AMC 8 · 2009 · #24

Easy mode Grade 4

Problem

Each letter $A$ , $B$ , $C$ , and $D$ stands for a single digit (a number from $0$ to $9$ ). Different letters can stand for the same digit or different digits.

Picture two simple column-addition and column-subtraction problems written with these letter-digits:

$\begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular}$ and $\begin{tabular}{ccc}&A&B\\ -&C&A\\ \hline &&A\end{tabular}$

The first one is an addition: the two-digit number $AB$ plus the two-digit number $CA$ gives the two-digit number $DA$ . The second is a subtraction: $AB$ minus $CA$ gives the one-digit number $A$ .

Both of these have to be true at the same time. What digit does $D$ stand for?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2009 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Four digits $A$, $B$, $C$, $D$ satisfy two cryptarithms at once. Adding the two-digit numbers $\overline{AB} + \overline{CA}$ gives the two-digit number $\overline{DA}$, and subtracting them gives the one-digit number $A$ (so the tens digit of the difference is $0$). Find the digit $D$.

Givens: $\overline{AB} + \overline{CA} = \overline{DA}$ (column form, no extra carry beyond the tens place); $\overline{AB} - \overline{CA} = A$ (a single digit, so the tens digit of the difference is $0$); $A$, $B$, $C$, $D$ are digits ($0$–$9$); Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$

Unknowns: The digit represented by $D$

Understand

Plan

Primary tool: #3 Write an Equation

Secondary: #7 Identify Subproblems

Cryptarithms are pure place-value puzzles. Tool #3 (Write an Equation) turns each column of the addition and subtraction into a small equation in the unknown digits (with carry/borrow bits). Tool #7 (Identify Subproblems) tells us to work column by column — start with the ones column, which forces $B$, then move to the tens column, which forces $A$, $C$, and finally $D$ — instead of trying to guess all four digits at once.

Execute — Answer: E

#3 Write an Equation 4.NBT.B.4 Step 1

Read the ones column of the addition: $B + A$ ends in $A$.
So $B + A \equiv A \pmod{10}$, which forces $B \equiv 0 \pmod{10}$.
Since $B$ is a digit, $B = 0$ and there is no carry into the tens column.

$$B + A \equiv A \pmod{10} \;\Rightarrow\; B = 0$$

💡 Adding $A$ to something and getting back $A$ in the ones place is exactly the Grade 4 "add multi-digit whole numbers using place value" idea — the only digit that vanishes under addition is $0$.

#3 Write an Equation 4.NBT.B.4 Step 2

Now use the ones column of the subtraction: $B - A$ ends in $A$.
Plugging in $B = 0$ gives $0 - A \equiv A \pmod{10}$.
Since $A \neq 0$, we must borrow $10$ from the tens column, turning the ones-column equation into $10 - A = A$.
Solving gives $A = 5$.

$$10 - A = A \;\Rightarrow\; 2A = 10 \;\Rightarrow\; A = 5$$

💡 Borrowing in column subtraction is the standard Grade 4 algorithm; the borrow turns the impossible "$0 - A = A$" into the solvable "$10 - A = A$".

#7 Identify Subproblems 4.NBT.B.4 Step 3

Use the tens column of the subtraction.
The borrow took $1$ away from the tens, so $(A - 1) - C = 0$.
With $A = 5$, this gives $C = A - 1 = 4$.

$$(A - 1) - C = 0 \;\Rightarrow\; C = A - 1 = 4$$

💡 Tool #7: once the ones column is solved, the tens column becomes its own small subproblem with one unknown.

#3 Write an Equation 4.NBT.B.4 Step 4

Finally, use the tens column of the addition.
There was no carry from the ones column (since $B + A = 0 + 5 = 5 < 10$), so $A + C = D$.
Substituting $A = 5$ and $C = 4$ gives $D = 9$.

$$D = A + C = 5 + 4 = 9 \;\Rightarrow\; \textbf{(E)}$$

💡 With $A$ and $C$ pinned down, the tens column of the addition reads off $D$ directly.

[1] #3 4.NBT.B.4 Read the ones column of the addition: $B + A$ ends in $A$. So $B + A \equiv A \p

[2] #3 4.NBT.B.4 Now use the ones column of the subtraction: $B - A$ ends in $A$. Plugging in $B

[3] #7 4.NBT.B.4 Use the tens column of the subtraction. The borrow took $1$ away from the tens,

[4] #3 4.NBT.B.4 Finally, use the tens column of the addition. There was no carry from the ones c

Review

Reasonableness: Plug the digits back in: $A = 5$, $B = 0$, $C = 4$, $D = 9$, so $\overline{AB} = 50$ and $\overline{CA} = 45$. Then $50 + 45 = 95 = \overline{DA}$ (tens digit $D = 9$, ones digit $A = 5$) and $50 - 45 = 5 = A$ (a single-digit answer with tens digit $0$). Both cryptarithms check out.

Alternative: Tool #6 (Guess and Check) on the addition column $A + C = D$: since the subtraction $\overline{AB} - \overline{CA}$ is a single digit, the two two-digit numbers differ by less than $10$, so $\overline{AB}$ and $\overline{CA}$ are very close. Trying $A = 5$ (the smallest answer choice for the related quantity) gives $\overline{AB} = 50$ and $\overline{CA} = 45$, whose difference $5$ is indeed $A$. The sum is $95$, so $D = 9$.

CCSS standards used (min grade 4)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Reading each column of the addition and subtraction (with its carry or borrow) as a small equation in the unknown digits, then solving column by column to find $B = 0$, $A = 5$, $C = 4$, and $D = 9$.)

⭐ This AMC 8 problem only needs the Grade 4 standard column-by-column add/subtract algorithm — once you read each column carefully, the digits fall out one at a time.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 4)

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