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| == Problem ==
| | #REDIRECT[[2003 AMC 12A Problems/Problem 5]] |
| The sum of the two 5-digit numbers <math>AMC10</math> and <math>AMC12</math> is <math>123422</math>. What is <math>A+M+C</math>?
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| <math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14 </math>
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| == Solution ==
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| <math>M+M</math> is odd, so <math>C+C</math> must carry over, and therefore, <math>C=7</math>.
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| <math>A+A=12</math>, so <math>A=6</math>.
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| <math>M+M+1=3</math>, so <math>M=1</math>.
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| <math>7+6+1=14\Longrightarrow\mathrm{(E)}</math>
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| == See Also ==
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| * [[2003 AMC 10A Problems]]
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| * [[2003 AMC 10A Problems/Problem 12 | Next Problem]]
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| * [[2003 AMC 10A Problems/Problem 10 | Previous Problem]]
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