AMC 10 · 2003 · #11

Grade 6 arithmetic
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
place-valuelinear-equations-one-vardigit-decomposition convert-to-algebrawork-backwards ↑ Prerequisites: place-value
📏 Medium solution 💡 2 insights
Problem
Two 5-digit numbers are written as AMC10 and AMC12, where A, M, and C are single digits that are the same in both numbers. Their sum is 123422. Find A+M+C.

Pick an answer.

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

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The two numbers look almost the same, so first use Tool #5 (Look for a Pattern) to spot what actually changes: only the very last digit, 0 versus 2. That means the second number is just 2 bigger than the first, so instead of two mystery numbers there is really only one. Tool #4 (Introduce a Variable) then names that single number N, turning the whole puzzle into one clean equation. Finally Tool #11 (Work Backwards) unwinds the equation — undo the +2, then undo the doubling — to recover N, and the digits of N hand you A, M, and C directly.

1STEP 1

See that the two numbers are twins

AMC10 and AMC12 share the first four digits A, M, C, 1; only the units digit differs, 0 versus 2, so AMC12 = AMC10 + 2.

AMC12 = AMC10 + 2
2STEP 2

Name the number, write one equation

Let N be the first number; the second is N + 2, so the given sum becomes 2N + 2 = 123422.

N + (N + 2) = 2N + 2 = 123422
3STEP 3

Undo the +2, then undo the doubling

Work backwards: remove the extra 2 to get 2N = 123420, then halve to get N = 61710.

2N = 123420 → N = 61710
4STEP 4

Read off the digits and add

Line 61710 up with A, M, C, 1, 0: A=6, M=1, C=7, so A+M+C = 14, choice (E).

A=6, M=1, C=7 → A+M+C = 14
Answer
14
Check the answer both ways. First, the found digits give A=6, M=1, C=7, so AMC10 = 61710 and AMC12 = 61712; their sum is 61710 + 61712 = 123422, matching the problem exactly. Second, the fixed last two digits 1 and 0 of AMC10 came out right, confirming nothing was mis-aligned. Since each of A, M, C is a single digit, their sum can be at most 27, so 14 is a sensible size and it appears as choice (E).
💡Key takeaway

The two numbers are identical except for the last digit, so one is just 2 more than the other; that makes the sum equal to twice the first number plus 2, and undoing those steps gives AMC10 = 61710, so A+M+C = 6+1+7 = 14.

  • See that the two numbers are twins
  • Name the number, write one equation
  • Undo the +2, then undo the doubling
  • Read off the digits and add

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