Difference between revisions of "2007 AMC 10A Problems/Problem 17"
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Revision as of 11:51, 4 July 2013
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
. These sum to
.
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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