1. $V$ each of the four rooms there are several items. Let $n \geqq 2$ be a natural number. We move one $n$-th of the items from the first room to the second. Then, one $n$-th of the (new) number of items is moved from the second room to the third. Similarly, from the third room to the fourth, and from the fourth back to the first. (Each time, we move whole items.) If you know that at the end there was the same number of items in each room, determine the least number of items that could have been in the second room at the beginning. For which $n$ can this happen?
SOLUTION. When analyzing the number of items after individual steps, we will proceed "backwards." First, let's show how the numbers of items in two rooms before the transfer can be determined from the numbers of items in the rooms after the transfer. Let's say that before the transfer from room $A$ to room $B$, there are $a$ and $b$ items in rooms $A$ and $B$, respectively. Let these numbers after the transfer be denoted by $a^{\prime}, b^{\prime}$. According to the problem statement, we have

$$
a^{\prime}=\frac{n-1}{n} a, \quad b^{\prime}=b+\frac{1}{n} a .
$$

From the first equation and then from the relation $a+b=a^{\prime}+b^{\prime}$, we find

$$
a=\frac{n}{n-1} a^{\prime}, \quad b=b^{\prime}-\frac{1}{n-1} a^{\prime} .
$$

Now, let $M$ be the number of items found in each of the four rooms at the end. By repeatedly applying the derived relation $\left(a^{\prime}, b^{\prime}\right) \rightarrow(a, b)$, we can work our way back to express the initial counts in terms of $M$ and $n$:

$$
\begin{array}{crrrr}
\text { At the end: } & M, & M, & M ; \\
\text { Before } 4 \rightarrow 1: & \frac{n-2}{n-1} M, & M, & M, & \frac{n}{n-1} M ; \\
\text { Before } 3 \rightarrow 4: & \frac{n-2}{n-1} M, & M, & \frac{n}{n-1} M, & M ; \\
\text { Before } 2 \rightarrow 3: & \frac{n-2}{n-1} M, & \frac{n}{n-1} M, & M, & M ; \\
\text { Before } 1 \rightarrow 2: & \frac{n(n-2)}{(n-1)^{2}} M, & \frac{(n-1)^{2}+1}{(n-1)^{2}} M, & M, & M .
\end{array}
$$

Since the number of items in the first room at the beginning was positive, we must have $n \geqq 3$. Now we can easily determine the minimum possible value of the expression

$$
V_{2}=\frac{(n-1)^{2}+1}{(n-1)^{2}} M
$$

The numerator and the denominator of the fraction differ by one, so the fraction cannot be simplified. For the result to be an integer, $M$ must be $k \cdot(n-1)^{2}$ for some $k$, and thus $V_{2}=k\left((n-1)^{2}+1\right)$.

For $n \geqq 3$, however, $(n-1)^{2}+1 \geqq 5$, so $V_{2} \geqq 5$. By choosing $n=3, k=1$, and $M=4$, we achieve the value $V_{2}=5$, and we can easily verify that the corresponding quadruple $(3,5,4,4)$ satisfies the conditions of the problem: after individual transfers from it, we get the quadruple $(2,6,4,4)$, then $(2,4,6,4)$, then $(2,4,4,6)$, and finally $(4,4,4,4)$. The sought minimum number of items $v$ in the second room is thus indeed 5 and can be achieved only for $n=3$, because for $n \geqq 4$, $V_{2} \geqq 3^{2}+1=10$.