Difference between revisions of "1985 AJHSME Problems/Problem 9"

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==See Also==
 
==See Also==
  
[[1985 AJHSME Problems]]
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{{AJHSME box|year=1985|num-b=8|num-a=10}}
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[[Category:Introductory Algebra Problems]]

Revision as of 14:30, 14 May 2009

Problem

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

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

Solution

First doing the subtraction, we get \[\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10}\]

We notice a lot of terms cancel. In fact, every term in the numerator except for the $1$ and every term in the denominator except for the $10$ will cancel, so the answer is $\frac{1}{10}$, or $\boxed{\text{A}}$

If you don't believe this, then rearrange the factors in the denominator to get \[\frac{1}{10}\times\frac{2}{2}\times\frac{3}{3}\times\cdots\times\frac{9}{9}\]

Everything except for the first term is $1$, so the product is $\frac{1}{10}$

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions
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