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2003 AMC 10A Problem 11

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2003 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: October 28, 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 start with the same three digits AMC. Written as place values, the sum is twice AMC hundreds, plus 10 plus 12, which gives AMC = 617.

Solution

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 $\boxed{\textbf{(E)}\ 14}$ .

Why this works

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.

Alternative approach

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 $6 + 1 + 7 = 14$ .

The trap

Treating AMC as A times M times C, or forgetting the last two digits 10 and 12 and just halving 123422.

Concise 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 $\boxed{\textbf{(E)}\ 14}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Number Theory, Chapter 9: Base Number Arithmetic
Introduction to Algebra, Chapter 7: Proportion (§7.4)

Techniques

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