Difference between revisions of "2006 AMC 8 Problems/Problem 24"
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==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{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array} </cmath> | <cmath> \begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array} </cmath> | ||
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<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> | ||
| − | ==Video | + | ==Video Solution by OmegaLearn== |
| + | https://youtu.be/7an5wU9Q5hk?t=3080 | ||
| + | |||
| + | ==Video Solution== | ||
https://youtu.be/sd4XopW76ps -Happytwin | https://youtu.be/sd4XopW76ps -Happytwin | ||
| − | https:// | + | https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ |
| − | https://www.youtube.com/ | + | https://www.youtube.com/watch?v=Y4DXkhYthhs ~David |
==Solution== | ==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{\textbf{(A)}\ 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>. | ||
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==Solution 2== | ==Solution 2== | ||
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Method 2: Bash it out to <math>waste</math> time | Method 2: Bash it out to <math>waste</math> time | ||
| − | <math>(100A+10B+A)(10C+D) = | + | <math>(100A+10B+A)(10C+D) = 1000C+100D+10C+D</math> |
| − | <math> | + | <math>1000AC+100BC+10AC+100AD+10BD+AD=1010C+101D</math> |
| − | |||
| − | |||
| − | |||
| − | <math> | + | <math>1010AC+100BC+101AD = 1010C + 101D</math> |
| − | <math> | + | <math>1010(A-1)(C) + 101(A-1)D + 100CB + 10BD=0</math> |
| + | <math>A=1</math> | ||
| + | ,<math>B=0</math> | ||
| − | And <math>0+1=1</math>, thus the answer is <math>A</math> | + | And <math>0+1=1</math>, thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math> |
==See Also== | ==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}} | {{MAA Notice}} | ||
Revision as of 19:44, 15 April 2023
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://youtu.be/sd4XopW76ps -Happytwin
https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ
https://www.youtube.com/watch?v=Y4DXkhYthhs ~David
Solution
, so 
. Therefore, 
and 
, so 
.
Solution 2
Method 1: Test 
Method 2: Bash it out to 
time
,
And 
, thus the answer is 
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. 
