A servant confesses to his master that he stole one liter of wine from a 250-liter barrel every day for 50 days and always replaced the missing amount with water. What percentage of wine was still in the barrel?
After the first theft, there were 2491 liters of wine left in the barrel. On the second day, the servant stole 11 liters of the mixture. In this, $\frac{249}{250} 1$ liter was wine, so in the barrel

$$
249-\frac{249}{250}=\frac{249^{2}}{250} 1 \text { liter of wine remained. }
$$

On the third day, in the 1 liter of mixture taken out, $\frac{249^{2}}{250^{2}} 1$ liter was wine, so

$$
\frac{249^{2}}{250}-\frac{249^{2}}{250^{2}}=\frac{249^{3}}{250^{2}} 1 \text { liter of wine remained. }
$$

Continuing this procedure, we find that on the fiftieth day

$$
\frac{249^{50}}{250^{49}} 1 \text { liter of wine remained in the barrel. }
$$

If the barrel contained $x \%$ of wine, then

$$
x=\frac{\frac{249^{50}}{250^{49}}}{250} 100=100\left(\frac{249}{250}\right)^{50}=81,846 \%
$$

(Kovács Gyula, Budapest.)

The problem was also solved by: Pauuz A., Sárközy P.