Difference between revisions of "2006 AMC 8 Problems/Problem 24"

Revision as of 19:38, 29 April 2012

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

\[\begin{tabular}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{tabular}\] (Error compiling LaTeX. Unknown error_msg)

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

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

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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 ==Solution==
-<math>CDCD = CD \cdot 101</math>, so <math>ABA = 101</math>. Therefore, <math>A = 1</math> and <math>B = 0</math>, so <math>A+B=1+0=\boxed{1}</math>.
+<math>CDCD = CD \cdot 101</math>, so <math>ABA = 101</math>. Therefore, <math>A = 1</math> and <math>B = 0</math>, so <math>A+B=1+0=\boxed{\textbf{(A)}\ 1}</math>.
 {{AMC8 box|year=2006|n=II|num-b=23|num-a=25}}