1951 AHSME Problems/Problem 31

Problem

A total of $28$ handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:

$\textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7$

Solution

The handshake equation is $h=\frac{n(n-1)}{2}$, where $n$ is the number of people and $h$ is the number of handshakes. There were $28$ handshakes, so $28=\frac{n(n-1)}{2}$ $56=n(n-1)$ The factors of $56$ are: $1, 2, 4, 7, 8, 14, 28, 56$. As we can see, only $7, 8$ fit the requirements $n(n-1)$ if $n$ was an integer. Therefore, the answer is $\textbf{(D)}\ 8$

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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