13.425 There are $n$ measuring cylinders with liquid. From the first cylinder, $1 / n$ of the liquid in it was poured into the second cylinder, then $1 / n$ of the liquid that ended up in the second cylinder after the transfer from the first cylinder was poured into the third cylinder, and so on. Finally, $1 / n$ of the liquid that ended up in the $n$-th cylinder after the transfer from the previous cylinder was poured back into the first cylinder. After this, each cylinder contained $a$ cm $^{3}$ of liquid. How much liquid was originally in each cylinder?
Solution. Let $x_{k}$ be the initial volume of liquid in the $k$-th beaker. After taking $\frac{1}{n}$ part from the $k$-th beaker (for $k \geq 2$), there remained, according to the problem, $a$ cm$^{3}$, which is $\frac{n-1}{n}$ of the volume that was there before the removal. Therefore, after the transfer from the $(k-1)$-th beaker to the $k$-th, there was $\frac{a n}{n-1}$ cm$^{3}$. Similarly, in the $(k-1)$-th beaker before the removal of liquid from it, there was $\frac{a n}{n-1}$ cm$^{3}$, i.e., $\frac{a}{n-1}$ cm$^{3}$ was transferred from it to the $k$-th beaker. (Note that these considerations are valid only for $k \geq 3$.) Hence, $\frac{a}{n-1}+x_{k}=\frac{a n}{n-1}$ for $k \geq 3 \Rightarrow x_{k}=a$ for $k \geq 3$. Additionally, $\frac{x_{1}}{n}+x_{2}=\frac{a n}{n-1}$ (after the first transfer) and $\frac{a}{n-1}+\frac{n-1}{n} x_{1}=a$ (after the last transfer), so $x_{1}=\frac{n(n-2) a}{(n-1)^{2}}, x_{2}=\frac{\left(n^{2}-2 n+2\right) a}{(n-1)^{2}}$. Answer: $x_{1}=\frac{a n(n-2)}{(n-1)^{2}} \mathrm{~cm}^{3}, x_{2}=\frac{a\left(n^{2}-2 n+2\right)}{(n-1)^{2}} \mathrm{~cm}^{3}, x_{3}=\ldots=x_{n}=a \mathrm{~cm}^{3}$.