2003 AMC 10A Problem 11

The two numbers share their first three digits, so treat those three digits as one number and use place value.

1. Call the three-digit number made by the digits $A, M, C$ by the name $N$. (For example, if the digits were $3, 4, 5$, then $N = 345$.)
2. Putting "$10$" after $N$ multiplies $N$ by $100$ and adds $10$. For example, $34510 = 345 \times 100 + 10$. So
$$AMC10 = 100N + 10, \qquad AMC12 = 100N + 12.$$
3. Add them and set the sum equal to $123422$:
$$200N + 22 = 123422.$$
4. Subtract $22$: $200N = 123400$. Divide by $200$: $N = 617$.
5. So $A = 6$, $M = 1$, $C = 7$, and
$$A + M + C = 6 + 1 + 7 = 14.$$

Check: $61710 + 61712 = 123422$. Correct.

The answer is $\boxed{\textbf{(E)}\ 14}$.

A row of digits is really a number built from place values, so splitting it into "front part times $100$, plus the last two digits" turns a digit puzzle into a simple equation. Whenever letters stand for digits, write the number in place-value form first.

The two numbers differ by $2$, so they sit one on each side of their average. The average is $123422 \div 2 = 61711$, so the numbers are $61710$ and $61712$. Reading off the first three digits gives $6 + 1 + 7 = 14$.