Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 2"

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== Problem ==
 
== Problem ==
The set <math>\displaystyle S</math> consists of all integers from <math>\displaystyle 1</math> to <math>\displaystyle 2007,</math> inclusive. For how many elements <math>\displaystyle n</math> in <math>\displaystyle S</math> is <math>\displaystyle f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?
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The [[set]] <math>S</math> consists of all [[integer]]s from <math>1</math> to <math>2007</math>, inclusive. For how many [[element]]s <math>n</math> in <math>S</math> is <math>f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?
  
 
==Solution==
 
==Solution==
{{solution}}
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<math>f(n) = \frac{2n^3+n^2-n-2}{n^2-1} = \frac{(n - 1)(2n^2 + 3n + 2)}{(n - 1)(n + 1)} = \frac{2n^2 + 3n + 2}{n + 1} = 2n + 1 + \frac1{n+1}</math>.  So in fact, there are 0 such elements of <math>S</math>.
  
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==See Also==
 
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{{Mock AIME box|year=2006-2007|n=2|num-b=1|num-a=3}}
*[[Mock AIME 2 2006-2007/Problem 1 | Previous Problem]]
 
 
 
*[[Mock AIME 2 2006-2007/Problem 3 | Next Problem]]
 
 
 
*[[Mock AIME 2 2006-2007]]
 

Latest revision as of 10:49, 4 April 2012

Problem

The set $S$ consists of all integers from $1$ to $2007$, inclusive. For how many elements $n$ in $S$ is $f(n) = \frac{2n^3+n^2-n-2}{n^2-1}$ an integer?

Solution

$f(n) = \frac{2n^3+n^2-n-2}{n^2-1} = \frac{(n - 1)(2n^2 + 3n + 2)}{(n - 1)(n + 1)} = \frac{2n^2 + 3n + 2}{n + 1} = 2n + 1 + \frac1{n+1}$. So in fact, there are 0 such elements of $S$.

See Also

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by
Problem 1
Followed by
Problem 3
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