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 by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=3080

Video Solution

https://www.youtube.com/watch?v=dQw4w9WgXcQ

https://www.youtube.com/watch?v=Y4DXkhYthhs ~David

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

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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