Difference between revisions of "2006 AIME A Problems/Problem 2"

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== Problem ==
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#REDIRECT [[2006 AIME I Problems/Problem 2]]
The lengths of the sides of a triangle with positive area are <math>\log_{10} 12</math>, <math>\log_{10} 75</math>, and <math>\log_{10} n</math>, where <math>n</math> is a positive integer. Find the number of possible values for <math>n</math>.
 
 
 
== Solution ==
 
By the [[Triangle Inequality]]:
 
 
 
<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
 
 
 
<math>\log_{10} 12n > \log_{10} 75 </math>
 
 
 
<math> 12n > 75 </math>
 
 
 
<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
 
 
 
Also:
 
 
 
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
 
 
 
<math>\log_{10} 12\cdot75 > \log_{10} n </math>
 
 
 
<math> n < 900 </math>
 
 
 
Combining these two inequalities:
 
 
 
<math> 6.25 < n < 900 </math>
 
 
 
The number of possible integer values for <math>n</math> is the number of integers over the interval <math>(6.25 , 900)</math> which is <math>893</math>
 
 
 
== See also ==
 
*[[2006 AIME II Problems]]
 

Latest revision as of 20:57, 5 June 2009