Difference between revisions of "2003 AMC 10A Problems/Problem 11"

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== Problem ==
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#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>?
 
 
 
<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14 </math>
 
 
 
== Solution ==
 
<math>AMC10+AMC12=123422</math>
 
 
 
<math>AMC00+AMC00=123400</math>
 
 
 
<math>AMC+AMC=1234</math>
 
 
 
<math>2\cdot AMC=1234</math>
 
 
 
<math>AMC=\frac{1234}{2}=617</math>
 
 
 
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>.
 
 
 
Therefore, <math>A+M+C = 6+1+7 = 14 \Rightarrow E </math>.
 
 
 
== See Also ==
 
*[[2003 AMC 10A Problems]]
 
*[[2003 AMC 10A Problems/Problem 10|Previous Problem]]
 
*[[2003 AMC 10A Problems/Problem 12|Next Problem]]
 
 
 
[[Category:Introductory Algebra Problems]]
 

Latest revision as of 13:35, 30 July 2011

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