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

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== See Also ==
 
== See Also ==
*[[2003 AMC 10A Problems]]
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{{AMC10 box|year=2003|ab=A|num-b=10|num-a=12}}
*[[2003 AMC 10A Problems/Problem 10|Previous Problem]]
 
*[[2003 AMC 10A Problems/Problem 12|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 10:20, 15 January 2008

Problem

The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$. What is $A+M+C$?

$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$

Solution

$AMC10+AMC12=123422$

$AMC00+AMC00=123400$

$AMC+AMC=1234$

$2\cdot AMC=1234$

$AMC=\frac{1234}{2}=617$

Since $A$, $M$, and $C$ are digits, $A=6$, $M=1$, $C=7$.

Therefore, $A+M+C = 6+1+7 = 14 \Rightarrow E$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions
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