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

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

Revision as of 21:35, 24 September 2020

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 $m^2n = 20^{20}$.


Video Solution

https://www.youtube.com/watch?v=x0QznvXcwHY

~IceMatrix


Video Solution 2

https://youtu.be/Va3MPyAULdU

~avn

See Also

2020 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 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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