[Number Theory]
Problem: Let $n$ be a four-digit number whose square root is three times the sum of the digits of $n$. Compute $n$.

Solution: Because $\sqrt{n}$ is a multiple of $3, n$ must be a multiple of 9 . Therefore the sum of the digits of $n$ is a multiple of 9 . Thus $\sqrt{n}$ must be a multiple of 27 , which implies that $n$ is a multiple of $27^{2}$. The only candidates to consider are $54^{2}(=2916)$ and $81^{2}(=6561)$, and only 2916 satisfies the desired conditions.

Final answer: 2916