2019/20 AHSMC Part 1 Problem 5

Suppose $2020^{2019} = xy$. Since $$2020 = 10 \cdot 202 = 2^2 \cdot 5 \cdot 101,$$ where $101$ is prime, we have $$xy = 2020^{2019} = 2^{4038} \cdot 5^{2019} \cdot 101^{2019}.$$ If $x$ and $y$ are relatively prime, each of $2^{4038}$, $5^{2019}$ and $101^{2019}$ is a factor of either $x$ or $y$. There are $2 \times 2 \times 2 = 8$ possibilities but since two products of the same numbers in a different order are considered the same, $4$ of these possibilities are double counted. Thus, there are $8 - 4 = 4$ possibilities. The answer is (C).