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> |
Revision as of 16:28, 17 January 2022
Problem
In the multiplication problem below ,
,
,
are different digits. What is
?
Video solution
https://youtu.be/sd4XopW76ps -Happytwin
https://youtu.be/7an5wU9Q5hk?t=3080
https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ
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.