Difference between revisions of "2006 AMC 8 Problems/Problem 24"
Brightknight (talk | contribs) |
Brightknight (talk | contribs) (→Solution) |
||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
| + | |||
| + | 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> | ||
| + | |||
| + | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9 </math> | ||
| + | |||
==Solution== | ==Solution== | ||
Revision as of 18:36, 29 April 2012
Problem
In the multiplication problem below 
, 
, 
, 
and are different digits. What is 
?
\[\begin{tabular}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{tabular}\] (Error compiling LaTeX. Unknown error_msg)
Solution
CDCD = CD*101, so ABA = 101. Therefore, A = 1 and B = 0, so A+B=1+0=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 | ||
