2005 AMC 12A Problem 8

Let

A,

M,

and

C

be digits with

\begin{aligned} &(100A + 10M + C) \\ &\quad {}\cdot (A + M + C) = 2005. \end{aligned}

What is

A?

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Solution

Since

A + M + C \le 9 + 9 + 9 = 27,

and

2005 = 5 \cdot 401,

the digit sum can only be

1

5.

It cannot be

1,

because then

100A+10M+C=2005>999.

Thus it must be the smaller nontrivial factor:

\begin{aligned} &100A + 10M + C = 401, \\ &\quad A + M + C = 5. \end{aligned}

Reading off the digits,

A = 4,

M = 0,

and

C = 1.

Thus, the correct answer is D.

Tagged: prime factorization · digits · bounding to limit cases