9.5. On a circle, $n>1$ points, called positions, are marked, dividing it into equal arcs. The positions are numbered clockwise from 0 to $n-1$. Vasya places a chip in one of them. Then the following actions, called moves, are repeated an unlimited number of times: Petya names some natural number, and Vasya moves the chip clockwise or counterclockwise by the number of positions specified by Petya. If at some point after Vasya's move, the chip lands in position number 0, Vasya loses, and Petya wins. For which $n$ can Petya always win, regardless of Vasya's moves?
Answer. For $n=2^{k}$, for all natural $k$.

Solution. 1) Let first $n=2^{k}$, for some natural $k$. A winning strategy for Petya is as follows: if after Vasya's move the chip is on a position with number $m=2^{a}(2 b+1)$, then Petya names the number $m=2^{a}$. After this, Vasya will move the chip either to the position with number $m=2^{a+1}(b+1)$, or to the position with number $m=2^{a+1} b$ (modulo $\left.n=2^{k}\right)$. In any case, the highest power of two dividing the position number, where the chip will be after Vasya's next move, increases by at least 1. Therefore, after no more than $k$ moves, the position number of the chip will be divisible by $n=2^{k}$, which is only possible for the position with number 0, and Petya will win.

2) Now let $n=2^{k} \cdot p, k \geq 0$, where $p=2 b+1>1$ is the largest odd divisor of $n$. Note that positions with numbers divisible by $p$ are bad for Vasya; if he places the chip on one of them first, Petya, using a strategy similar to the one described in point 1), naming the number $m=2^{a} \cdot p$ each time, will win. Therefore, Vasya must first place the chip on any position with number $x$ not divisible by $p$. Suppose Petya then names any natural number $s$ from the range from 1 to $n-1$. One of the numbers $x-s$ and $x+s$ does not divide by $p$, otherwise their sum, equal to $2 x$, would divide by $p$, and thus $x$ would divide by $p$, which contradicts his choice. Therefore, on each move, Vasya can place the chip on a position whose number does not divide by $p$, and it will never end up on the position with number 0. Thus, for $n$ that are not powers of two, Petya cannot win.

Grading criteria. ($\cdot$) A winning strategy for Petya in the case $n=2^{k}$ is provided: 3 points. ($\cdot$) A strategy for Vasya that allows avoiding a loss in the case $n \neq 2^{k}$ is provided: 4 points. In it: the choice of the first position: 2 points; proof of the possibility of choosing the correct direction for each subsequent move: 2 points. Points are not deducted for the lack of a strict consideration of position numbers modulo $n$ when transitioning through 0.