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2003 AMC 10A Problem 11

Problem 11 of 25IntermediateArithmeticLogic

The sum of the two 55-digit numbers AMC10AMC10 and AMC12AMC12 is 123422.123422. What is A+M+C?A + M + C?

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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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Tagged: cryptarithm · place value

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