2022/23 AHSMC Part 1 Problem 12

If $m = 3k$ then $m^m = (m^k)^3$ hence $m^m$ is a perfect cube. If $m = 3k+1$, then $m^m = (m^k)^3 m$ which is a perfect cube if $m$ is a perfect cube. If $m = 3k+2$, then $m^m = (m^k)^3 m^2$, which is a perfect cube if $m$ is a perfect cube. Between $1$ and $300$ there are $100$ multiples of $3$, and four numbers of the form $3k+1$ or $3k+2$ which are perfect cubes ($1$, $8 = 2^3$, $64 = 4^3$, and $125 = 5^3$). Therefore there are $104$ integers $m$ having the requested properties. The answer is (D).