2003 AMC 10A Problems/Problem 11: Difference between revisions

Revision as of 10:20, 15 January 2008

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

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

$AMC10+AMC12=123422$

$AMC00+AMC00=123400$

$AMC+AMC=1234$

$2\cdot AMC=1234$

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

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

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

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 10 Problems and Solutions

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 == See Also ==
-*[[2003 AMC 10A Problems]]
+{{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]]