also give the exact time?

An economical carpenter had a block of wood measuring eight inches long by four inches wide by three and three-quarter inches deep. How many pieces, each measuring two and a half inches by one inch and a half by one inch and a quarter, could he cut out of it? It is all a question of how you cut them out. Most people would have more waste material left over than is necessary. How many pieces could you get out of the block?

Four merry tramps bought, borrowed, found, or in some other manner obtained possession of a box of biscuits, which they agreed to divide equally amongst themselves at breakfast next morning. In the night, while the others were fast asleep under the greenwood tree, one man approached the box, devoured exactly a quarter of the number of biscuits, except the odd one left over, which he threw as a bribe to their dog. Later in the night a second man awoke and hit on the same idea, taking a quarter of what remained and giving the odd biscuit to the dog. The third and fourth men did precisely the same in turn, taking a quarter of what they found and giving the odd biscuit to the dog. In the morning they divided what remained equally amongst them, and again gave the odd biscuit to the animal. Every man noticed the reduction in the contents of the box, but, believing himself to be alone responsible, made no comments. What is the smallest possible number of biscuits that there could have been in the box when they first acquired it?

SOLUTIONS

THE CANTERBURY PUZZLES

The 8 cheeses can be removed in 33 moves, 10 cheeses in 49 moves, and 21 cheeses in 321 moves. I will give my general method of solution in the cases of 3, 4, and 5 stools.

Write out the following table to any required length:--

Stools. Number of Cheeses.

Number of Moves.

The first row contains the natural numbers. The second row is found by adding the natural numbers together from the beginning. The numbers in the third row are obtained by adding together the numbers in the second row from the beginning. The fourth row contains the successive powers of 2, less 1. The next series is found by doubling in turn each number of that series and adding the number that stands above the place where you write the result. The last row is obtained in the same way. This table will at once give solutions for any number of cheeses with three stools, for triangular numbers with four stools, and for pyramidal numbers with five stools. In these cases there is always only one method of solution--that is, of piling the cheeses.

In the case of three stools, the first and fourth rows tell us that 4 cheeses may be removed in 15 moves, 5 in 31, 7 in 127. The second and fifth rows show that, with four stools, 10 may be removed in 49, and 21 in 321 moves. Also, with five stools, we find from the third and sixth rows that 20 cheeses require 111 moves, and 35 cheeses 351 moves. But we also learn from the table the necessary method of piling. Thus, with four stools and 10 cheeses, the previous column shows that we must make piles of 6 and 3, which will take 17 and 7 moves respectively--that is, we first pile the six smallest cheeses in 17 moves on one stool; then we pile the next 3 cheeses on another stool in 7 moves; then remove the largest cheese in 1 move; then replace the 3 in 7 moves; and finally replace the 6 in 17: making in all the necessary 49 moves. Similarly we are told that with five stools 35 cheeses must form piles of 20, 10, and 4, which will respectively take 111, 49, and 15 moves.

If the number of cheeses in the case of four stools is not triangular, and in the case of five stools pyramidal, then there will be more than one way of making the piles, and subsidiary tables will be required. This is the case with the Reve's 8 cheeses. But I will leave the reader to work out for himself the extension of the problem.

The diagram on page 165 will show how the Pardoner started from the large black town and visited all the other towns once, and once only, in fifteen straight pilgrimages.

The way to arrange the sacks of flour is as follows:--2, 78, 156, 39, 4. Here each pair when multiplied by its single neighbour makes the number in the middle, and only five of the sacks need be moved. There are just three other ways in which they might have been arranged , but they all require the moving of seven sacks.

The Knight declared that as many as 575 squares could be marked off on his shield, with a rose at every corner. How this result is achieved may be realized by reference to the accompanying diagram:--Join A, B, C, and D, and there are 66 squares of this size to be formed; the size A, E, F, G gives 48; A, H, I, J, 32; B, K, L, M, 19; B, N, O, P, 10; B, Q, R, S, 4; E, T, F, C, 57; I, U, V, P, 33; H, W, X, J, 15; K, Y, Z, M, 3; E, a, b, D, 82; H, d, M, D, 56; H, e, f, G, 42; K, g, f, C, 32; N, h, z, F, 24; K, h, m, b, 14; K, O, S, D, 16; K, n, p, G, 10; K, q, r, J, 6; Q, t, p, C, 4; Q, u, r, i, 2. The total number is thus 575. These groups have been treated as if each of them represented a different sized square. This is correct, with the one exception that the squares of the form B, N, O, P are exactly the same size as those of the form K, h, m, b.

