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'Weekly Problem 10 - 2013' printed from http://nrich.maths.org/

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After the first mouse has eaten, $\frac{2}{3}$ of the cheese remains. After the second mouse has eaten, $\frac{2}{3}$ of $\frac{2}{3}$, that is $\frac{4}{9}$, of the cheese remains. Finally, after the third mouse has eaten, $\frac{2}{3}$ of $\frac{4}{9}$, that is $\frac{8}{27}$ of the cheese remains. So the mice ate $\frac{19}{27}$ of the cheese.

This problem is taken from the UKMT Mathematical Challenges.

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