]

11.--_The Nun's Puzzle._

[Illustration]

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 x 3^3 x 5 x 7
Powers = 3 3 1 1
Therefore 4 x 4 x 2 x 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 b