$AMC10+AMC12=123422$
$AMC00+AMC00=123400$
$AMC+AMC=1234$
$2\cdot AMC=1234$
$AMC=\frac{1234}{2}=617$
Since $A$, $M$, and $C$ are digits, $A=6$, $M=1$, $C=7$. 
Therefore, $A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14}$.
The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$. What is $A+M+C$? 
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$