What is the number of people that leave the meeting? In a business meeting, each person shakes hands with each other person, with the exception of Mr. L. Since Mr. L arrives after some people have left, he shakes hands only with those present. If the total number of handshakes is exactly 100, how many people left the meeting before Mr. L arrived? (Nobody shakes hands with the same person more than once.)
The question is from CMI2011 UG entrance exam paper. 
 A: Let $n$ be the total number of people at the meeting before Mr. L arrived. Then the number of handshakes will be $\dfrac{n(n-1)}{2}$. Suppose $m$ people left before Mr. L arrived .So he shakes $n-m$ hands. Then
\begin{align*}
\frac{n(n-1)}{2}+n-m & =100\\
n^2+n-2m & = 200\\
m & = \frac{n(n+1)}{2}-100.
\end{align*}
But $0 < m < n$ (assuming that at least one person left). So we want
\begin{align*}
0 & < \frac{n(n+1)}{2}-100 < n\\
200 & < n(n+1) < 2n+\color{red}{200}.
\end{align*}
edit:
I had made a typo because of which my initial conclusion was incorrect:
The first inequality suggests that $n \geq 14$ but the second inequality suggests $n \leq 14$. So $n=14$ is the answer. This yields $m=5$.
A: First, we need to find the number of people in the room before Mr. L arrived, call it $N$. $N$ is the greatest integer such that ${N \choose 2} \leq 100$. This is because ${{N} \choose 2}+(N-1) < {{N+1} \choose 2}$. We find then that $N=14$ since ${14 \choose 2}+9=100$. That is, 5 people left the room, since Mr. L only shook hands with 9.
