THE SCULPTOR'S PROBLEM.
(
Money Puzzles)
An ancient sculptor was commissioned to supply two statues, each on a
cubical pedestal. It is with these pedestals that we are concerned. They
were of unequal sizes, as will be seen in the illustration, and when the
time arrived for payment a dispute arose as to whether the agreement was
based on lineal or cubical measurement. But as soon as they came to
measure the two pedestals the matter was at once settled, because,
curiously enough, the number of lineal feet was exactly the same as the
number of cubical feet. The puzzle is to find the dimensions for two
pedestals having this peculiarity, in the smallest possible figures. You
see, if the two pedestals, for example, measure respectively 3 ft. and 1
ft. on every side, then the lineal measurement would be 4 ft. and the
cubical contents 28 ft., which are not the same, so these measurements
will not do.
Answer:
A little thought will make it clear that the answer must be fractional,
and that in one case the numerator will be greater and in the other case
less than the denominator. As a matter of fact, the height of the larger
cube must be 8/7 ft., and of the smaller 3/7 ft., if we are to have the
answer in the smallest possible figures. Here the lineal measurement is
11/7 ft.--that is, 1+4/7 ft. What are the cubic contents of the two
cubes? First 8/7 x 3/7 x 8/7 = 512/343, and secondly 3/7 x 3/7 x 3/7 =
27/343. Add these together and the result is 539/343, which reduces to
11/7 or 1+4/7 ft. We thus see that the answers in cubic feet and lineal
feet are precisely the same.
The germ of the idea is to be found in the works of Diophantus of
Alexandria, who wrote about the beginning of the fourth century. These
fractional numbers appear in triads, and are obtained from three
generators, a, b, c, where a is the largest and c the smallest.
Then ab + c squared = denominator, and a squared - c squared, b squared - c squared, and a squared - b squared will be
the three numerators. Thus, using the generators 3, 2, 1, we get 8/7,
3/7, 5/7 and we can pair the first and second, as in the above
solution, or the first and third for a second solution. The
denominator must always be a prime number of the form 6n + 1, or
composed of such primes. Thus you can have 13, 19, etc., as
denominators, but not 25, 55, 187, etc.
When the principle is understood there is no difficulty in writing down
the dimensions of as many sets of cubes as the most exacting collector
may require. If the reader would like one, for example, with plenty of
nines, perhaps the following would satisfy him: 99999999/99990001 and
19999/99990001.
