Difference between revisions of "1984 AJHSME Problem 9"

(Created page with "== Problem == The product of the 9 factors <math>\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =</math> <math>\text{(A)}\ \...")
 
(Please delete. The AJHSME didn't exist in 1984.)
 
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== Problem ==
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{{delete | AJHSME did not exist in 1984}}
The product of the 9 factors <math>\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =</math>
 
 
 
<math>\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}</math>
 
 
 
== Solution ==
 
This product simplifies to:
 
<cmath>\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.</cmath>
 
Numerators and denominators cancel to yield the answer: <math>\boxed{\text{(A)} \frac{1}{10}}.</math>
 

Latest revision as of 20:59, 3 May 2023

This article has been proposed for deletion. The reason given is: AJHSME did not exist in 1984.

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