The puzzle propounded by the jovial host of the "Tabard" Inn of Southwark had proved more popular than any other of the whole collection. "I see, my merry masters," he cried, "that I have sorely twisted thy brains by my little piece of craft. Yet it is but a simple matter for me to put a true pint of fine old ale in each of these two measures, albeit one is of five pints and the other of three pints, without using any other measure whatever."

The host of the "Tabard" Inn thereupon proceeded to explain to the pilgrims how this apparently impossible task could be done. He first filled the 5-pint and 3-pint measures, and then, turning the tap, allowed the barrel to run to waste--a proceeding against which the company protested; but the wily man showed that he was aware that the cask did not contain much more than eight pints of ale. The contents, however, do not affect the solution of the puzzle. He then closed the tap and emptied the 3-pint into the barrel; filled the 3-pint from the 5-pint; emptied the 3-pint into the barrel; transferred the two pints from the 5-pint to the 3-pint; filled the 5-pint from the barrel, leaving one pint now in the barrel; filled 3-pint from 5-pint; allowed the company to drink the contents of the 3-pint; filled the 3-pint from the 5-pint, leaving one pint now in the 5-pint; drank the contents of the 3-pint; and finally drew off one pint from the barrel into the 3-pint. He had thus obtained the required one pint of ale in each measure, to the great astonishment of the admiring crowd of pilgrims.

The illustration shows how the square is to be cut into four pieces, and how these pieces are to be put together again to make a magic square. It will be found that the four columns, four rows, and two long diagonals now add up to 34 in every case.

The piece of tapestry had to be cut along the lines into three pieces so as to fit together and form a perfect square, with the pattern properly matched. It was also stipulated in effect that one of the three pieces must be as small as possible. The illustration shows how to make the cuts and how to put the pieces together, while one of the pieces contains only twelve of the little squares.

The carpenter said that he made a box whose internal dimensions were exactly the same as the original block of wood--that is, 3 feet by 1 foot by 1 foot. He then placed the carved pillar in this box and filled up all the vacant space with a fine, dry sand, which he carefully shook down until he could get no more into the box. Then he removed the pillar, taking great care not to lose any of the sand, which, on being shaken down alone in the box, filled a space equal to one cubic foot. This was, therefore, the quantity of wood that had been cut away.

The illustration will show how three of the arrows were removed each to a neighbouring square on the signboard of the "Chequers" Inn, so that still no arrow was in line with another. The black dots indicate the squares on which the three arrows originally stood.

As there are eighteen cards bearing the letters "CANTERBURY PILGRIMS," write the numbers 1 to 18 in a circle, as shown in the diagram. Then write the first letter C against 1, and each successive letter against the second number that happens to be vacant. This has been done as far as the second R. If the reader completes the process by placing Y against 2, P against 6, I against 10, and so on, he will get the letters all placed in the following order:--CYASNPTREIRMBLUIRG, which is the required arrangement for the cards, C being at the top of the pack and G at the bottom.

This puzzle amounts to finding the smallest possible number that has exactly sixty-four divisors, counting 1 and the number itself as divisors. The least number is 7,560. The pilgrims might, therefore, have ridden in single file, two and two, three and three, four and four, and so on, in exactly sixty-four different ways, the last manner being in a single row of 7,560.

The Merchant was careful to say that they were going over a common, and not to mention its size, for it certainly would not be possible along an ordinary road!

The fewest possible moves for getting the prisoners into their dungeons in the required numerical order are twenty-six. The men move in the following order:--1, 2, 3, 1, 2, 6, 5, 3, 1, 2, 6, 5, 3, 1, 2, 4, 8, 7, 1, 2, 4, 8, 7, 4, 5, 6. As there are never more than one vacant dungeon to be moved into, there can be no ambiguity in the notation.

in the fewest possible moves.

See also solution to No. 94.

