2003 AMC 10A Problems/Problem 11

Revision as of 19:09, 4 November 2006 by I_like_pie (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

$M+M$ is odd, so $C+C$ must carry over, and therefore, $C=7$.

$A+A=12$, so $A=6$.

$M+M+1=3$, so $M=1$.

$7+6+1=14\Longrightarrow\mathrm{(E)}$


See Also

Invalid username
Login to AoPS