The Canterbury Puzzles, and Other Curious Problems

Can you also give the exact time?

113 (#pgepubid00250).—Cutting a Wood Block

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?

114 (#pgepubid00251).—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?

SOLUTIONS

THE CANTERBURY PUZZLES

1 (#pgepubid00010).—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:—

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 (#pgepubid00011).—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 (#pgepubid00012).—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.

4 (#pgepubid00013).—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 (#pgepubid00014).—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 (#pgepubid00015).—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 (#pgepubid00016).—Clerk of Oxenford's Puzzle

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.

8 (#pgepubid00017).—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 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.

9 (#pgepubid00018).—The Carpenter's Puzzle

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.

10 (#pgepubid00019).—The Puzzle of the Squire's Yeoman

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.

11 (#pgepubid00020).—The Nun's Puzzle

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.

12 (#pgepubid00021).—The Merchant's Puzzle

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!

To find how many different numbers will divide a given number, N, let N = a

b

c

…, where a, b, c … are prime numbers. Then the number of divisors will be (p + 1) (q + 1) (r + 1) …, which includes as divisors 1 and N itself. Thus in the case of my puzzle—

To find the smallest number that has a given number of divisors we must proceed by trial. But it is important sometimes to note whether or not the condition is that there shall be a given number of divisors and no more. For example, the smallest number that has seven divisors and no more is 64, while 24 has eight divisors, and might equally fulfil the conditions. The stipulation as to "no more" was not necessary in the case of my puzzle, for no smaller number has more than sixty-four divisors.

13 (#pgepubid00022).—The Man of Law's Puzzle

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.

The diagram may be simplified by my "buttons and string" method, fully explained in A. in M., p. 230. It then takes one of the simple forms of A or B, and the solution is much easier. In A we use counters; in B we can employ rooks on a corner of a chessboard. In both cases we have to get the order

in the fewest possible moves.

See also solution to No. 94 (#pgepubid00231).

14 (#pgepubid00023).—The Weaver's Puzzle

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.