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

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==Problem==
 
==Problem==
 
Find the number of ordered pairs of positive integers <math>(m,n)</math> such that <math>{m^2n = 20 ^{20}}</math>.
 
Find the number of ordered pairs of positive integers <math>(m,n)</math> such that <math>{m^2n = 20 ^{20}}</math>.
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==Solution==
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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>. <math>n</math> will be assigned by default. There are <math>21\times11=231</math> ways to select a perfect square factor of <math>20^20</math>, thus our answer is <math>\mbox{231}</math>.
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~superagh

Revision as of 16:39, 7 June 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 $20^20$, $m^2$. $n$ will be assigned by default. There are $21\times11=231$ ways to select a perfect square factor of $20^20$, thus our answer is $\mbox{231}$. ~superagh