Question

no rush, just a brain teaser to keep u all sharp (psh, like u need it brainaics!):

If you were at a party, and everyone shook hands with every other person, and there were 66 handshakes in total... then how many people were at the party?

I'm off to a concert, will post the answer tomoz, slaters all and good luck ;)

Answer 1

There is n+1 people and the number of handshakes is 1+2+3+4+5+6...
The sum is therefore
\[\frac{ n(n+1) }{ 2 }\]
Each handshake is counted twice so we divide by 2
We solve that = 66 and get
n(n+1)=132
n^2+n-132=0 and solve for the quadratic equation
(n+12)(n-11)=0
Since the number of people cannot be -12 then it is 11
So the final answer is 11 I think

Answer 2

nope..
u r so close... but nt close enough :-)
if there were "n" amount of people and u were one of them
u have to do
(n - 1) handshakes
then the other person will have to do ( n-2) ... and so on until the amount of handshakes will be 1

so... this is an arithmetic progression..
1 , 2, 3, 4, ...... ( n-1)
and (n-1) is the "n" th term...

so...
Sn = n( a + l) /2
66 = (n-1)( 1 + n-1 ) /2
132 = n(n-1)
n^2-n-132=0
(n-12)(n+11)=0


so...
the amount of people is 12 not 11 .....