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 = ( 1 - \frac 12)( 1 + \frac 12 + \frac 14 + \frac 18 + \frac 1{16} ) ( 1 - \frac 13 ) | + | <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>= \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>. |
Revision as of 18:26, 12 October 2007
Problem
Let
![$A = \left( 1 + \frac 12 + \frac 14 + \frac 18 + \frac 1{16} \right) \left( 1 + \frac 13 + \frac 19\right) \left( 1 + \frac 15\right) \left( 1 + \frac 17\right) \left( 1 + \frac 1{11} \right) \left( 1 + \frac 1{13}\right),$](http://latex.artofproblemsolving.com/e/4/8/e486e74b04f4bd209f89e7fbcfddf5da13e61844.png)
![$B = \left( 1 - \frac 12\right)^{-1} \left( 1 - \frac 13 \right)^{-1} \left(1 - \frac 15\right)^{-1} \left(1 - \frac 17\right)^{-1} \left(1-\frac 1{11}\right)^{-1} \left(1 - \frac 1{13}\right)^{-1},$](http://latex.artofproblemsolving.com/d/5/e/d5e23478e31b53adb822e6d2aad7363ec9fc66bf.png)
and
![$C = 1 + \frac 12 + \frac 13 + \frac 14 + \frac 15 + \frac 16 + \frac 17 + \frac 18 + \frac 19 + \frac 1{10} + \frac 1{11} + \frac 1{12} + \frac 1{13} + \frac 1{14} + \frac 1{15} +\frac 1{16}.$](http://latex.artofproblemsolving.com/2/6/9/269f02ea71e5d27d5957d40740300477d0a605db.png)
Then which of the following inequalities is true?
![$\mathrm{(A) \ } A > B > C \qquad \mathrm{(B) \ } B > A > C \qquad \mathrm{(C) \ } C > B > A \qquad \mathrm{(D) \ } C > A > B \qquad \mathrm{(E) \ } B > C > A$](http://latex.artofproblemsolving.com/9/8/e/98e41e3f9d1b5077255187c05cbfc2846c13db22.png)
Solution
If you imagine expanding out the expression for , you can see that every term in
will appear once, along with plenty of others. (Think of the prime factorizations and you can figure out which products give the terms of
.) Since all terms are positive,
.
$\left(1 - \frac 15\right) \left( 1 + \frac 15) \left(1 - \frac 17\right)$ (Error compiling LaTeX. Unknown error_msg)
so
.
Putting these two facts together, .