Question

At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

Answer 1

HI!!

Answer 2

if i remember correctly, if there are \(n\) people there are \[\binom{n}{2}\] handshakes

Answer 3

set \[\frac{n(n-1)}{2}=66\] and solve for \(n\)

Answer 4

This is a bit confusing...

Answer 5

i can see that

Answer 6

if \[\frac{n(n-1)}{2}=66\] then \[n(n-1)=132\]

Answer 7

see if you can think of two consecutive numbers whose product is 132

Answer 8

you can probably guess it quickly

Answer 9

Notice the sequence: 1,3,6,10,15
Let's see how they are related:

Note that the difference between the number of handshakes for each step is equal to the number of people doing the handshaking:

$$
a_1-a_0=1-0=1\\
a_2-a_1=3-1=2\\
a_3-a_2=6-3=3\\
a_4-a_3=10-6=4\\
a_5-a_4=15-10=5\\
\cdots\\
a_n-a_{n-1}=n
$$
If you add everything on the left side of the equal sign and everything on the right side of the equal sign, you get:

$$
a_n-a_0=\cfrac{n(n+1)}{2}\\
a_n=\cfrac{n(n+1)}{2}\\
$$
We want \(a_n=66\):

$$
66=\cfrac{n(n+1)}{2}\\\\
2\times 66=n^2+n\\
n^2+n-132=0\\
\implies n = 11
$$
So there will be 66 handshakes if there are 11 people at the party.