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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.

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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 $m=5$.

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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.

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