Difference between revisions of "2003 AMC 10A Problems/Problem 15"
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Therefore, the desired probability is <math>\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}</math> | Therefore, the desired probability is <math>\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}</math> | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/UfzS5griBic | ||
| + | |||
| + | ~savannahsolver | ||
==Controversy== | ==Controversy== | ||
Revision as of 12:19, 14 June 2021
Problem
What is the probability that an integer in the set 
is divisible by 
and not divisible by 
?
Solution
There are 
integers in the set.
Since every 
integer is divisible by , there are 
integers divisible by in the set.
To be divisible by both and , a number must be divisible by 
.
Since every 
integer is divisible by 
, there are 
integers divisible by both and in the set.
So there are 
integers in this set that are divisible by and not divisible by .
Therefore, the desired probability is 
Video Solution by WhyMath
~savannahsolver
Controversy
Due to the wording of the question, it may be taken as "Find the probability that an integer in said set is divisible by 2 and not 3 EXISTS". One example would be 2, which is not a multiple of 3, thus the probability is 1. But because 1 is not an option, we can assume that it was not meant like that.
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
