Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
m (→‎Solution 2 (Official MAA))
 
(27 intermediate revisions by 15 users not shown)
Line 3: Line 3:
  
 
==Solution==
 
==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>\boxed{231}</math>.
+
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>.
 +
 
 +
 
 +
<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
−
<br><br>
+
 
−
[[Category:Introductory Number Theory Problems]]
+
~TheBeast5520
 +
 
 +
==Solution 2 (Official MAA)==
 +
Because <math>20^{20}=2^{40}5^{20}</math>, if <math>m^2n = 20^{20}</math>, there must be nonnegative integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> such that
 +
<math>m = 2^a5^b</math> and <math>n = 2^c5^d</math>. Then
 +
<cmath>2a + c = 40</cmath>
 +
and
 +
<cmath>2b+d = 20</cmath>
 +
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>.
 +
 
 +
== 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 <math>\boxed{231}</math>.
 +
 
 +
https://purplecomet.org/views/data/2020HSSolutions.pdf
 +
 
 +
~Lopkiloinm
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/Gs27CPxRiTA
 +
 
 +
~savannahsolver
 +
 
 
==See Also==
 
==See Also==
 
{{AIME box|year=2020|n=II|before=First Problem|num-a=2}}
 
{{AIME box|year=2020|n=II|before=First Problem|num-a=2}}
 +
[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:59, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_1&oldid=216208"