AMC 8 · 2006 · #24

Grade 6 number-theory

Archetype Small-Domain Constrained Search

digit-constraintsdigit-decompositionplace-valuelogical-deduction pattern-recognitiondigit-constraints ↑ Prerequisites: place-valuemulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

In the multiplication problem below $A$ , $B$ , $C$ , $D$ are different digits. What is $A+B$ ?

$\begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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 $A + B$.

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 Look for a Pattern 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.

$$\overline{CDCD} = (10C + D) \cdot 100 + (10C + D) = (10C + D) \cdot 101 = \overline{CD} \cdot 101$$

💡 Repeating a 2-digit block in a 4-digit slot multiplies the block by $101$ — the same Grade 5 place-value idea that makes $\overline{aa} = a \cdot 11$.

#13 Convert to Algebra 6.EE.A.2 Step 2

Rewrite the multiplication line as an equation and substitute the pattern from Step 1.

$$\overline{ABA} \times \overline{CD} = \overline{CDCD} = \overline{CD} \cdot 101$$

💡 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.

#13 Convert to Algebra 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.

$$\overline{ABA} = \dfrac{\overline{CD} \cdot 101}{\overline{CD}} = 101$$

💡 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.

#5 Look for a Pattern 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 $C = 2, D = 3$.)

$$A + B = 1 + 0 = 1 \;\Longrightarrow\; \textbf{(A)}$$

💡 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.

[1] #5 5.NBT.A.1 Spot the repetition in $\overline{CDCD}$. The 4-digit number reads as the 2-digi

[2] #13 6.EE.A.2 Rewrite the multiplication line as an equation and substitute the pattern from S

[3] #13 6.EE.B.7 Cancel the common factor $\overline{CD}$ from both sides. Because $C \neq 0$, th

[4] #5 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.2 Read 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.1 Recognize 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.2 Write, 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.7 Solve 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

More like this