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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 22"

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If you imagine expanding out the [[expression]] for <math>A</math>, you can see that every term in <math>C</math> will appear once, along with plenty of others.  (Think of the [[prime factorization]]s and you can figure out which products give the terms of <math>C</math>.)  Since all terms are [[positive]], <math>A > C</math>.
 
If you imagine expanding out the [[expression]] for <math>A</math>, you can see that every term in <math>C</math> will appear once, along with plenty of others.  (Think of the [[prime factorization]]s and you can figure out which products give the terms of <math>C</math>.)  Since all terms are [[positive]], <math>A > C</math>.
  
−
<math>\frac AB = \left( 1 - \frac 12\right)\left( 1 + \frac 12 + \frac 14 + \frac 18 + \frac 1{16} \right) \left( 1 - \frac 13 \right)</math> <math>\left( 1 + \frac 13 + \frac 19\right)</math> <math>\left(1 - \frac 15\right) \left( 1 + \frac 15) \left(1 - \frac 17\right)</math> <math>\left( 1 + \frac 17\right) \left(1-\frac 1{11}\right)</math> <math>\left( 1 + \frac 1{11} \right)</math> <math>\left(1 - \frac 1{13}\right)</math>  <math>\left( 1 + \frac 1{13}\right)</math>  
+
<math>\frac AB = \left( 1 - \frac 12\right)\left( 1 + \frac 12 + \frac 14 + \frac 18 + \frac 1{16} \right) \left( 1 - \frac 13 \right)</math> <math>\left( 1 + \frac 13 + \frac 19\right)</math> <math>\left(1 - \frac 15\right) \left( 1 + \frac 15\right) \left(1 - \frac 17\right)</math> <math>\left( 1 + \frac 17\right) \left(1-\frac 1{11}\right)</math> <math>\left( 1 + \frac 1{11} \right)</math> <math>\left(1 - \frac 1{13}\right)</math>  <math>\left( 1 + \frac 1{13}\right)</math>  
  
 
<math>= \left(1 - \frac1{32}\right)\left(1 - \frac1{27}\right)\left(1 - \frac1{25}\right) \left(1-\frac1{49}\right)\left(1-\frac1{121}\right)\left(1-\frac1{169}\right) < 1</math> so <math>A < B</math>.
 
<math>= \left(1 - \frac1{32}\right)\left(1 - \frac1{27}\right)\left(1 - \frac1{25}\right) \left(1-\frac1{49}\right)\left(1-\frac1{121}\right)\left(1-\frac1{169}\right) < 1</math> so <math>A < B</math>.

Latest revision as of 01:42, 9 November 2007

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

If you imagine expanding out the expression for $A$, you can see that every term in $C$ will appear once, along with plenty of others. (Think of the prime factorizations and you can figure out which products give the terms of $C$.) Since all terms are positive, $A > C$.

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

$= \left(1 - \frac1{32}\right)\left(1 - \frac1{27}\right)\left(1 - \frac1{25}\right) \left(1-\frac1{49}\right)\left(1-\frac1{121}\right)\left(1-\frac1{169}\right) < 1$ so $A < B$.

Putting these two facts together, $B > A > C \Longrightarrow \mathrm{(B) \ }$.

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