Let A, M, and C be digits with \beginaligned &(100A + 10M + C) \ &\quad × (A + M + C) = 2005. \endaligned What is A?

2005 AMC 12A Problem 8 — Solution

2005 AMC 12A Problem 8

Attempt Problem 8 of the 2005 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12A solutions, or check the answer key.

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Let $A,$ $M,$ and $C$ be digits with $\begin{aligned} &(100A + 10M + C) \\ &\quad {}\cdot (A + M + C) = 2005. \end{aligned}$ What is $A?$

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:prime factorization digits bounding to limit cases

Difficulty rating: 1350

Solution:

Since $A + M + C \le 9 + 9 + 9 = 27,$ and $2005 = 5 \cdot 401,$ the digit sum can only be $1$ or $5.$ 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.

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