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2010 AMC 10B Problems/Problem 8

Revision as of 13:53, 24 January 2011 by Jexmudder (talk | contribs)

We see how many common integer factors 48 and 64 share. 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 $\boxed{\mathrm{(E)} 5}$ possibilities for the ticket price.

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