Difference between revisions of "2006 AMC 8 Problems/Problem 24"
Brightknight (talk | contribs) (→Solution) |
|||
| (46 intermediate revisions by 20 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | In the multiplication problem below <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> | + | In the multiplication problem below <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> are different digits. What is <math>A+B</math>? |
| − | <cmath> \begin{ | + | <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> | ||
| − | ==Solution== | + | ==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 | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/IC_2SxI821c | ||
| + | |||
| + | ==Solution 1== | ||
| + | |||
| + | <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>. | ||
| + | |||
| + | ==See Also== | ||
{{AMC8 box|year=2006|n=II|num-b=23|num-a=25}} | {{AMC8 box|year=2006|n=II|num-b=23|num-a=25}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:11, 8 November 2024
Contents
Problem
In the multiplication problem below 
, 
, 
, 
are different digits. What is 
?
![\[\begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}\]](http://latex.artofproblemsolving.com/c/9/8/c98496cee54a981529919a464508835b9321f88e.png)
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
Video Solution by WhyMath
Solution 1
, so 
. Therefore, 
and 
, so 
.
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
