AMC 8 · 2003 · #14

Easy mode Grade 4

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Problem

Look at this addition puzzle. Each letter stands for a different digit ( $0$ through $9$ ).

$\setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$

So $\text{TWO} + \text{TWO} = \text{FOUR}$ . Same letter means same digit. Different letters mean different digits.

We are told two things: $T = 7$ , and $O$ is an even number.

With those clues, $W$ can only be one value. What is $W$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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

#7 Identify Subproblems 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 $1$.

$$F = 1$$

💡 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 $1$.

#7 Identify Subproblems 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 $F = 1$.

$14 + c_2 = 10 + O \;\Rightarrow\; O = 4 + c_2$. So $O = 4$ (if $c_2 = 0$) or $O = 5$ (if $c_2 = 1$). The given $O$ even forces $O = 4$ and $c_2 = 0$.

💡 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.

#7 Identify Subproblems 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.

$4 + 4 = 8 < 10 \;\Rightarrow\; R = 8$ and $c_1 = 0$.

💡 No regrouping needed here — the column sum is a single digit, so $R$ is exactly that digit and nothing carries left.

#3 Eliminate Possibilities 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 $\{T, F, O, R\} = \{7, 1, 4, 8\}$.

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

💡 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.

[1] #7 4.NBT.A.2 Subproblem 1 — thousands column. The sum has one more digit than the addends, so

[2] #7 4.NBT.B.4 Subproblem 2 — hundreds column. $T + T + c_2 = 10F + O$ where $c_2$ is the carry

[3] #7 4.NBT.B.4 Subproblem 3 — units column. With $O = 4$, compute $O + O$ to find $R$ and the c

[4] #3 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.2 Read 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.4 Fluently 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.4 Find 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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