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| − | ==Problem==
| + | #redirect [[2010 AMC 12B Problems/Problem 16]] |
| − | Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?
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| − | <math>\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}</math>
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| − | ==Solution 1==
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| − | 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==
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| − | 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.
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| − | ==See Also==
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| − | {{AMC10 box|year=2010|ab=B|num-b=17|num-a=19}}
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