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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>AMC10+AMC12=123422</math>
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| − | <math>AMC00+AMC00=123400</math>
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| − | <math>AMC+AMC=1234</math>
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| − | <math>2\cdot AMC=1234</math>
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| − | <math>AMC=\frac{1234}{2}=617</math>
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| − | Since <math>A</math>, <math>M</math>, and <math>C</math> are digits, <math>A=6</math>, <math>M=1</math>, <math>C=7</math>.
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| − | Therefore, <math>A+M+C = 6+1+7 = 14 \Rightarrow E </math>.
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| − | == See Also ==
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| − | {{AMC10 box|year=2003|ab=A|num-b=10|num-a=12}}
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| − | [[Category:Introductory Algebra Problems]] | |
