Difference between revisions of "2010 AMC 10B Problems/Problem 18"

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Revision as of 23:07, 13 July 2011 (view source) Yilun (talk \| contribs) ← Older edit		Latest revision as of 19:45, 26 May 2020 (view source) Jbala (talk \| contribs) (Changed redirect target from 2010 AMC 12B Problems/ Problem 16 to 2010 AMC 12B Problems/Problem 16) (Tag: Redirect target changed)
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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>	+	#redirect [[2010 AMC 12B Problems/Problem 16]]