2007 AMC 10A Problems/Problem 17
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Contents
[hide]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
Video Solution
~savannahsolver
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |
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