TITLE: How many members were there in the club?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: We start with the same equation:
$n(n-1)/2 = 17.5 + (n-4)k$ where $k$ is a half-integer; $2k \in \mathbb{Z}$
$n^2-n-35=(n-4)2k$
Thus $(n-4) | (n^2-n-35) = (n-4)(n+3) - 23$
Thus $(n-4) | 23$ and so $n=5$ or $n=27$
Now $n=5$ is impossible otherwise 17.5 is an impossible score
Therefore $n=27$