2010 AMC 10B Problems/Problem 18
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Contents
Problem
Positive integers ,
, and
are randomly and independently selected with replacement from the set
. What is the probability that
is divisible by
?
Solution 1
First we factor into
. For
to be divisible by three we can either have
be a multiple of 3 or
be a multiple of three. Adding the probability of these two being divisible by 3 we get that the probability is
Solution 2
We look at the probability of each term being 3. 1/3 for the first term, 1/9 for the second term, and 1/27 for the third term. So the solution is 13/27 or E.
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AMC 10 Problems and Solutions |
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