A barrel contains $v$ liters of wine. From this, we successively draw $a_{1}, a_{2}, \ldots, a_{n}$ liters, always replacing the removed amount with water. How much wine remains in the barrel after the $n$-th draw?
After the first measurement, $b_{1}=v-a_{1}$ liters of wine remain, and to this, $a_{1}$ liters of water are added. In the resulting mixture, $\frac{b_{1}}{v}=\frac{v-a_{1}}{v}$ liters of wine will be in each liter, and thus during the second measurement, $\frac{v-a_{1}}{v} a_{2}=\frac{b_{1}}{v} a_{2}$ liters of wine are removed; therefore, the amount of wine remaining in the mixture is

$$
b_{2}=b_{1}-\frac{b_{1}}{v} a_{2}=\frac{b_{1}\left(v-a_{2}\right)}{v} \quad \text { liters of wine. }
$$

Now, in 1 liter of the mixture, there is $\frac{b_{2}}{v}$ and in $a_{3}$ liters, there is $\frac{b_{2}}{v} a_{3}$ liters of wine, so after the third measurement,

$$
b_{3}=b_{2}-\frac{b_{2}}{v} a_{3}=\frac{b_{2}\left(v-a_{3}\right)}{v} \quad \text { liters of wine }
$$

remain in the mixture, and so on. After the n-th measurement,

$$
b_{n}=\frac{b_{n-1}\left(v-a_{n}\right)}{v}
$$

liters of wine remain in the mixture. According to this,

$$
b_{1} b_{2} b_{3} \ldots b_{n-1} b_{n}=\left(v-a_{1}\right) \frac{b_{1}\left(v-a_{2}\right)}{v} \cdot \frac{b_{2}\left(v-a_{3}\right)}{v} \ldots \frac{b_{n-1}\left(v-a_{n}\right)}{v}
$$

and thus

$$
b_{n}=\frac{\left(v-a_{1}\right)\left(v-a_{2}\right)\left(v-a_{3}\right) \ldots\left(v-a_{n}\right)}{v^{n-1}}
$$

NB. If $a_{1}=a_{2}=\ldots=a_{n}=a$, then $b_{n}=(v-a)\left(\frac{v-a}{v}\right)^{n-1} ;$ this is the n-th term of a geometric progression where the first term is $v-a$ and the common ratio is $\frac{v-a}{v}$.

Akos Fehervary (Esztergomi rg. VI. o. Bp. II.)