2003 AMC 10A Problem 11

Attempt Problem 11 of the 2003 AMC 10A 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 10A solutions, or check the answer key.

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

11.

The sum of the two 55-digit numbers AMC10AMC10 and AMC12AMC12 is 123422.123422. What is A+M+C?A + M + C?
1010
1111
1212
1313
1414
Answer: E
Concepts:cryptarithmplace value
Small hint:
Write each number as 100⋅AMC‾100 \cdot \overline{AMC} plus its last two digits
Big hint:
Their sum is 200⋅AMC‾+22=123422200 \cdot \overline{AMC} + 22 = 123422
Solution:
The two numbers equal 100⋅AMC‾+10100 \cdot \overline{AMC} + 10 and 100⋅AMC‾+12,100 \cdot \overline{AMC} + 12, so their sum is 200⋅AMC‾+22=123422.200 \cdot \overline{AMC} + 22 = 123422.
Then 200⋅AMC‾=123400,200 \cdot \overline{AMC} = 123400, so AMC‾=617.\overline{AMC} = 617.
Therefore A+M+C=6+1+7=14.A + M + C = 6 + 1 + 7 = 14.
Thus, the correct answer is E.
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