The illustration shows clearly how the Weaver cut his square of beautiful cloth into four pieces of exactly the same size and shape, so that each piece contained an embroidered lion and castle unmutilated in any way.

The number that the Sompnour confided to the Wife of Bath was twenty-nine, and she was told to begin her count at the Doctor of Physic, who will be seen in the illustration standing the second on her right. The first count of twenty-nine falls on the Shipman, who steps out of the ring. The second count falls on the Doctor, who next steps out. The remaining three counts fall respectively on the Cook, the Sompnour, and the Miller. The ladies would, therefore, have been left in possession had it not been for the unfortunate error of the good Wife. Any multiple of 2,520 added to 29 would also have served the same purpose, beginning the count at the Doctor.

The Monk might have placed dogs in the kennels in two thousand nine hundred and twenty-six different ways, so that there should be ten dogs on every side. The number of dogs might vary from twenty to forty, and as long as the Monk kept his animals within these limits the thing was always possible.

The Abbot of Chertsey was quite correct. The curiously-shaped cross may be cut into four pieces that will fit together and form a perfect square. How this is done is shown in the illustration.

Here we have indeed a knotty problem. Our text-books tell us that all spheres are similar, and that similar solids are as the cubes of corresponding lengths. Therefore, as the circumferences of the two phials were one foot and two feet respectively and the cubes of one and two added together make nine, what we have to find is two other numbers whose cubes added together make nine. These numbers clearly must be fractional. Now, this little question has really engaged the attention of learned men for two hundred and fifty years; but although Peter de Fermat showed in the seventeenth century how an answer may be found in two fractions with a denominator of no fewer than twenty-one figures, not only are all the published answers, by his method, that I have seen inaccurate, but nobody has ever published the much smaller result that I now print. The cubes of and added together make exactly nine, and therefore these fractions of a foot are the measurements of the circumferences of the two phials that the Doctor required to contain the same quantity of liquid as those produced. An eminent actuary and another correspondent have taken the trouble to cube out these numbers, and they both find my result quite correct.

If the phials were one foot and three feet in circumference respectively, then an answer would be that the cubes of and added together make exactly 28. See also No. 61, "The Silver Cubes."

Some years ago I published a solution for the case of

of which Legendre gave at some length a "proof" of impossibility; but I have since found that Lucas anticipated me in a communication to Sylvester.

The illustration shows how the sixteen trees might have been planted so as to form as many as fifteen straight rows with four trees in every row. This is in excess of what was for a long time believed to be the maximum number of rows possible; and though with our present knowledge I cannot rigorously demonstrate that fifteen rows cannot be beaten, I have a strong "pious opinion" that it is the highest number of rows obtainable.

The answer to this puzzle is shown in the illustration, where the numbers on the sixteen bottles all add up to 30 in the ten straight directions. The trick consists in the fact that, although the six bottles in which the flowers have been placed are not removed, yet the sixteen need not occupy exactly the same position on the table as before. The square is, in fact, formed one step further to the left.

The portrait may be drawn in a single line because it contains only two points at which an odd number of lines meet, but it is absolutely necessary to begin at one of these points and end at the other. One point is near the outer extremity of the King's left eye; the other is below it on the left cheek.

The five hundred silver pennies might have been placed in the four bags, in accordance with the stated conditions, in exactly 894,348 different ways. If there had been a thousand coins there would be 7,049,112 ways. It is a difficult problem in the partition of numbers. I have a single formula for the solution of any number of coins in the case of four bags, but it was extremely hard to construct, and the best method is to find the twelve separate formulas for the different congruences to the modulus 12.

A very little examination of the original drawing will have shown the reader that, as he will have at first read the conditions, the puzzle is quite impossible of solution. We have therefore to look for some loophole in the actual conditions as they were worded. If the Parson could get round the source of the river, he could then cross every bridge once and once only on his way to church, as shown in the annexed illustration. That this was not prohibited we shall soon find. Though the plan showed all the bridges in his parish, it only showed "part of" the parish itself. It is not stated that the river did not take its rise in the parish, and since it leads to the only possible solution, we must assume that it did. The answer would be, therefore, as shown. It should be noted that we are clearly prevented from considering the possibility of getting round the mouth of the river, because we are told it "joined the sea some hundred miles to the south," while no parish ever extended a hundred miles!

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