Difference between revisions of "2010 AMC 10B Problems/Problem 8"

 
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We see how many common integer factors 48 and 64 share.
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#redirect [[2010 AMC 12B Problems/Problem 3]]
Of the factors of 48 - 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48; only 1, 2, 4, 8, and 16 are factors of 64.
 
So there are <math>\boxed{\mathrm {(E)} 5}</math> possibilities for the ticket price.
 

Latest revision as of 20:36, 26 May 2020