2003 AMC 10B Problems/Problem 14
Problem
Given that where both and are positive integers, find the smallest possible value for .
Solution
is not a perfect power, so the smallest possible value of is .
Video Solution by WhyMath
~savannahsolver
See Also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AMC 10 Problems and Solutions |
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