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>\displaystyle S</math> consists of all [[integer]]s from <math>\displaystyle 1</math> to <math>\displaystyle 2007</math>, inclusive. For how many [[element]]s <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?
  
 
==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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Revision as of 12:28, 15 September 2006

Problem

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