2020 AIME I Problems/Problem 10

Note: Please do not post problems here until after the AIME.

Problem

Solution

Taking inspiration from $4^4 \mid 10^{10}$ we are inspired to take $n$ to be $p^2$ , the lowest prime not dividng $210$ , or $11 \implies n = 121$ . Now, there are $242$ factors of $11$ , so $11^{242} \mid m^m$ , and then $m = 11k$ for $k \geq 22$ . Now, $\gcd(m+n, 210) = \gcd(11+k,210) = 1$ . Noting $k = 26$ is the minimal that satisfies this, we get $(n,m) = (121,286)$ . Thus, it is easy to verify this is minimal and we get $\boxed{407}$ .

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2020 AIME I Problems/Problem 10

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