AMC 8 · 2006 · #24
Grade 6 number-theoryProblem
In the multiplication problem below A, B, C, D are different digits. What is A+B?
×CADBCCADD
Pick an answer.
AMC 8 2006 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 multiplication $\overline{ABA} \times \overline{CD} = \overline{CDCD}$, the letters $A, B, C, D$ are four different digits. Find $A + B$.
Givens: $\overline{ABA}$ is a 3-digit number whose hundreds and ones digits are both $A$; $\overline{CD}$ is a 2-digit number with tens digit $C$ and ones digit $D$; $\overline{CDCD}$ is a 4-digit number whose digits are $C, D, C, D$ in that order; $A, B, C, D$ are four different digits; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $9$
Unknowns: The value of $A + B$
Understand
Restated: In the multiplication $\overline{ABA} \times \overline{CD} = \overline{CDCD}$, the letters $A, B, C, D$ are four different digits. Find $A + B$.
Givens: $\overline{ABA}$ is a 3-digit number whose hundreds and ones digits are both $A$; $\overline{CD}$ is a 2-digit number with tens digit $C$ and ones digit $D$; $\overline{CDCD}$ is a 4-digit number whose digits are $C, D, C, D$ in that order; $A, B, C, D$ are four different digits; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $9$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #13 Convert to Algebra
The number $\overline{CDCD}$ is just the two-digit block $\overline{CD}$ written twice. Tool #5 (Look for a Pattern) spots that this repetition is the same trick as $\overline{abab} = \overline{ab} \times 101$ (similar to how $\overline{aa} = a \times 11$). Once we name that pattern, Tool #13 (Convert to Algebra) turns the multiplication line into a one-line equation $\overline{ABA} \times \overline{CD} = \overline{CD} \times 101$, which we can divide by $\overline{CD}$ to read $\overline{ABA}$ directly. No guessing needed.
Execute — Answer: A
5.NBT.A.1 Step 1 - Spot the repetition in $\overline{CDCD}$.
- The 4-digit number reads as the 2-digit block $\overline{CD}$ copied into the thousands-hundreds slot and again into the tens-ones slot.
- Use place value to expand it.
💡 Repeating a 2-digit block in a 4-digit slot multiplies the block by $101$ — the same Grade 5 place-value idea that makes $\overline{aa} = a \cdot 11$.
6.EE.A.2 Step 2 Rewrite the multiplication line as an equation and substitute the pattern from Step 1.
💡 Naming the unknown 3-digit number as $\overline{ABA}$ and rewriting the column-multiplication as a single equation is the Grade 6 "variables in expressions" move.
6.EE.B.7 Step 3 - Cancel the common factor $\overline{CD}$ from both sides.
- Because $C \neq 0$, the number $\overline{CD}$ is between $10$ and $99$, never zero, so division is safe.
💡 Dividing both sides by the same nonzero quantity is the Grade 6 one-step equation move — and it makes the answer fall out without solving for $C$ or $D$ at all.
4.NBT.A.2 Step 4 - Read off $A$ and $B$ from $\overline{ABA} = 101$, then add.
- The digits of $101$ are $1$, $0$, $1$, so $A = 1$ and $B = 0$.
- (As a quick check that all four letters can still be different: $C$ and $D$ just need to be two different digits drawn from $\{2, 3, 4, 5, 6, 7, 8, 9\}$ with $C \neq 0$ — plenty of valid choices exist, for example $C = 2, D = 3$.)
💡 Reading individual digits out of a 3-digit number is Grade 4 place-value, the same skill used to read $101$ as one hundred and one.
5.NBT.A.1 Spot the repetition in $\overline{CDCD}$. The 4-digit number reads as the 2-digi 6.EE.A.2 Rewrite the multiplication line as an equation and substitute the pattern from S 6.EE.B.7 Cancel the common factor $\overline{CD}$ from both sides. Because $C \neq 0$, th 4.NBT.A.2 Read off $A$ and $B$ from $\overline{ABA} = 101$, then add. The digits of $101$ Review
Reasonableness: Test the answer with a concrete choice of $C$ and $D$. Pick $C = 2, D = 3$, so $\overline{CD} = 23$. Then $\overline{ABA} \times 23 = 101 \times 23 = 2323 = \overline{CDCD}$, exactly as required. Digits used: $A = 1, B = 0, C = 2, D = 3$ — all four different. The product checks out, and $A + B = 1$ matches choice (A). Try $C = 5, D = 7$ to be sure: $101 \times 57 = 5757$ — same pattern, same $A + B$. The answer is independent of the specific $C, D$ chosen, which is exactly why the problem can be solved without finding them.
Alternative: Tool #6 (Guess and Check) on the answer choices. Since $\overline{ABA}$ is a palindrome 3-digit number that must divide some $\overline{CDCD}$ exactly, test $A = 1$ first (smallest choice): with $B = 0$, $\overline{ABA} = 101$. Then $101 \times \overline{CD} = \overline{CDCD}$ for every two-digit $\overline{CD}$, since multiplying by $101$ shifts and adds. The pattern works on the first guess, so $A + B = 1$ confirms (A).
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers using base-ten numerals; recognize that each place is ten times the place to its right (Reading the digits of $101$ to identify $A = 1$ and $B = 0$ from $\overline{ABA} = 101$.)5.NBT.A.1Recognize that in a multi-digit number, a digit in one place represents ten times what it represents in the place to its right (Expanding $\overline{CDCD}$ as $(10C+D) \cdot 100 + (10C+D) = (10C+D) \cdot 101$ via place-value reasoning.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Translating the column-multiplication diagram into the algebraic equation $\overline{ABA} \times \overline{CD} = \overline{CD} \cdot 101$.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form $px = q$ (Dividing both sides of $\overline{ABA} \times \overline{CD} = 101 \cdot \overline{CD}$ by the nonzero factor $\overline{CD}$ to isolate $\overline{ABA} = 101$.)
⭐ When a block of digits repeats — like $\overline{CDCD}$ being $\overline{CD}$ written twice — it always factors out as that block times $101$. Spot that pattern and the multiplication puzzle collapses to a Grade 6 one-step equation.
⭐ When a block of digits repeats — like $\overline{CDCD}$ being $\overline{CD}$ written twice — it always factors out as that block times $101$. Spot that pattern and the multiplication puzzle collapses to a Grade 6 one-step equation.
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