AMC 8 · 2003 · #14

Grade 4 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

digit-constraintsplace-valuedigit-decompositionmulti-digit-arithmeticparity caseworksystematic-enumerationdigit-constraints ↑ Prerequisites: place-valuemulti-digit-arithmeticdigit-constraints

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Problem

In the column addition TWO + TWO = FOUR, each letter is a different digit 0-9. Given T = 7 and O even, find the only possible value of W.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

A cryptarithm is one big puzzle, but column addition hands us four ready-made subproblems (Tool #7): the thousands column, the hundreds column, the units column, and the tens column. Solve them in the order that locks in the most letters first — thousands → hundreds → units → tens. Once T, F, O, R are pinned down, the tens column reduces to choosing W from a tiny list. There Tool #3 (Eliminate Possibilities) takes over: test W = 0, 1, 2, 3, 4 against the "all digits different" rule and cross out the bad ones.

1STEP 1

Thousands column: the extra leading digit is a carry, and doubling a three-digit number caps that carry, so F = 1.

F = 1

2STEP 2

Hundreds column: 14 + c₂ = 10 + O gives O = 4 + c₂, and the even-O condition selects O = 4 with carry c₂ = 0.

14 + c₂ = 10 + O → O = 4 + c₂. So O = 4 (if c₂ = 0) or O = 5 (if c₂ = 1). The given O even forces O = 4 and c₂ = 0.

3STEP 3

Units column: O = 4 gives O + O = 8, still under ten, so R = 8 and nothing carries into the tens (c₁ = 0).

4 + 4 = 8 ＜ 10 → R = 8 and c₁ = 0.

4STEP 4

Tens column: with both carries 0, W + W = U and 2W under ten; only W = 3 gives a U (6) that no other letter already uses.

W = 0 → U = 0 = W (collision). W = 1 = F (used). W = 2 → U = 4 = O (used). W = 3 → U = 6, which is new and unused. W = 4 = O (used). Only W = 3 survives, so the answer is (D).

Answer

Plug the full assignment T = 7, W = 3, O = 4, F = 1, U = 6, R = 8 back in: 734 + 734 = 1468, which matches FOUR = 1468. All six letters use six different digits {1, 3, 4, 6, 7, 8}, T = 7, and O = 4 is even — every condition holds. The other answer choices fail the "all different" rule (or push U ≥ 10), confirming W = 3 is the unique survivor.

💡Key takeaway

A cryptarithm is just column addition in disguise. Solve one column at a time, lock in F, then O, then R, and the tens column hands you a short list for W — only W = 3 keeps every letter on a different digit.

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