Difference between revisions of "1985 AJHSME Problem 9"

(Solution)
(In-depth Solution by Boundless Brain!)
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<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>
 
<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>
  
== In-depth Solution by Boundless Brain!==
+
== In-depth Solution by BoundlessBrain!==
 
https://youtu.be/yBrLXnjasgg
 
https://youtu.be/yBrLXnjasgg
  

Revision as of 23:02, 29 June 2023

Problem

The product of the 9 factors $\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =$

$\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}$

In-depth Solution by BoundlessBrain!

https://youtu.be/yBrLXnjasgg

Solution

This product simplifies to: \[\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.\] Numerators and denominators cancel to yield the answer: $\boxed{\text{(A)} \frac{1}{10}}.$