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

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==Solution==
 
==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>. 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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In this problem, we want to find the number of ordered pairs <math>(m, n)</math> such that <math>m^2n = 20^{20}</math>. Let <math>x = m^2</math>. Therefore, we want two numbers, <math>x</math> and <math>n</math>, such that their product is <math>20^{20}</math> and <math>x</math> is a perfect square. Note that there is exactly one valid <math>n</math> for a unique <math>x</math>, which is <math>\tfrac{20^{20}}{x}</math>. This reduces the problem to finding the number of unique perfect square factors of <math>20^{20}</math>.
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<math>20^{20} = 2^{40} \cdot 5^{20} = \left(2^2\right)^{20}\cdot\left(5^2\right)^{10}.</math> Therefore, the answer is <math>21 \cdot 11 = \boxed{231}.</math>
  
 
~superagh
 
~superagh
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<br><br>
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~TheBeast5520
  
 
==Solution 2 (Official MAA)==
 
==Solution 2 (Official MAA)==

Revision as of 02:32, 6 March 2021

Problem

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

Solution

In this problem, we want to find the number of ordered pairs $(m, n)$ such that $m^2n = 20^{20}$. Let $x = m^2$. Therefore, we want two numbers, $x$ and $n$, such that their product is $20^{20}$ and $x$ is a perfect square. Note that there is exactly one valid $n$ for a unique $x$, which is $\tfrac{20^{20}}{x}$. This reduces the problem to finding the number of unique perfect square factors of $20^{20}$.

$20^{20} = 2^{40} \cdot 5^{20} = \left(2^2\right)^{20}\cdot\left(5^2\right)^{10}.$ Therefore, the answer is $21 \cdot 11 = \boxed{231}.$

~superagh ~TheBeast5520

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

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 (Problems • Answer Key • Resources)
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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