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Difference between revisions of "2010 AMC 10B Problems/Problem 18"
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First we factor <math>abc+bc+c</math> into <math>a(b(c+1)+1)</math>. For <math>a(b(c+1)+1)</math> to be divisable 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 divisable by 3 we get that the probability is <math> frac\{13}{27}</math>
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First we factor <math>abc+bc+c</math> into <math>a(b(c+1)+1)</math>. For <math>a(b(c+1)+1)</math> to be divisable 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 divisable by 3 we get that the probability is <math> \frac{13}{27}</math>

Revision as of 23:07, 13 July 2011

First we factor $abc+bc+c$ into $a(b(c+1)+1)$. For $a(b(c+1)+1)$ to be divisable by three we can either have $a$ be a multiple of 3 or $b(c+1)+1$ be a multiple of three. Adding the probability of these two being divisable by 3 we get that the probability is $\frac{13}{27}$

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