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

(Purple Comet Math Meet April 2020)
(Video Solution)
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The first equation has <math>21</math> solutions corresponding to <math>a = 0,1,2,\dots,20</math>, and the second equation has <math>11</math> solutions corresponding to <math>b = 0,1,2,\dots,10</math>. Therefore there are a total of <math>21\cdot11 = 231</math> ordered pairs <math>(m,n)</math> such that <math>m^2n = 20^{20}</math>.
 
The first equation has <math>21</math> solutions corresponding to <math>a = 0,1,2,\dots,20</math>, and the second equation has <math>11</math> solutions corresponding to <math>b = 0,1,2,\dots,10</math>. Therefore there are a total of <math>21\cdot11 = 231</math> ordered pairs <math>(m,n)</math> such that <math>m^2n = 20^{20}</math>.
  
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== Video Solution ==
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https://youtu.be/zfChnbMGLVQ?t=4612
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~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==
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~IceMatrix
 
~IceMatrix
 
  
 
==Video Solution 2==
 
==Video Solution 2==

Revision as of 02:26, 17 January 2021

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://youtu.be/zfChnbMGLVQ?t=4612

~ pi_is_3.14

Video Solution

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

~IceMatrix

Video Solution 2

https://youtu.be/Va3MPyAULdU

~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 $\boxed{231}$.

https://purplecomet.org/views/data/2020HSSolutions.pdf

~Lopkiloinm

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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