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### Maths Handshake Puzzle

Maths Handshake Puzzle - 4 September

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

1. 12 People:

#12 shakes w/ #1-11 (11 shakes)
#11 shakes w/ #1-10 (10 shakes)
#10 shakes w/ #1-9 (9 shakes)...

11+10+9+8+7+6+5+4+3+2+1+0=66 total handshakes

1. correct nC2 = 66 , and n comes out to be 12.
n*(n-1) = 66*2 and 12 satisfies this relation

2. ans: 13 members
for ex: if 4 mems are there then 6 shakes... i.e. (n-1*(n))/2 so here n-1*n/2=66, then n is 13

3. This comment has been removed by the author.

4. ((13-1)*13))/2 = 78
((12-1)*12))/2 = 66

Still going with 12 :)

Also, check the parentheses: ((n-1)(n))/2

As written, (n-1*(n))/2 simplifies to (n-n)/2, and thence to, well, 0.

5. nc2= n*(n-1)/2=66
n=12

6. 1+2+3+....+(n-1) = 66; total persons = n
n(n-1)/2=66 => n(n-1) = 132 => n = 12
Explaination: If there are n people, the first person
will shake hand with n-1 persons, the second with n-2, the third with n-3, and the (n-1)th person with n-(n-1)=1
person; ie, the last person. Hence, total handshakes are (n-1)+(n-2)+(n-3)+....+3+2+1

7. 12 for sure..

8. 12 people....
coz,

n = total persons
totally 66 handshakes,
2 persons = 1 handshake
and each person shakes hand with n - 1 persons

==> Half of ( n(n-1) ) = 66
==> n(n-1)/2 = 66
==> n = 12 or -11

==> n = 12

1. where do does half came?

9. 12, it's so simple

10. Here's the script i made, just save as anything.bat, 14 here, :
@echo off
set people=1
set handshakes=0
set cnt=0
:LOOP
set /A people+=1
set /A cnt+=1
set /A handshakes+=%cnt%
echo.handshakes:%handshakes%
if "%handshakes%" NEQ "66" Goto :LOOP
echo.Handshakes:%handshakes%
echo.People:%people%
pause > nul
exit /b

11. 12

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.

12. it's just nC2=66
so n=12

13. This comment has been removed by the author.

14. (X-1)+(X-2)+(X-3)+......+{X-(X-1)}=66
So X=12

15. It's basically 11+10+9+8+7+6+5+4+3+2+1+0 which equals 66
The answer is 12. Cmon people it was so easy.

16. 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.