1985 AHSME Problems/Problem 12
Problem
Let ,
and
be distinct prime numbers, where
is not considered a prime. Which of the following is the smallest positive perfect cube having
as a divisor?
Solution
A number of the form will be a perfect cube precisely when
,
, and
are multiples of 3. (Clearly, since we are looking for the smallest possible perfect cube, we can assume that it has no other prime factors.) Furthermore, for it to be a multiple of
, we must have
,
, and
. The smallest multiple of
that is at least
is
, the smallest that is at least
is again
, and the smallest that is at least
is
. Hence the smallest possible perfect cube with
as a divisor is
.
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 13 | |
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