AMC 8 · 2009 · #24
Easy mode Grade 4Problem
Each letter , , , and stands for a single digit (a number from to ). 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 plus the two-digit number gives the two-digit number . The second is a subtraction: minus gives the one-digit number .
Both of these have to be true at the same time. What digit does stand for?
Pick an answer.
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
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) $9$
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
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.
💡 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 $0$.
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 $A = 5$.
💡 Borrowing in column subtraction is the standard Grade 4 algorithm; the borrow turns the impossible "$0 - A = A$" into the solvable "$10 - A = A$".
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 $C = A - 1 = 4$.
💡 Tool #7: once the ones column is solved, the tens column becomes its own small subproblem with one unknown.
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 $D = 9$.
💡 With $A$ and $C$ pinned down, the tens column of the addition reads off $D$ directly.
4.NBT.B.4 Read the ones column of the addition: $B + A$ ends in $A$. So $B + A \equiv A \p 4.NBT.B.4 Now use the ones column of the subtraction: $B - A$ ends in $A$. Plugging in $B 4.NBT.B.4 Use the tens column of the subtraction. The borrow took $1$ away from the tens, 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.4Fluently 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.
⭐ 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.