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

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In the multiplication problem below <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and  are different digits. What is <math>A+B</math>?  
 
In the multiplication problem below <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and  are different digits. What is <math>A+B</math>?  
  
<cmath> \begin{tabular}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{tabular} </cmath>
+
<cmath> \begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array} </cmath>
  
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9 </math>
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9 </math>

Revision as of 20:05, 10 March 2015

Problem

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

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

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

Solution

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

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
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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