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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 (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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