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

Revision as of 23:33, 10 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 $2a + c = 40$ 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://www.youtube.com/watch?v=x0QznvXcwHY

~IceMatrix

Video Solution 2

https://youtu.be/Va3MPyAULdU

~avn

@@ Line 21: / Line 21: @@
 ~IceMatrix
+==Video Solution 2==
+https://youtu.be/Va3MPyAULdU
+~avn
 ==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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