Ball Game

Let $n$ be the number of people in the group. Starting from Celia and going clockwise, we label each person with the numbers $0, 1, 2, \dots, n - 1,$ so that after $k$ passes, the person with the number $k \! \pmod{n}$ has the ball.

From the given information, we know that Paul's number is $6 \! \pmod{n}$ and Celia's number is $15 \! \pmod{n}.$ Since we let Celia be at position 0, we must have $15 \equiv 0 \! \pmod{n}.$ If Paul were next to Celia, then we also have $6 \equiv \pm 1 \! \pmod{n},$ since the positions next to Celia are 1 and $n - 1 \equiv -1 \! \pmod{n}.$

We have a system of congruences, and our question is whether there exists $n$ that satisfies them. Indeed, $n = 5$ satisfies both congruences: $15 \equiv 0 \! \pmod{5},$ and $6 \equiv 1 \! \pmod{5}.$ Thus, we conclude that yes , it is possible for Celia and Paul to be sitting next to each other.

There is probably a nice general way to solve such problems using group theory or something, but I just tried a few arrangements until I found something that worked.

If Paul is immediately to Celia's left with 3 other people it works out as described. I believe that is the unique solution here.

Paul gets the ball on pass 1, then (with 5 people total) on pass number 6. Celia gets it 4 passes later, then again in 5 more passes for a total of 9 after Paul has it.

Now I have to sit down, as my head still seems to be spinning.

After 15 passes, Celia has the ball again. This implies that the number of people at the table is a factor of 15: 1, 3, 5, or 15. If we check these possibilities it's obvious that if there are five people at the table, Paul is at Celia's left. We don't really need modular arithmetic here.

obviously possible.Paul is sitting right to Celia.

Problem states: The ball starts with Celia and after $6$ passes it's with Paul. If Celia sits next to the Paul, that means that the ball would have been with Celia after $5$ or $7$ passes from the beginning of the game. So, we can conclude that one round would consist of $5$ or $7$ passes. If it consists of $5$ passes, then after every $5th$ pass $(5, 10, 15,...)$ the ball will be with Celia. And the problem states that after $9$ passes after initial $6$ , the ball is again with Celia. Since $6+9=15$ , it is possible that Celia and Paul sit next to each other.

There are $N > 2$ people around the table. (And $N = 3$ would be pushing "number of people"...)

The ball goes from Celia to Celia in $6 + 9 = 15$ moves. Thus $15$ is a multiple of $N$ . This means $N = 3, 5, 15$ .

We can rule out $N = 3$ because then the first six passes would also end up at Celia.

So consider $N = 5$ . After five moves, Celia has the ball; after six moves (we are told), Paul has it. Therefore Paul sits to the left of Celia in this case.

1. Lets assume Celia at position 1
2. Only way Paul could be next to celia if he is at position 2 or (n-1)
3. Let's try to find range of "n"
4. Since the ball passes to Paul after 6th pass, hence there can't be more than 5 guys in this round table i.e., total 7 guys
5. Now we will make cases for n = 1,2,3,4,5 ( n $\neq$ 0 because question says "with a number of other people")

Case 1: when n = 1 i.e., total 3 people

6 Passes - 1 $-->$ 2, 2 $-->$ 3, 3 $-->$ 1, 1 $-->$ 2, 2 $-->$ 3, 3 $-->$ 1

As we can see in the 6 passes that ball hasn't been to position 2 or 3. Hence no need to check for 9 passes and this ain't answer

Case 2: when n = 2 i.e., total 4 people

6 Passes - 1 $-->$ 2, 2 $-->$ 3, 3 $-->$ 4, 4 $-->$ 1, 1 $-->$ 2, 2 $-->$ 3

As we can see in the 6 passes that ball hasn't been to position 2 or 4. Hence no need to check for 9 passes and this ain't answer

Case 3: when n = 3 i.e., total 5 people

6 passes - 1 $-->$ 2, 2 $-->$ 3, 3 $-->$ 4, 4 $-->$ 5, 5 $-->$ 1, 1 $-->$ 2

As we can see in the 6 passes that ball is at position so maybe this can be the answer, let's check for 9 passes after this

9 Passes - 2 $-->$ 3, 3 $-->$ 4, 4 $-->$ 5, 5 $-->$ 1, 1 $-->$ 2, 2 $-->$ 3, 3 $-->$ 4, 4 $-->$ 5, 5 $-->$ 1

Now this is what we were looking for i.e., ball is to Paul position 2 at 6th pass and back to Celia position 1 at 9th pass

So, this is what we were finding for. I don't think its worthy anymore to solve for n = 4,5 XD

The information in the problem may be extended to include the fact that Celia gets the ball returned to her after 15 passes. This fact alone enables only three possibilities for the number of people sitting around the round table. Either there are 3, 5 or 15 people, otherwise the ball cannot come back to Celia in the right number of moves. If there were 3 people sitting around the table then Celia would also get the ball after 6 passes, which is contradicted by the information given in the problem. If there were 15 people and Paul got the ball after 6 passes then he would not be sitting next to Celia. The answer may still be "yes" if the situation for 5 people satisfies all of the conditions given in the problem. For 5 people, with Paul sitting to the left of Celia then Celia would get the ball again after 5, 10 and 15 passes and Paul would get the ball after 1, 6 and 11 passes. In this arrangement (i.e. 5 people with Paul to the left of Celia), Paul gets the ball after 6 passes and Celia gets the ball after 15 passes, so the answer is yes!.

Additional people could have joined the game after it started, adding more "passes" to the rotation.

Five people around the table, Celia, Paul, P3,P4,P5. [key > = pass, {n} move number ] Celia > Paul {1} Paul > P1 {2} P1 > P2 {3} P2 > P3 {4} P3 > Celia {5} and Celia > Paul {6}. It continues round the table Paul > P1 {7} P1 > P2 {8} P2 > P3 {9} P3 > Celia {10} Celia > Paul {11} Paul > P1 {12} P1 > P2 {13} P2 > P3 {14} and finally P3 > Celia {15} . So Paul receives in on the sixth pass and Celia has it on the fifteenth pass.