AMC 8 · 2009 · #24

Grade 4 logic
place-valuedigit-decompositionlogical-deductionmulti-digit-arithmetic identify-subproblemscasework ↑ Prerequisites: place-valuemulti-digit-arithmetic
📏 Medium solution 💡 4 insights
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Problem
Four digits A, B, C, D satisfy two cryptarithms at once. Adding the two-digit numbers AB + CA gives the two-digit number DA, and subtracting them gives the one-digit number A (so the tens digit of the difference is 0). Find the digit D.

Pick an answer.

(A)
5
(B)
6
(C)
7
(D)
8
(E)
9

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Write an Equation

Cryptarithms are pure place-value puzzles. Tool #3 (Write an Equation) turns each column of the addition and subtraction into a small equation in the unknown digits (with carry/borrow bits). Tool #7 (Identify Subproblems) tells us to work column by column — start with the ones column, which forces B, then move to the tens column, which forces A, C, and finally D — instead of trying to guess all four digits at once.

1STEP 1

Ones column of the addition: B + A ends in A, so B = 0 — and no carry into the tens column.

B + A ≡ A (mod 10) → B = 0
2STEP 2

Subtraction ones column: with B = 0, the 0 - A needs a borrow, so 10 - A = A and A = 5.

10 - A = A → 2A = 10 → A = 5
3STEP 3

Subtraction tens column: the borrow took 1 from the tens, so (A - 1) - C = 0 and C = 4.

(A - 1) - C = 0 → C = A - 1 = 4
4STEP 4

Addition tens column: no carry from the ones, so D = A + C = 5 + 4 = 9.

D = A + C = 5 + 4 = 9 → (E)
Answer
9
Plug the digits back in: A = 5, B = 0, C = 4, D = 9, so AB = 50 and CA = 45. Then 50 + 45 = 95 = DA (tens digit D = 9, ones digit A = 5) and 50 - 45 = 5 = A (a single-digit answer with tens digit 0). Both cryptarithms check out.
💡Key takeaway

This AMC 8 problem only needs the Grade 4 standard column-by-column add/subtract algorithm — once you read each column carefully, the digits fall out one at a time.