Difference between revisions of "2010 AMC 10B Problems/Problem 18"
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==Solution 1== | ==Solution 1== | ||
| − | First we factor <math>abc+ | + | First we factor <math>abc+ab+a</math> into <math>a(b(c+1)+1)</math>. For <math>a(b(c+1)+1)</math> to be divisible by three we can either have <math>a</math> be a multiple of 3 or <math>b(c+1)+1</math> be a multiple of three. Adding the probability of these two being divisible by 3 we get that the probability is <math>\boxed{\textbf{(E)}\ \frac{13}{27}}</math> |
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==Solution 2== | ==Solution 2== | ||
Revision as of 21:27, 2 February 2012
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 | |
| 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 | ||
