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At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Answers
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lets say there are n persons
first person shakes hand with everyone else: n-1 times(n-1 persons)
second person shakes hand with everyone else(not with 1st as its already done): n-2 times
3rd person shakes hands with remaining persons: n-3
So total handshakes will be = (n-1) + (n-2) + (n-3) +…… 0;
= (n-1)*(n-1+1)/2 = (n-1)*n/2 = 66
= n^2 -n = 132
=(n-12)(n+11) = 0;
= n = 12 OR n =-11
-11 is ruled out so the answer is 12 persons.
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Answer:
12
Step-by-step 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 (AB) .
If we have 3 people, there are 3 handshakes (AB,AC,BC) .
If we have 4 people, there are 6 handshakes (AB,AC,AD,BC,BD,CD).
If we have 5 people, there are 10 handshakes
(AB,AC,AD,AE,BC,BD,BE,CD,CE,DE).
See that we can express the number of handshakes as the sum of consecutive positive integers, starting with 1, i.e.
1+2+3+...+(n−1)
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⇒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⇒
⇒n−1=11⇒n=12