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]]
Let [[set]] <math> \mathcal{A} </math> be a 90-[[element]] [[subset]] of <math> \{1,2,3,\ldots,100\}, </math> and let <math> S </math> be the sum of the elements of <math> \mathcal{A}. </math> Find the number of possible values of <math> S. </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]]
 
 
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Algebra Problems]]
 

Latest revision as of 21:57, 5 June 2009