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 = _a_^p _b_^q _c_^r ..., 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-- 7,560 = 2^3 × 3^3 × 5 × 7 Powers = 3 3 1 1 Therefore 4 × 4 × 2 × 2 = 64 divisors. 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.--_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. [Illustration] 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 [Illustration: {1 2 3} {4 5 6} {7 8 } ] in the fewest possible moves. See also solution to No. 94. 14.--_The Weaver's Puzzle._