2010 AMC 10B Problems/Problem 18: Difference between revisions
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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 | 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 
into 
. For to be divisable by three we can either have 
be a multiple of 3 or 
be a multiple of three. Adding the probability of these two being divisable by 3 we get that the probability is 
