AMC 10 · 2014 · #10

Grade 6 number-theorylogic
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
digit-constraintsplace-valuemulti-digit-arithmeticlogical-deduction logical-deduction ↑ Prerequisites: place-value
📏 Medium solution 💡 3 insights
Problem
In a column addition, the five-digit number ABBCB is added to the five-digit number BCADA, and the sum is the five-digit number DBDDD. Here A, B, C, D stand for four distinct digits, and each letter always means the same digit. Count how many different digits D could possibly be.

Pick an answer.

(A)
2
(B)
4
(C)
7
(D)
8
(E)
9

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

A cryptarithm is solved by naming the hidden quantities and letting the columns speak. Tool #4 (Introduce a Variable) adds one carry digit c₁,…,c₄ between the columns, turning the picture into five exact equations. Tool #7 (Identify Subproblems) treats each column as its own tiny equation, and comparing two of them collapses the carries to zero and pins C=0. Once the puzzle reduces to A+B=D, Tool #14 (Extreme Principle) finds the smallest and largest D can be, and everything between is reachable — so counting the values is just measuring that range.

1STEP 1

Name a carry for each column

Line the columns up by place value and let c₁,c₂,c₃,c₄ be the carries out of units, tens, hundreds, thousands — each one 0 or 1.

units: A+B=D+10c₁, tens: C+D+c₁=D+10c₂→ C+c₁=10c₂, ; hundreds: A+B+c₂=D+10c₃, thousands: B+C+c₃=B+10c₄→ C+c₃=10c₄, ten-thousands: A+B+c₄=D
2STEP 2

Compare units and hundreds

Units and hundreds both add A+B, so subtracting leaves c₂ = 10(c₃-c₁): a single carry that is a multiple of 10, hence c₂ = 0 and c₃ = c₁.

(A+B+c₂)-(A+B)=(D+10c₃)-(D+10c₁) → c₂=10(c₃-c₁) → c₂=0, c₃=c₁
3STEP 3

Force every carry to zero and C to zero

The tens column now reads C+c₁=0, and digits are never negative, so C = 0 and c₁=0; then c₃=0 and the thousands column gives c₄=0.

C+c₁=0→ C=0, c₁=0; c₃=c₁=0; C+c₃=10c₄→ c₄=0
4STEP 4

Reduce the whole puzzle to A + B = D

With no carries the units, hundreds and ten-thousands columns all say A+B=D with A+B ≤ 9, and distinctness then comes free once A ≠ B.

A+B=D, A,B ≥ 1, A ≠ B, C=0, D=A+B ≤ 9
5STEP 5

Sweep D from smallest to largest

Smallest is 1+2=3, largest is 4+5=9, and every value between is a reachable sum, so D runs 3 through 9 — 7 values, choice (C).

D_min=1+2=3, D_max=9, D∈{3,4,5,6,7,8,9} → 7 values → (C)
Answer
7
Test one case fully. Take A=1, B=2, so C=0 and D=A+B=3. Then ABBCB=12202 and BCADA=20131, and 12202+20131=32333, which is exactly DBDDD=32333 with D=3, B=2. The pattern holds, and the four digits 1,2,0,3 are distinct. Since the same A+B=D rule works for every valid pair, and D runs over the seven integers 3 through 9, the count of 7 is consistent with answer (C).
💡Key takeaway

Give each hidden carry a name; the columns then force C=0 and A+B=D, so D can be any digit from 3 to 9 — seven values in all.

  • Name a carry for each column
  • Compare units and hundreds
  • Force every carry to zero and C to zero
  • Reduce the whole puzzle to A + B = D
  • Sweep D from smallest to largest

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