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2003 AMC 10A Problems/Problem 11

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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

$AMC10+AMC12=123422$

$AMC00+AMC00=123400$

$2\cdot AMC=1234$

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

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 (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2003_AMC_10A_Problems/Problem_11"

Category: Introductory Algebra Problems

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

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