e boxes were united they would form a square
number. What is the smallest number of doubloons that there could have
been in any one of the boxes?

132.--THE NINE TREASURE BOXES.

The following puzzle will illustrate the importance on occasions of
being able to fix the minimum and maximum limits of a required number.
This can very frequently be done. For example, it has not yet been
ascertained in how many different ways the knight's tour can be
performed on the chess board; but we know that it is fewer than the
number of combinations of 168 things taken 63 at a time and is greater
than 31,054,144--for the latter is the number of routes of a particular
type. Or, to take a more familiar case, if you ask a man how many coins
he has in his pocket, he may tell you that he has not the slightest
idea. But on further questioning you will get out of him some such
statement as the following: "Yes, I am positive that I have more than
three coins, and equally certain that there are not so many as
twenty-five." Now, the knowledge that a certain number lies between 2
and 12 in my puzzle will enable the solver to find the exact answer;
without that information there would be an infinite number of answers,
from which it would be impossible to select the correct one.

This is another puzzle received from my friend Don Manuel Rodriguez, the
cranky miser of New Castile. On New Year's Eve in 1879 he showed me nine
treasure boxes, and after informing me that every box contained a square
number of golden doubloons, and that the difference between the contents
of A and B was the same as between B and C, D and E, E and F, G and H,
or H and I, he requested me to tell him the number of coins in every one
of the boxes. At first I thought this was impossible, as there would be
an infinite number of different answers, but on consideration I found
that this was not the case. I discovered that while every box contained
coins, the contents of A, B, C increased in weight in alphabetical
order; so did D, E, F; and so did G, H, I; but D or E need not be
heavier than C, nor G or H heavier than F. It was also perfectly certain
that box A could not contain more than a dozen coins at the outside;
there might not be half that number, but I was positive that there were
not more than twelve. With this knowledge I was able to arrive at the
correct answer.

In short, we have to discover nine square numbers such that A, B, C; and
D, E, F; and G, H, I are thr