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

## Solution 3

Consider the place values of the digits of $AMC10$ and $AMC12$.

When we add $AMC10$ and $AMC12$, $C + C$ must result in a units digit of $4$, meaning $C$ is either $2$ or $7$. Since $M$ is odd, this means a ten was carried over to the next place value from $C + C$, and thus $C = 7$ (as $7 + 7 = 14$ and the ten is carried over). Now, we know 3 is the units digit of $M + M + 1$, so $M$ is either $1$ or $6$. Again, we must look at the digit before $M$, or $A$. $A$ is even, so $M$ must be less than $5$, or else the ten would be carried over. Ergo, $M$ is $1$. Nothing is carried over, so we have $A + A = 12$, and $A = 6$. Therefore, the sum of $A$, $M$, and $C$ is $6 + 1 + 7 = \boxed{\mathrm{(E)}\ 14}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 