# 2003 AMC 12A Problems/Problem 5

The following problem is from both the 2003 AMC 12A #5 and 2003 AMC 10A #11, so both problems redirect to this page.

## 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, $A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14}$.

## Solution 2

We know that $AMC12$ is $2$ more than $AMC10$. We set up $AMC10=x$ and $AMC12=x+2$. We have $x+x+2=123422$. Solving for $x$, we get $x=61710$. Therefore, the sum $A+M+C= \boxed{\mathrm{(E)}\ 14}$.

## See Also

 2003 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 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 All AMC 10 Problems and Solutions
 2003 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 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 All AMC 12 Problems and Solutions

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