problem: The sum of the two $5$-digit numbers $AMC10$ and $AMC12$ is $123422.$ What is $A + M + C?$
solution: Since the last two digits of $AMC10$ and $AMC12$ sum to $22,$ we have \[
  AMC + AMC = 2(AMC) = 1234.
  \] Hence $AMC=617,$ so $A=6,$ $M=1,$ $C=7,$ and $A+M+C = 6+1+7 = \boxed{14}.$