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

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

Guest Feb 25, 2015

#3
+91451
+5

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) =  $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$

so

$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$

$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$

Obviously there is not a neg number of people so there must be 12 people.

Melody  Feb 25, 2015
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#1
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can you even solve that

Guest Feb 25, 2015
#2
+81007
+5

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

CPhill  Feb 25, 2015
#3
+91451
+5

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) =  $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$

so

$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$

$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$

Obviously there is not a neg number of people so there must be 12 people.

Melody  Feb 25, 2015

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