# 2006 AMC 8 Problems/Problem 24

## Problem

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

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

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9$

## Video Solution

https://youtu.be/sd4XopW76ps -Happytwin

## Solution 1

$CDCD = CD \cdot 101$, so $ABA = 101$. Therefore, $A = 1$ and $B = 0$, so $A+B=1+0=\boxed{\textbf{(A)}\ 1}$.

## Solution 2

Method 1: Test $examples.$

Method 2: Bash it out to $waste$ time

$(100A+10B+A)(10C+D) = 1000C+100D+10C+D$ $1000AC+100BC+10AC+100AD+10BD+AD=1010C+101D$

$1010AC+100BC+101AD = 1010C + 101D$

$1010(A-1)(C) + 101(A-1)D + 100CB + 10BD=0$

$A=1$ and $B=0$.

$0+1=1$, thus the answer is $\boxed{\textbf{(A)}\ 1}$

## Solution 3

Because $DA=D$, $A$ must be $1$. Writing it out, we can see that $1B1 \cdot CD =0D0D +C0C0$ So, $B$ must be $0$. $1+0=1$. Thus, our answer is $\boxed{\textbf{(A)}\ 1}$. - J.L.L (Feel free to edit)

## Solution 4

We know that $A$ is 1 because after you multiply the first column $A$ and $D$ you get $D$. Noticing that the value of $CD$ does not matter as long it is a $2$ digit number, let's give the value of the $2$ digit number $CD$ $10$. After doing some multiplication using the traditional method, our product is $1B10$. We know that our end product has to be $CDCD$, so since our value of $CD$ is 10 our product should be $1010$. Therefore, $B$ is 0 because $B$ is in the spot of $0$. We are not done as the problem is asking for the value of $A+B$ which is just $\boxed{\textbf{(A)}\ 1}$.

- LearnForEver