AMC 8 · 2003 · #14
Grade 4 number-theoryProblem
In this addition problem, each letter stands for a different digit.
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If T = 7 and the letter O represents an even number, what is the only possible value for W?
Pick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $W$.
Givens: Column addition: $TWO + TWO = FOUR$; Each letter stands for a different digit; $T = 7$; $O$ is an even digit; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The digit $W$
Understand
Restated: 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 $W$.
Givens: Column addition: $TWO + TWO = FOUR$; Each letter stands for a different digit; $T = 7$; $O$ is an even digit; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
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 $\to$ hundreds $\to$ units $\to$ 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.
Execute — Answer: D
4.NBT.A.2 Step 1 - Subproblem 1 — thousands column.
- The sum has one more digit than the addends, so $F$ is the carry out of the hundreds column.
- Since the hundreds column sum is at most $T + T + 1 = 7 + 7 + 1 = 15$, that carry must be $1$.
💡 Grade 4 place-value reasoning: the digit that "falls out the top" of a column is the carry, and doubling a $3$-digit number can carry at most $1$.
4.NBT.B.4 Step 2 - Subproblem 2 — hundreds column.
- $T + T + c_2 = 10F + O$ where $c_2$ is the carry from the tens column ($c_2 \in \{0, 1\}$).
- Plug in $T = 7$ and $F = 1$.
💡 Grade 4 standard-algorithm addition: a column's digit equals the column sum minus $10 \cdot$ (carry out). The even/odd filter picks one case.
4.NBT.B.4 Step 3 - Subproblem 3 — units column.
- With $O = 4$, compute $O + O$ to find $R$ and the carry $c_1$ into the tens column.
💡 No regrouping needed here — the column sum is a single digit, so $R$ is exactly that digit and nothing carries left.
4.OA.B.4 Step 4 - Subproblem 4 — tens column.
- With $c_1 = 0$ in and $c_2 = 0$ out, the tens column gives $W + W = U$ with $2W < 10$, so $W \in \{0, 1, 2, 3, 4\}$.
- Now apply Tool #3 (Eliminate) using "all letters are different digits" and the already-used set $\{T, F, O, R\} = \{7, 1, 4, 8\}$.
💡 Grade 4 "reason about whole numbers": $W$ must double to a single digit and must keep every letter on a different value. Five candidates, one survivor.
4.NBT.A.2 Subproblem 1 — thousands column. The sum has one more digit than the addends, so 4.NBT.B.4 Subproblem 2 — hundreds column. $T + T + c_2 = 10F + O$ where $c_2$ is the carry 4.NBT.B.4 Subproblem 3 — units column. With $O = 4$, compute $O + O$ to find $R$ and the c 4.OA.B.4 Subproblem 4 — tens column. With $c_1 = 0$ in and $c_2 = 0$ out, the tens column Review
Reasonableness: 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 $U \ge 10$), confirming $W = 3$ is the unique survivor.
Alternative: Tool #6 (Guess and Check) on the answer choices is even faster on a timed test: write $T = 7$ and just try each of $W = 0, 1, 2, 3, 4$, compute $2 \times \overline{7W4} = 2 \times (700 + 10W + 4)$, and look for a result of the form $1\,4\,U\,8$ with all six digits different. Only $W = 3$ gives $1468$, matching (D).
CCSS standards used (min grade 4)
4.NBT.A.2Read and write multi-digit whole numbers using base-ten numerals; recognize place value (Identifying that the leading $F$ is the carry out of the hundreds column and bounding it to $F = 1$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Column-by-column reasoning $T + T + c_2 = 10F + O$ and $O + O = 10 c_1 + R$ to pin down $O = 4$ and $R = 8$.)4.OA.B.4Find factor pairs and recognize multiples; reason about whole numbers in a given range (Listing $W \in \{0, 1, 2, 3, 4\}$ from $2W < 10$ and eliminating candidates that double to a used digit or collide with $W$ itself.)
⭐ 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.
⭐ 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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