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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 22"
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Quickly simplifying the factors and cancelling gives us that <math>A=124/55</math> and <math>B=1001/192</math>. It is fairly obvious that <math>C</math> will be about <math>2</math>. Therefore <math>B>A>C</math>.
 
Quickly simplifying the factors and cancelling gives us that <math>A=124/55</math> and <math>B=1001/192</math>. It is fairly obvious that <math>C</math> will be about <math>2</math>. Therefore <math>B>A>C</math>.
  
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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 21|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 23|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 10:50, 23 July 2006

Problem

Let

$A = \left( 1 + \frac 12 + \frac 14 + \frac 18 + \frac 1{16} \right) \left( 1 + \frac 13 + \frac 19\right) \left( 1 + \frac 15\right) \left( 1 + \frac 17\right) \left( 1 + \frac 1{11} \right) \left( 1 + \frac 1{13}\right),$
$B = \left( 1 - \frac 12\right)^{-1} \left( 1 - \frac 13 \right)^{-1} \left(1 - \frac 15\right)^{-1} \left(1 - \frac 17\right)^{-1} \left(1-\frac 1{11}\right)^{-1} \left(1 - \frac 1{13}\right)^{-1},$

and

$C = 1 + \frac 12 + \frac 13 + \frac 14 + \frac 15 + \frac 16 + \frac 17 + \frac 18 + \frac 19 + \frac 1{10} + \frac 1{11} + \frac 1{12} + \frac 1{13} + \frac 1{14} + \frac 1{15} +\frac 1{16}.$

Then which of the following inequalities is true?

$\mathrm{(A) \ } A > B > C \qquad \mathrm{(B) \ } B > A > C \qquad \mathrm{(C) \ } C > B > A \qquad \mathrm{(D) \ } C > A > B \qquad \mathrm{(E) \ } B > C > A$

Solution

Quickly simplifying the factors and cancelling gives us that $A=124/55$ and $B=1001/192$. It is fairly obvious that $C$ will be about $2$. Therefore $B>A>C$.

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