2020 AIME II Problems/Problem 1

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 $\mbox{231}$ .

~superagh The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2020 AIME II Problems/Problem 1

Problem

Solution