The Canterbury Puzzles Part 19

114.--_The Tramps and the Biscuits._

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?

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SOLUTIONS

THE CANTERBURY PUZZLES

1.--_The Reve's Puzzle._

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.

3 1 2 3 4 5 6 7 Natural Numbers.

4 1 3 6 10 15 21 28 Triangular Numbers.

5 1 4 10 20 35 56 84 Triangular Pyramids.

Number of Moves.

3 1 3 7 15 31 63 127 4 1 5 17 49 129 321 769 5 1 7 31 111 351 1023 2815

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.

2.--_The Pardoner's Puzzle._

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.

See No. 320, "The Rook's Tour," in _A. in M._

3.--_The Miller's Puzzle._

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 (4, 39, 156, 78, 2; or 3, 58, 174, 29, 6; or 6, 29, 174, 58, 3), but they all require the moving of seven sacks.

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4.--_The Knight's Puzzle._

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.

5.--_The Wife of Bath's Riddles._

The good lady explained that a bung that is made fast in a barrel is like another bung that is falling out of a barrel because one of them is _in secure_ and the other is also _insecure_. The little relationship poser is readily understood when we are told that the parental command came from the father (who was also in the room) and not from the mother.

6.--_The Host's Puzzle._

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.

7.--_Clerk of Oxenford's Puzzle._

The ill.u.s.tration 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.

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8.--_The Tapiser's Puzzle._

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 ill.u.s.tration 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.

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