2003 AMC 10A Problem 11 — Solution | LIVE by Po-Shen Loh

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2003 AMC 10A Problem 11

Below is the professionally curated solution for Problem 11 of the 2003 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:cryptarithm place value

Difficulty rating: 1310

11.

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

$10$

$11$

$12$

$13$

$14$

Solution:

The two numbers equal $100 \cdot \overline{AMC} + 10$ and $100 \cdot \overline{AMC} + 12,$ so their sum is $200 \cdot \overline{AMC} + 22 = 123422.$

Then $200 \cdot \overline{AMC} = 123400,$ so $\overline{AMC} = 617.$

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

Thus, the correct answer is E.

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