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 $a+b$.

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

Solution

\[3^8\cdot5^2 = (3^4)^2\cdot5^2 = (3^4\cdot5)^2 = 405^2\]

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

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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