## Task 2 - 020932

A task from the year 1494:

At the top of a tree, which is 60 ells high, sits a mouse, and at the bottom on the ground is a cat. The mouse climbs down $\frac{1}{2}$ ell each day and climbs back up $\frac{1}{6}$ ell each night. The cat climbs up 1 ell each day and climbs back down $\frac{1}{4} \times 4$ ells each night.

After how many days does the cat reach the mouse?
Let $x$ be the number of days. Over the course of a day (including the night), the mouse approaches the cat by $\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$ ell.

At the same time, the cat approaches the mouse by $1-\frac{1}{4}=\frac{3}{4}$ ell. In total, the cat and the mouse get $\frac{13}{12}$ ell closer. This is true for the first $x-1$ days. On the last day, the night should not be included.

On this last day, the mouse approaches the cat by $\frac{1}{2}$ ell and the cat approaches the mouse by 1 ell. Therefore, the following must hold:

$$
\frac{13}{12}(x-a)+\frac{3}{2} \geq 60
$$

which is true for $x \geq 55$. On the 55th day, the cat reaches the mouse.