Let $A,M$, and $C$ be digits with
\[(100A+10M+C)(A+M+C) = 2005\]
What is $A$?
$(\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 3 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 5$
Clearly the two quantities are both integers, so we check the [prime factorization](https://artofproblemsolving.com/wiki/index.php/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 $\mathrm{(D)}$.