2003 AMC 10B Problems/Problem 14

Problem

Given that $3^8\cdot5^2=a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$ .

$\textbf{(A) } 25 \qquad\textbf{(B) } 34 \qquad\textbf{(C) } 351 \qquad\textbf{(D) } 407 \qquad\textbf{(E) } 900$

$3^8\cdot5^2$

$=3^2\cdot3^2\cdot3^2\cdot3^2\cdot5^2$

$=(3\cdot3\cdot3\cdot3\cdot5)^2$

$=405^2$

$405$ is not a perfect power, so the smallest possible value of $a+b$ is $405+2=\boxed{407\textbf{ (D)}}$ .

2003 AMC 10B (Problems • Answer Key • Resources)
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