At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
+118696 At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = k−12(1+(k−1))=k(k−1)2
so
k(k−1)2=66k(k−1)=132k2−k−132=0
k2−k−132=0⇒{k=12k=−11}
Obviously there is not a neg number of people so there must be 12 people.
+130458 We can solve this by this "formula"
n(n-1)/ 2 = 66 multiply by 2 on each side
n(n-1) = 132 simplify and rearrange
n^2 - n - 132 = 0 factor
(n - 12) (n+11) = 0 and setting each factor to 0, we have that n = 12 and n = -11
Reject the negative
n = 12 people
We can see this easily
The 12th person shakes 11 hands
The 11th person shakes 10 hands, etc.
So...... the sum of
11 + 10 + 9 +......+ 3 + 2 + 1 = 66
+118696 At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = k−12(1+(k−1))=k(k−1)2
so
k(k−1)2=66k(k−1)=132k2−k−132=0
k2−k−132=0⇒{k=12k=−11}
Obviously there is not a neg number of people so there must be 12 people.
