2006 AMC 8 Problems/Problem 24

Problem

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

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

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

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

Method 1: Test examples Method 2: When you do bash:

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

And now 0+1=2. I mean, 1. So the answer is \boxed{\textbf{(A)}

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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