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2003 AMC 12A Problem 5

Problem 5 of 25EasierNumber Theory

The sum of the two 55-digit numbers AMC10‾\overline{AMC10} and AMC12‾\overline{AMC12} is 123422.123422. What is A+M+C?A + M + C?

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Solution

Write AMC10‾=100⋅AMC‾+10\overline{AMC10}=100\cdot\overline{AMC}+10 and AMC12‾=100⋅AMC‾+12.\overline{AMC12}=100\cdot\overline{AMC}+12. Their sum is 200⋅AMC‾+22=123422,200\cdot\overline{AMC}+22=123422, so AMC‾=617.\overline{AMC}=617. Then A+M+C=6+1+7=14.A+M+C=6+1+7=14. Thus, the correct answer is E.

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Concepts: place value · cryptarithm

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.