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

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<cmath>2b+d = 20</cmath>
 
<cmath>2b+d = 20</cmath>
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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>.
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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 = \boxed{231}</math> ordered pairs <math>(m,n)</math> such that <math>m^2n = 20^{20}</math>.
  
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== Video Solution by OmegaLearn==
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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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https://youtu.be/zfChnbMGLVQ?t=4612
+
https://www.youtube.com/watch?v=VA1lReSkGXU
  
−
~ pi_is_3.14
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~ North America Math Contest Go Go Go
  
 
==Video Solution==
 
==Video Solution==
Line 45: Line 49:
  
 
~Lopkiloinm
 
~Lopkiloinm
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==Video Solution by WhyMath==
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https://youtu.be/Gs27CPxRiTA
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 13:59, 25 February 2024

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 $21\cdot11 = \boxed{231}$ ordered pairs $(m,n)$ such that $m^2n = 20^{20}$.

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=VA1lReSkGXU

~ North America Math Contest Go Go Go

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

Video Solution by WhyMath

https://youtu.be/Gs27CPxRiTA

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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