beginning the
count at the Doctor.

17.--_The Monk's Puzzle._

The Monk might have placed dogs in the kennels in two thousand nine
hundred and twenty-six different ways, so that there should be ten dogs
on every side. The number of dogs might vary from twenty to forty, and
as long as the Monk kept his animals within these limits the thing was
always possible.

The general solution to this puzzle is difficult. I find that for _n_
dogs on every side of the square, the number of different ways is (_n_^4
+ 10_n_^3 + 38_n_^2 + 62_n_ + 33) / 48, where _n_ is odd, and ((_n_^4 +
10_n_^3 + 38_n_^2 + 68_n_) / 48) + 1, where _n_ is even, if we count only
those arrangements that are fundamentally different. But if we count all
reversals and reflections as different, as the Monk himself did, then _n_
dogs (odd or even) may be placed in ((_n_^4 + 6_n_^3 + 14_n_^2 + 15_n_) /
6) + 1 ways. In order that there may be _n_ dogs on every side, the
number must not be less than 2_n_ nor greater than 4_n_, but it may be
any number within these limits.

An extension of the principle involved in this puzzle is given in No. 42,
"The Riddle of the Pilgrims." See also "The Eight Villas" and "A
Dormitory Puzzle" in _A. in M._

18.--_The Shipman's Puzzle._

There are just two hundred and sixty-four different ways in which the
ship _Magdalen_ might have made her ten annual voyages without ever going
over the same course twice in a year. Every year she must necessarily end
her tenth voyage at the island from which she first set out.

[Illustration]

19.--_The Puzzle of the Prioress._

The Abbot of Chertsey was quite correct. The curiously-shaped cross may
be cut into four pieces that will fit together and form a perfect
square. How this is done is shown in the illustration.

See also p. 31 in _A. in M._

20.--_The Puzzle of the Doctor of Physic._

Here we have indeed a knotty problem. Our text-books tell us that all
spheres are similar, and that similar solids are as the cubes of
corresponding lengths. Therefore, as the circumferences of the two phials
were one foot and two feet respectively and the cubes of one and two
added together make nine, what we have to find is two other numbers whose
cubes added together make nine. These numbers clearly must be fractional.
Now, this little question has really engaged the attention of learned men
for two hundred and fifty years; but although Peter de Fermat showed in
the