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

Revision as of 22:04, 7 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

@@ Line 9: / Line 9: @@
 [[Category:Introductory Number Theory Problems]]
 ==See Also==
-{{AIME box|year=2020|n=II|num-b=First Problem|num-a=2}}
+{{AIME box|year=2020|n=II|before=First Problem|num-a=2}}
 {{MAA Notice}}

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

Page

Toolbox

Search

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

Revision as of 22:04, 7 June 2020

Problem

Solution

See Also