# Difference between revisions of "2020 AIME II Problems/Problem 1"

## Problem

Find the number of ordered pairs of positive integers $(m,n)$ such that ${m^2n = 20 ^{20}}$.

## Solution

First, we find the prime factorization of $20^{20}$, which is $2^{40}\times5^{20}$. The equation ${m^2n = 20 ^{20}}$ tells us that we want to select a perfect square factor of $20^{20}$, $m^2$. The $n$ might throw you off here, but it's actually kind of irrelevant because once $m$ is selected, the remaining factor will already be assigned as $\frac{20^{20}}{m^2}$. There are $21\cdot11=231$ ways to select a perfect square factor of $20^{20}$, thus our answer is $\boxed{231}$.

~superagh

## Solution 2 (Official MAA)

Because $20^{20}=2^{40}5^{20}$, if $m^2n = 20^{20}$, there must be nonnegative integers $a$, $b$, $c$, and $d$ such that $m = 2^a5^b$ and $n = 2^c5^d$. Then $$2a + c = 40$$ and $$2b+d = 20$$ The first equation has $21$ solutions corresponding to $a = 0,1,2,\dots,20$, and the second equation has $11$ solutions corresponding to $b = 0,1,2,\dots,10$. Therefore there are a total of $21\cdot11 = 231$ ordered pairs $(m,n)$ such that $m^2n = 20^{20}$.

~ pi_is_3.14

~IceMatrix

~avn

## Purple Comet Math Meet April 2020

Notice, that this was the exact same problem (with different wording of course) as Purple Comet HS problem 3 and remembering the answer, put $\boxed{231}$.

~Lopkiloinm

## See Also

 2020 AIME II (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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