2005 AMC 12A Problem 8

Below is the professionally curated solution for Problem 8 of the 2005 AMC 12A, 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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Concepts:prime factorization digits bounding to limit cases

Difficulty rating: 1350

Let $A,$ $M,$ and $C$ be digits with $(100A + 10M + C)(A + M + C) = 2005.$ What is $A?$

$1$

$2$

$3$

$4$

$5$

Solution:

Since $A + M + C \le 9 + 9 + 9 = 27,$ and $2005 = 5 \cdot 401,$ the digit sum must be the smaller factor: $100A + 10M + C = 401, \quad A + M + C = 5.$

Reading off the digits, $A = 4,$ $M = 0,$ and $C = 1.$

Thus, the correct answer is D.

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