Question

Brainteasers! Medal for the first to answer correct!

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

This is just for fun, I already know the answer. First one to answer correctly gets a medal!

Answer 1

let me think first

Answer 2

@amistre64
@hartnn
@iPwnBunnies
Can you figure it out? :)

Answer 3

dont forget you cant shake your own hand

Answer 4

I am working on it now

Answer 5

I would guess but like, I would only think of the 66 hands that were shaken, unless there is a number of people who had already shaken hands. i think i'm thinking too hard

Answer 6

I'm not seeing how it can be 66, if everybody shook each other's hands...

Answer 7

yea true

Answer 8

now that you mention it.

Answer 9

and then its asking how many people were at the party, and everyone has 2 hands, so if you half 66 you get 33 so i think 33 o_o

Answer 10

Ok, this formula will help:
\[\frac{ n(n+1) }{ 2 } = 66\]
:)

Answer 11

LOGIC has failed me!!!!! T_T

Answer 12

lol

Answer 13

Are you saying people shake hands twice. ._.

Answer 14

66

Answer 15

Hey people are weird like that.

Answer 16

Think of it this way, you have 1 person who shakes everyone's hand, yet everyone does that, and in total, 66 hands were shaken, how many people were at this party :) its obviously not 66 because the total is going to be bigger than the number of people there since each person shook more than one.

Answer 17

You guys want the answer?

Answer 18

yes yes

Answer 19

12

Answer 20

12

Answer 21

Ok, the answer is 12 :)

Answer 22

Ah, I see it now.

Answer 23

Explanation:
In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.
Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.
This is the quadratic equation n2+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.

Since 66 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +... etc. until the total is 66. The last number that you entered (11) is n.

Answer 24

yep :)

Answer 25

Fun brainteaser site :)
http://dailybrainteaser.blogspot.com/2012/12/hard-math-problems.html
And don't worry, I couldn't figure it out either lol.

Answer 26

assume 1 person shakes 65 hands and leaves the room
then 1 person shakes 64 hands and leaves the room
then 1 person shakes 63 hands and leaves the room
...
when 1 person is left, the shake no hands and just leave ... how many handshakes were given?

Answer 27

lol, i read the question backwards, but im sure that can be worked around to match this question

Answer 28

So 66 people must have started in the room because that 1 person started shaking hands with 65 people, right?

Answer 29

yes, but i read this as 66 people in a room, how many handshakes; but the sequence is arithmatic either way

Answer 30

Then instead of the formula:
\[\frac{ n(n+1) }{ 2 } = 66\]
It will be:
\[\frac{ n(n-1) }{ 2 } = 66\]
right?

Answer 31

this is a good question - i'd see it before. Its pretty clever really.

Answer 32

if there are n in the room, then n-1 handshakes
then n-2 hand shakes
then n-3 hand shakes
...
then 1 handshake
then 0 handshakes

add up to 66 shakes altogether

Answer 33

let me chk\[\frac{(n-1)(n-1+1)}{2}=66\]
\[\frac{n(n-1)}{2}=66\]
yes

Answer 34

So then:
\[n = 12\]
:)

Answer 35

solving the quadratic gets us n=-11 or n=12

Answer 36

yep

Answer 37

And since there cannot be negative people, the answer is 12 lol