THE ARTILLERYMEN'S DILEMMA.
(
Money Puzzles)
[Illustration: [Pyramid of cannon-balls]]
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"All cannon-balls are to be piled in square pyramids," was the order
issued to the regiment. This was done. Then came the further order, "All
pyramids are to contain a square number of balls." Whereupon the trouble
arose. "It can't be done," said the major. "Look at this pyramid, for
example; there are sixteen balls at the base, then nine, then four, then
one at the top, making thirty balls in all. But there must be six more
balls, or five fewer, to make a square number." "It _must_ be done,"
insisted the general. "All you have to do is to put the right number of
balls in your pyramids." "I've got it!" said a lieutenant, the
mathematical genius of the regiment. "Lay the balls out singly." "Bosh!"
exclaimed the general. "You can't _pile_ one ball into a pyramid!" Is it
really possible to obey both orders?
Answer:
We were required to find the smallest number of cannon balls that we
could lay on the ground to form a perfect square, and could pile into a
square pyramid. I will try to make the matter clear to the merest
novice.
1 2 3 4 5 6 7
1 3 6 10 15 21 28
1 4 10 20 35 56 84
1 5 14 30 55 91 140
Here in the first row we place in regular order the natural numbers.
Each number in the second row represents the sum of the numbers in the
row above, from the beginning to the number just over it. Thus 1, 2, 3,
4, added together, make 10. The third row is formed in exactly the same
way as the second. In the fourth row every number is formed by adding
together the number just above it and the preceding number. Thus 4 and
10 make 14, 20 and 35 make 55. Now, all the numbers in the second row
are triangular numbers, which means that these numbers of cannon balls
may be laid out on the ground so as to form equilateral triangles. The
numbers in the third row will all form our triangular pyramids, while
the numbers in the fourth row will all form square pyramids.
Thus the very process of forming the above numbers shows us that every
square pyramid is the sum of two triangular pyramids, one of which has
the same number of balls in the side at the base, and the other one ball
fewer. If we continue the above table to twenty-four places, we shall
reach the number 4,900 in the fourth row. As this number is the square
of 70, we can lay out the balls in a square, and can form a square
pyramid with them. This manner of writing out the series until we come
to a square number does not appeal to the mathematical mind, but it
serves to show how the answer to the particular puzzle may be easily
arrived at by anybody. As a matter of fact, I confess my failure to
discover any number other than 4,900 that fulfils the conditions, nor
have I found any rigid proof that this is the only answer. The problem
is a difficult one, and the second answer, if it exists (which I do not
believe), certainly runs into big figures.
For the benefit of more advanced mathematicians I will add that the
general expression for square pyramid numbers is (2n cubed + 3n squared + n)/6.
For this expression to be also a square number (the special case of 1
excepted) it is necessary that n = p squared - 1 = 6t squared, where 2p squared - 1 = q squared
(the "Pellian Equation"). In the case of our solution above, n = 24, p =
5, t = 2, q = 7.
Informational
