(solved by Lucas Boczkowski). Find all $n \in\{1,2, \ldots, 999\}$ such that $n^{2}$ is equal to the cube of the sum of the digits of $n$.
. For $n^{2}$ to be a cube, $n$ must itself be a cube. Like $n27^{3}$. However, the sum of the digits of $n$ is at most $9+9+9=27$. For $n \in\{1,8,27,64,125\}$, we manually verify that 1 and 27 have the required property and that this is not the case for the others. The solutions are therefore $n=1$ and $n=27$.