# At a party 66 handshakes occurred. Each person shook hands exactly once with each of the other people present. How many people were present?

12

#### Explanation:

Let's start with small numbers of people and handshakes and move from there. I'll represent people with letters to show the handshakes:

If we have 2 people, there is 1 handshake $\left(A B\right)$.

If we have 3 people, there are 3 handshakes $\left(A B , A C , B C\right)$.

If we have 4 people, there are 6 handshakes $\left(A B , A C , A D , B C , B D , C D\right)$.

If we have 5 people, there are 10 handshakes $\left(A B , A C , A D , A E , B C , B D , B E , C D , C E , D E\right)$.

See that we can express the number of handshakes as the sum of consecutive positive integers, starting with 1, i.e. $1 + 2 + 3 + \ldots + \left(n - 1\right)$ and the number of people present is $n$

Let's test this with 5 people. We have $1 + 2 + 3 + 4 = 10$ handshakes. $n - 1 = 4 \implies n = 5$ which is the number of people.

So what we need to do is add up to 66 and we'll be able to find the number of people:

$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66 \implies$

$\implies n - 1 = 11 \implies n = 12$