How to find the |S|?

  • Thread starter Greychu
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  • #1
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Homework Statement


Let S be the sum of all integer values of n such that [tex]\frac {n^2+12n-43} {n+6} [/tex] is an integer. What is the value of |S|.

Homework Equations


Since it's sum, S = [tex]\frac {n} {2} \ (2a+(n-1)d) [/tex] where a is the first term.
The [tex]\frac {n^2+12n-43} {n+6} [/tex] = x, where x is an integer

The Attempt at a Solution



[tex]\frac {n^2+12n-43} {n+6} [/tex]= x
[tex] {n^2+(12-x)n-43-6x}[/tex] = 0


basically I have no idea "Let S be the sum of all integer values of n" means?
Is it means that S = n or otherwise? Need clarification for this.
 

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Answers and Replies

  • #2
473
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You need to find out for what values of ##n## the expression ##\frac {n^2+12n-43} {n+6} ## is an integer.

It's not an arithmetic progression.

Try defining ##m=n+6## and then express ##n^2+12n-43## in terms of ##m##.
 
  • #3
HallsofIvy
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I would complete the square in the numerator. From that it turns out that the fraction is an integer for only a very small number of values of n!​
 
  • #4
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I think I got it by completing the square.
it will becomes [tex]\frac {(n+6)^2 - 79}{n+6}[/tex]

Since 79 is prime number,
Solving n + 6 = ± 1 and n + 6 = ± 79 will gives rise to 4 integers, which is -5, -7, 73 and -85.
Hence, |S| = |-5-7+73-85| = 24

Thanks for pointing out the completing the square. It helps.
 

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