ists in taking away a fifth of the contents of
the jug--that is, one-fifth of a pint of water mixed with one-fifth of a
quarter of a pint of wine. We thus leave behind in the jug four-fifths of
a quarter of a pint of wine--that is, one-fifth of a pint--while we
transfer from the jug to the bottle an equal quantity (one-fifth of a
pint) of water.

45.--_The Riddle of the Cellarer._

There were 100 pints of wine in the cask, and on thirty occasions John
the Cellarer had stolen a pint and replaced it with a pint of water.
After the first theft the wine left in the cask would be 99 pints; after
the second theft the wine in the cask would be 9801/100 pints (the square
of 99 divided by 100); after the third theft there would remain
970299/10000 (the cube of 99 divided by the square of 100); after the
fourth theft there would remain the fourth power of 99 divided by the
cube of 100; and after the thirtieth theft there would remain in the cask
the thirtieth power of 99 divided by the twenty-ninth power of 100. This
by the ordinary method of calculation gives us a number composed of 59
figures to be divided by a number composed of 58 figures! But by the use
of logarithms it may be quickly ascertained that the required quantity is
very nearly 73-97/100 pints of wine left in the cask. Consequently the
cellarer stole nearly 26.03 pints. The monks doubtless omitted the answer
for the reason that they had no tables of logarithms, and did not care to
face the task of making that long and tedious calculation in order to get
the quantity "to a nicety," as the wily cellarer had stipulated.

By a simplified process of calculation, I have ascertained that the exact
quantity of wine stolen would be

26.0299626611719577269984907683285057747323737647323555652999

pints. A man who would involve the monastery in a fraction of fifty-eight
decimals deserved severe punishment.

46.--_The Riddle of the Crusaders._

The correct answer is that there would have been 602,176 Crusaders, who
could form themselves into a square 776 by 776; and after the stranger
joined their ranks, they could form 113 squares of 5,329 men--that is, 73
by 73. Or 113 x 73^2 - 1 = 776^2. This is a particular case of the
so-called "Pellian Equation," respecting which see _A. in M._, p. 164.

47.--_The Riddle of St. Edmondsbury._

The reader is aware that there are prime numbers and composite whole
numbers. Now, 1,111,111 cannot be a prime number,