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How to find the |S|?

  1. Jan 6, 2015 #1
    1. The problem statement, all variables and given/known data
    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|.

    2. Relevant 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

    3. 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.
     

    Attached Files:

  2. jcsd
  3. Jan 6, 2015 #2
    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##.
     
  4. Jan 6, 2015 #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!​
     
  5. Jan 6, 2015 #4
    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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