My solution Note that no three collisions occur simultaneously since at all times there are only $2$ distinct angular speeds, and only two particles can collide simultaneously (basically the first collision has only two particles, so by the end of it only one has a different angular velocity, so the second is also two particles and so on) Let us for now ignore the condition that the particles are distinct. Let $\omega$ be the angular speed. Then, we can ignore collisions. The next time the angles between the particles becomes $\theta = \frac{360^{\circ}}{n}$ is after the 'retrogade particle' (moving in the opposite direction), returns to its spot, and since both its original spot and the retrogade particle itself move in opposite directions with angular speed $\omega$, the next time the angles subtended at the centre of the circle becomes same is at time $\frac{360^{\circ}}{2\omega}$. In this process, $n-1$ collisions take place. Next now we use the condition that the particles are distinct. so, their relative order stays same due to collisions. the first particle transfers to the second particle (first particle is the initial retrogade particle) to the second one, and then travels in the same direction as all the other particles. So, by the time $\frac{360^{\circ}}{2\omega}$ it rotates by an angle of $\frac{360^{\circ}(n-2)}{2n}$ (due to relative order and angles subtended at the centre being same, all other particles rotate by the same angle in the same direction). So, we need to find the least $k$ such that $\frac{(n-2)k}{2n}$ is an integer, and $s = (n-1)k$. So, we will get If $n$ is odd, then $s = 2n(n-1)$ If $n \equiv 2 \pmod{4}$, then $s = \frac{n(n-1)}{2}$ If $n$ is divisible by $4$, then $s = n(n-1)$.

Official wording: everythingpi3141592 wrote: There are $n\ge 3$ particles on a circle situated at the vertices of a regular $n$-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before. Let $s$ be the smallest number of collisions after which all particles return to their original positions. Find $s$. Proposed by N.V. Tejaswi The following stronger version of the problem is also true: Suppose $k$ particles are moving clockwise and $n-k$ are moving anti-clockwise, then the smallest number of collisions after which all particles return to their original positions is $$\frac{2nk(n-k)}{\gcd(2n,n-2k)}.$$

Rijul saini wrote: Official wording: everythingpi3141592 wrote: There are $n\ge 3$ particles on a circle situated at the vertices of a regular $n$-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before. Let $s$ be the smallest number of collisions after which all particles return to their original positions. Find $s$. Proposed by N.V. Tejaswi The following stronger version of the problem is also true: Suppose $k$ particles are moving clockwise and $n-k$ are moving anti-clockwise, then the smallest number of collisions after which all particles return to their original positions is $$\frac{2nk(n-k)}{\gcd(2n,n-2k)}.$$ When you say this are you implying it's the same for each arrangement or that over all arrangements this is the lcm of the minimums?

Rijul saini wrote: Official wording: everythingpi3141592 wrote: There are $n\ge 3$ particles on a circle situated at the vertices of a regular $n$-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before. Let $s$ be the smallest number of collisions after which all particles return to their original positions. Find $s$. Proposed by N.V. Tejaswi The following stronger version of the problem is also true: Suppose $k$ particles are moving clockwise and $n-k$ are moving anti-clockwise, then the smallest number of collisions after which all particles return to their original positions is $$\frac{2nk(n-k)}{\gcd(2n,n-2k)}.$$ Tiny correction: The $k$ particles have to be contiguous (or at least have an aperiodic arrangement).

What was the point of puttting this prob the main idea of the prob also exists in the ant problem in pranav sriram?

idkk wrote: What was the point of puttting this prob the main idea of the prob also exists in the ant problem in pranav sriram? So that people like me spend 4.5 hours getting 9/10. Also, the word is "putting". Please fix your t-key.