2020 AIME I Problems/Problem 10

Problem

Let $m$ and $n$ be positive integers satisfying the conditions

$\quad\bullet\ \gcd(m+n,210)=1,$

$\quad\bullet\ m^m$ is a multiple of $n^n,$ and

$\quad\bullet\ m$ is not a multiple of $n.$

Find the least possible value of $m+n.$

Solution 1

Taking inspiration from $4^4 \mid 10^{10}$ we are inspired to take $n$ to be $p^2$, the lowest prime not dividing $210$, 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 $\boxed{407}$. ~awang11

Solution 2

Assume for the sake of contradiction that $n$ is a multiple of a single digit prime number, then $m$ must also be a multiple of that single digit prime number to accommodate for $n^n | m^m$. However that means that $m+n$ is divisible by that single digit prime number, which violates $\gcd(m+n,210) = 1$, so contradiction.

$n$ is also not 1 because then $m$ would be a multiple of it.

Thus, $n$ is a multiple of 11 and/or 13 and/or 17 and/or...

Assume for the sake of contradiction that $n$ has at most 1 power of 11, at most 1 power of 13...and so on... Then, for $n^n | m^m$ to be satisfied, $m$ must contain at least the same prime factors that $n$ has. This tells us that for the primes where $n$ has one power of, $m$ also has at least one power, and since this holds true for all the primes of $n$, $n|m$. Contradiction.

Thus $n$ needs more than one power of some prime. The obvious smallest possible value of $n$ now is $11^2 =121$. Since $121^{121}=11^{242}$, we need $m$ to be a multiple of 11 at least $242$ that is not divisible by $121$ and most importantly, $\gcd(m+n,210) = 1$. $242$ is divisible by $121$, out. $253+121$ is divisible by 2, out. $264+121$ is divisible by 5, out. $275+121$ is divisible by 2, out. $286+121=37\cdot 11$ and satisfies all the conditions in the given problem, and the next case $n=169$ will give us at least $169\cdot 3$, so we get $\boxed{407}$.

Video Solution

https://youtu.be/Z47NRwNB-D0


Video Solution

https://www.youtube.com/watch?v=FQSiQChGjpI&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=7 ~ MathEx

See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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