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Talk:2012 AMC 10A Problems/Problem 20

Revision as of 15:18, 10 February 2012 by Bobcroucher (talk | contribs)

I get $\frac{49}{256}$ for this problem. I used your same logic, but noted that the center square can start out white or black, giving us: $\frac{2\cdot7\cdot7}{2^9}$.

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