The members of a chess club took part in a round robin competition in which each plays every one else once. All members scored the same number of points, except four juniors whose total scored $17.5$. How many members were there in the club? Assume that for each win a player scores $1$ point, draw $0.5$ points and zero for losing.
My attempt is
$$
\binom n 2 = 17.5 + (n-4)k
$$
I can't proceed further. Can you tell me how to proceed or any other way to solve this?
Best Answer
Suppose there are $n$ members. That means there are $\cfrac{n(n-1)}{2}$ matches (and points) altogether.
We are given that $4$ members scored $17.5$ points between them, so the remaining $n-4$ scored $\cfrac{n(n-1)}{2}-17.5=\cfrac {n(n-1)-35}2$ points.
So each of the $n-4$ players (who scored the same) scored $\cfrac {n^2-n-35}{2(n-4)}$ points. Now the possible points scores come in units of $0.5$ so this implies that $\cfrac {n^2-n-35}{(n-4)}$ is an integer. You should be able to do something from there.
Further hint