Difference between revisions of "1951 AHSME Problems/Problem 31"

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<math> \textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7 </math>
 
<math> \textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7 </math>
 
[[1951 AHSME Problems/Problem 31|Solution]]
 
  
 
==Solution==
 
==Solution==
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== See Also ==
 
== See Also ==
{{AHSME 50p box|year=1951|num-b=24|num-a=26}}  
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{{AHSME 50p box|year=1951|num-b=30|num-a=32}}  
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:26, 5 July 2013

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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All AHSME Problems and Solutions

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