2010 AMC 12B Problems/Problem 16
Problem 16
Positive integers , , and are randomly and independently selected with replacement from the set . What is the probability that is divisible by ?
Solution 1
We group this into groups of , because .
If , we are done. There is a probability of that that happens.
Otherwise, we have , which means that . So either or which will lead to the property being true. There are a chance for each bundle of cases to be true. Thus, the total for the cases is . But we have to multiply by because this only happens with a chance. So the total is actually .
The grand total is
Solution 2 (Minor change from Solution 1)
Just like solution 1, we see that there is a chance of and chance of
Now, we can just use PIE (Principals of Inclusion and Exclusion) to get our answer to be
-Conantwiz2023
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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