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−  == Problem ==
 +  #REDIRECT[[2003 AMC 12A Problems/Problem 5]] 
−  The sum of the two 5digit 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 ==
 
−  <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 ==
 
−  * [[2003 AMC 10A Problems]]
 
−  * [[2003 AMC 10A Problems/Problem 12  Next Problem]]
 
−  * [[2003 AMC 10A Problems/Problem 10  Previous Problem]]
 