Problem 14 · 2003 AMC 8
Hard
Logic & Word Problems
place-valuecasework
In this addition problem, each letter stands for a different digit.
T W O + T W O ------- F O U R
If T = 7 and the letter O represents an even number, what is the only possible value for W?
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Answer: D — W = 3.
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Hint 1 of 2
TWO + TWO is just TWO doubled, so every column is a digit being doubled (maybe plus a carry) — that's a strong constraint.
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Hint 2 of 2
Start at the most-constrained column. With T = 7, the hundreds give 14 (+ carry), forcing the leading F = 1 and pinning O; then "O is even" finishes it.
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Approach: double TWO column by column, starting where it's pinned
- TWO + TWO = 2 × TWO, and doubling a 3-digit number can only push it to 4 digits by carrying a 1 — so F = 1, the only digit a carry can produce.
- Hundreds column is the most pinned: 7 + 7 = 14, plus a possible carry from the tens, lands the hundreds digit O at 4 or 5. Since O is even, O = 4 — and that means nothing carried in, so the tens column did not overflow.
- Units: O + O = 4 + 4 = 8, so R = 8 with no carry. The tens column is therefore just W + W = U, with no carry either way.
- Now place W: 2W = U must stay under 10 (no overflow) and U must avoid the used digits 7, 1, 4, 8. W = 0→U = 0 (clashes with W and is reused), W = 1, 4 taken, W = 2→U = 4 (taken). Only W = 3 works, giving U = 6. W = 3.
- You'll see this again: in cryptarithms, attack the column with the fewest unknowns first (often the leading carry, which is almost always 1), and let each forced digit cascade into the next column.
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