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

m (wik, i don't think this falls under combinatorics..)
m (rv)
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== Problem ==
 
== Problem ==
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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>
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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 ==
 
== Solution ==
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By the [[Triangle Inequality]]:
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The smallest <math>S</math> is <math>1+2+ \ldots +90 = 91 \cdot 45 = 4095</math>. The largest <math>S</math> is <math>11+12+ \ldots +100=111\cdot 45=4995</math>. All numbers between <math>4095</math> and <math>4995</math> are possible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>.
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<div style="text-align:center;">
 
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<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
 
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<math>\log_{10} 12n > \log_{10} 75 </math>
 
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<math> 12n > 75 </math>  
 
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<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
 
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</div>
 
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Also:
 
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<div style="text-align:center;">
 
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<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
 
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<math>\log_{10} 12\cdot75 > \log_{10} n </math>
 
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<math> n < 900 </math>  
 
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</div>
 
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Combining these two inequalities:
 
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<cmath> 6.25 < n < 900 </cmath>
 
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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 ==
 
== See also ==
 
{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
  
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[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Combinatorics Problems]]
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[[Category:Intermediate Algebra Problems]]
 

Revision as of 19:23, 25 September 2007

Problem

Let set $\mathcal{A}$ be a 90-element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$

Solution

The smallest $S$ is $1+2+ \ldots +90 = 91 \cdot 45 = 4095$. The largest $S$ is $11+12+ \ldots +100=111\cdot 45=4995$. All numbers between $4095$ and $4995$ are possible values of S, so the number of possible values of S is $4995-4095+1=901$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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