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2003 AMC 10A Problem 11

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2003 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Bases & Digits · difficulty 2 of 5 · about 2 min · position P11-15

The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$ ?

A)
$10$
B)
$11$
C)
$12$
D)
$13$
E)
$14$

Answer

Key insight

Both numbers are 100 times the three-digit number AMC plus 10 or 12, so 200*AMC + 22 = 123422 and AMC = 617.

Solution

Let $N$ be the three-digit number with digits $A, M, C$ . By place value,
$AMC10 = 100N + 10, \qquad AMC12 = 100N + 12.$
Adding, $200N + 22 = 123422$ , so $200N = 123400$ and $N = 617$ .

Thus $A = 6$ , $M = 1$ , $C = 7$ , and $A + M + C = 6 + 1 + 7 = 14$ .

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

Why this works

A string of digits is a number, and splitting it at a place-value boundary ( $100 \cdot \text{(front)} + \text{(back)}$ ) turns a cryptarithm into ordinary algebra. When two such numbers share their leading digits, their sum is just twice the shared part plus the known tails.

Alternative approach

The two numbers average $123422 / 2 = 61711$ , which must be $AMC11$ . Reading off the digits gives $6 + 1 + 7 = 14$ .

The trap

Reading AMC as a product of digits, or forgetting the trailing 10 and 12 and dividing 123422 by 200.

Common mistakes

Reading AMC as a product of digits, or forgetting the trailing 10 and 12 and dividing 123422 by 200.
Doing column addition with carries and slipping on the units digit ( $C + C$ ends in $4$ so $C$ is $2$ or $7$ ; the carry decides).

Techniques

Set up the equation/formula and compute; no special trick needed