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)}\ \...")
 
(Problem)
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== 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)}\ \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 ==
 
== Solution ==
 
This product simplifies to:
 
This product simplifies to:
 
<cmath>\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.</cmath>
 
<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>
 
Numerators and denominators cancel to yield the answer: <math>\boxed{\text{(A)} \frac{1}{10}}.</math>

Revision as of 20:43, 31 January 2021

Solution

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