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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 $\frac{8}{27}$ of the cheese remains.

This problem is taken from the UKMT Mathematical Challenges.
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