2003 AMC 10B Problems/Problem 14
Contents
[hide]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 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.