AMC12 2005 12A, Problem 8

Let $A,M$, and $C$ be digits with \[(100A+10M+C)(A+M+C) = 2005\] What is $A$?

Solution: Clearly the two quantities are both integers, so we check the prime factorization of $2005 = 5 \cdot 401$. It is easy to see now that $(A,M,C) = (4,0,1)$ works, so the answer is $$. $\fbox{4}$.

Answer: 4