2003 AMC 12A Problem 5 — Solution | LIVE by Po-Shen Loh

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2003 AMC 12A Problem 5

Attempt Problem 5 of the 2003 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 2003 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

The sum of the two $5$ -digit numbers $\overline{AMC10}$ and $\overline{AMC12}$ is $123422.$ What is $A + M + C?$

$10$

$11$

$12$

$13$

$14$

Answer: E

Concepts:place value cryptarithm

Difficulty rating: 1200

Solution:

Write $\overline{AMC10}=100\cdot\overline{AMC}+10$ and $\overline{AMC12}=100\cdot\overline{AMC}+12.$

Their sum is $200\cdot\overline{AMC}+22=123422,$ so $\overline{AMC}=617.$

Then $A+M+C=6+1+7=14.$

Thus, the correct answer is E.

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