2005 AMC 12A Problems/Problem 8

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Problem

Let $A,M$, and $C$ be digits with

\[(100A+10M+C)(A+M+C) = 2005\]

What is $A$?

$(\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 3 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 5$

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 $\mathrm{(D)}$.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AMC 12 Problems and Solutions
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