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

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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
 
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
 
<math>m = 2^a5^b</math> and <math>n = 2^c5^d</math>. Then
−
\begin{align*}
+
&\begin{align*}
 
2a + c &= 40\text{~~and}\\
 
2a + c &= 40\text{~~and}\\
 
2b+d &= 20.
 
2b+d &= 20.
−
\end{align*}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>.
+
\end{align*} &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>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:41, 8 June 2020

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 $20^{20}$, $m^2$. $n$ will be assigned by default. There are $21\times11=231$ 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 &\begin{align*} 2a + c &= 40\text{~~and}\\ 2b+d &= 20. \end{align*} &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://www.youtube.com/watch?v=x0QznvXcwHY

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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