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>. | ||
| + | |||
| + | ==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>. <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>. | ||
| + | ~superagh | ||
Revision as of 17:39, 7 June 2020
Problem
Find the number of ordered pairs of positive integers 
such that 
.
Solution
First, we find the prime factorization of 
, which is 
. The equation tells us that we want to select a perfect square factor of , 
. 
will be assigned by default. There are 
ways to select a perfect square factor of , thus our answer is 
.
~superagh
