$\mathrm{Na}$ our planned cottage, we brought the cat Vilda. On Monday, she caught $\frac{1}{2}$ of all the mice, on Tuesday $\frac{1}{3}$ of the remaining, on Wednesday $\frac{1}{4}$ of those left after Tuesday's hunt, and on Thursday only $\frac{1}{5}$ of the remainder. On Friday, the remaining mice preferred to move away. How many mice were originally at the cottage if on Friday, two more mice moved away than Vilda caught on Tuesday? Don't forget to verify that each day's catch was an integer number of mice.

(M. Volfová, M. Dillingerová)
On Monday, the cat caught $\frac{1}{2}$ of all the mice.

On Tuesday, it caught $\frac{1}{3}$ of the remaining $\frac{1}{2}$, i.e., $\frac{1}{6}$ of all the mice; $\frac{1}{2}-\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$ of all the mice remained.

On Wednesday, it caught $\frac{1}{4}$ of $\frac{1}{3}$, i.e., $\frac{1}{12}$ of all the mice; $\frac{1}{3}-\frac{1}{12}=\frac{3}{12}=\frac{1}{4}$ of all the mice remained.

On Thursday, it caught $\frac{1}{5}$ of $\frac{1}{4}$, i.e., $\frac{1}{20}$ of all the mice; $\frac{1}{4}-\frac{1}{20}=\frac{4}{20}=\frac{1}{5}$ of all the mice remained (and they moved away on Friday).

According to the problem, $\frac{1}{5}$ of all the mice is two more than what was caught on Tuesday, i.e., more than $\frac{1}{6}$ of all the mice. Two mice thus represent the difference between $\frac{1}{5}$ and $\frac{1}{6}$ of all the mice. Since $\frac{1}{5}-\frac{1}{6}=\frac{1}{30}$, two mice represent $\frac{1}{30}$ of all the mice, so originally there were $2 \cdot 30=60$ mice on the cottage.

Verification: On Monday, 30 mice were caught, leaving 30; on Tuesday, 10 were caught, leaving 20; on Wednesday, 5 were caught, leaving 15; on Thursday, 3 were caught, leaving 12, which is indeed 2 more than the Tuesday catch.

Alternative solution. If we denote the original number of all the mice as $x$, all the considerations above can be expressed using this unknown. The final consideration can then be solved by the equation as follows:

$$
\begin{aligned}
\frac{1}{5} x & =\frac{1}{6} x+2, \\
6 x & =5 x+60, \\
x & =60 .
\end{aligned}
$$

Evaluation. 1 point for the fractions representing the catches on Tuesday, Wednesday, and Thursday; 1 point for the fraction representing the number of mice that moved away; 1 point for the final result; 1 point for verifying that all catches are whole numbers.