Difference between revisions of "2006 AMC 8 Problems/Problem 24"
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==Solution 3== | ==Solution 3== | ||
− | Because <math>DA=D</math>, <math>A</math> must be 1. Writing it out, we can see that | + | Because <math>DA=D</math>, <math>A</math> must be <math>1</math>. Writing it out, we can see that |
<math>1B1 | <math>1B1 | ||
X CD | X CD | ||
=0D0D | =0D0D | ||
− | + | +C0C0</math> | |
− | So, <math>B</math> must be 0. 1+0=1. Thus, our answer is | + | So, <math>B</math> must be <math>0</math>. <math>1+0=1</math>. Thus, our 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 12:39, 9 September 2023
Contents
Problem
In the multiplication problem below ,
,
,
are different digits. What is
?
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
Solution 3
Because ,
must be
. Writing it out, we can see that
So,
must be
.
. Thus, our 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.