A PROBLEM IN SQUARES.
(
Money Puzzles)
We possess three square boards. The surface of the first contains five
square feet more than the second, and the second contains five square
feet more than the third. Can you give exact measurements for the sides
of the boards? If you can solve this little puzzle, then try to find
three squares in arithmetical progression, with a common difference of 7
and also of 13.
Answer:
The sides of the three boards measure 31 in., 41 in., and 49 in. The
common difference of area is exactly five square feet. Three numbers
whose squares are in A.P., with a common difference of 7, are 113/120,
337/120, 463/120; and with a common difference of 13 are 80929/19380,
106921/19380, and 127729/19380. In the case of whole square numbers the
common difference will always be divisible by 24, so it is obvious that
our squares must be fractional. Readers should now try to solve the case
where the common difference is 23. It is rather a hard nut.
