How many members were there in the club? 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?
 A: 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

 Rewrite this as $\cfrac {(n-4)(n+3)-23}{(n-4)}$ 

A: 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 $n=27$
A: While the answer already given is basically correct, I'd like to add a couple of observations.
Your starting point $${n \choose 2} = 17.5 + (n-4)k$$is correct.  
From it, we can deduce a couple of things.  
First of, for all $n \in \mathbb N$, the outcome of ${n \choose 2}$ is in $\mathbb N$ as well, and so should be $ 17.5 + (n-4)k$.
Since $k$ is scored in half points, this means that $2k \in \mathbb N$. But as $n \in \mathbb N$, it's $k$ that must supply that $0.5$ to make the outcome of the right-hand side a natural number. This means that $k \notin \mathbb N$ and $2k$ is odd, or if you prefer, $k \in\mathbb Q \ |\ 2k=2m-1, m \in \mathbb N$.
Now we can also deduce that $n$ is odd, since we need an odd times $k$ to get that $0.5$ we need.  
So the chess club has an odd number of members, who all played an odd number of draws (except for the four juniors).
We're ready now to solve the equation you gave.
$$\begin{align}
{n \choose 2} & = 17.5 + (n-4) \cdot k \\
\frac{n(n-1)}{2} & = 17.5 + (n-4) \cdot k \\
n^2 - n & = 35 + (n-4) \cdot 2k \\
\frac{n^2 - n - 35}{n-4} & = 2k \\
\frac{(n-4)(n+3) - 23}{n-4} & = 2k \\
n + 3 - \frac{23}{n-4} & = 2k \\
\end{align}$$ with $2k$ and $n$ being odd, as stated earlier.
Therefore, $n-4 = 1\vee n-4=23$. If $n-4=1$ then $n=5$. However, ${n \choose 2} > 17.5$. Since ${5 \choose 2} \ngtr 17.5$, the only option left is $n-4 = 23 \iff n=27$.
This in turn gives $k=14.5$.
