Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 22"
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== Problem == | == Problem == | ||
+ | Let | ||
− | <center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ } </math></center> | + | <center><math> 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), </math></center> |
+ | |||
+ | <center> <math> 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}, </math></center> | ||
+ | |||
+ | and | ||
+ | |||
+ | <center><math> 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}. </math></center> | ||
+ | |||
+ | Then which of the following inequalities is true? | ||
+ | |||
+ | <center><math> \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 </math></center> | ||
== Solution == | == Solution == | ||
+ | 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>. | ||
== See also == | == See also == | ||
* [[University of South Carolina High School Math Contest/1993 Exam]] | * [[University of South Carolina High School Math Contest/1993 Exam]] |
Revision as of 20:05, 22 July 2006
Problem
Let
and
Then which of the following inequalities is true?
Solution
Quickly simplifying the factors and cancelling gives us that and . It is fairly obvious that will be about . Therefore .