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 $x=6170$. Therefore, the sum $A+M+C= \boxed{\mathrm{(E)}\ 14}$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions
2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 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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