2003 AMC 10B Problem 14 — Solution | LIVE by Po-Shen Loh

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2003 AMC 10B Problem 14

Attempt Problem 14 of the 2003 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

Given that $3^8 \cdot 5^2 = a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b.$

$25$

$34$

$351$

$407$

$900$

Answer: D

Concepts:prime factorization perfect power optimization

Difficulty rating: 1390

Solution:

Because $a$ must be divisible by $5,$ and $3^8 \cdot 5^2$ is divisible by $5^2$ but not $5^3,$ we need $b\le 2.$

Taking $b=2$ gives $a=\sqrt{3^8 \cdot 5^2}=3^4 \cdot 5=405,$ so $a+b=407.$ This beats $b=1,$ which gives $a+b=164{,}026.$

Thus, the correct answer is D.

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