Difference between revisions of "2006 AMC 8 Problems/Problem 24"
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Because <math>DA=D</math>, <math>A</math> must be <math>1</math>. 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 | ||
− | + | \cdot CD | |
=0D0D | =0D0D | ||
+C0C0</math> | +C0C0</math> |
Revision as of 23:35, 17 January 2024
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/watch?v=Y4DXkhYthhs ~David
Solution 1
, 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
.
- J.L.L (Feel free to edit)
yoooooooooooooooooooooo
Solution 4
We know that is 1 because after you multiply the first column
and
you get
. Noticing that the value of
does not matter as long it is a
digit number, let's give the value of the
digit number
. After doing some multiplication using the traditional method, our product is
. We know that our end product has to be
, so since our value of
is 10 our product should be
. Therefore,
is 0 because
is in the spot of
. We are not done as the problem is asking for the value of
which is just
.
- LearnForEver
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.