[11].—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].—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 = ap bq cr ..., 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—