2007 AMC 10A Problems/Problem 17
Contents
Problem
Suppose that and
are positive integers such that
. What is the minimum possible value of
?
Solution
must be a perfect cube, so each power of a prime in the factorization for
must be divisible by
. Thus the minimum value of
is
, which makes
. The minimum possible value for the sum of
and
is
Solution 2
First, we need to prime factorize .
=
. We need
to be in the form
. Therefore, the smallest
is
.
= 45, and since
, our answer is
=
~Arcticturn
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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