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Arithmetic/Divisibility homework problem

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve for n in the following equations:

    [tex]n-2\mid n^{2}-n+3[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Should i use the euclidean division to solve it? Thank you before hand.

    Edit: When i used euclidean division i got n+1 and a remainder of 5 but i don't know how to get n.
     
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 23, 2012 #2
    Re: Arithmetic/Divisibility

    Start by using the definition of divides to produce an equation. This equation will introduce a new variable, say, k. Solve for n using the quadratic formula (which will involve k) and then choose values of k so that n is an integer (I assume n is an integer in this problem).

    See what you come up with!
     
  4. Apr 23, 2012 #3
    Re: Arithmetic/Divisibility

    So that means you should get: n^2-n+3=k(n-2) am i right?
     
  5. Apr 23, 2012 #4

    Mark44

    Staff: Mentor

    Re: Arithmetic/Divisibility

    Right.
     
  6. Apr 23, 2012 #5
    Re: Arithmetic/Divisibility

    Yes now how can i continue from there?
     
  7. Apr 23, 2012 #6
    Re: Arithmetic/Divisibility

    Ever heard of quadratic equations?
     
  8. Apr 23, 2012 #7
    Re: Arithmetic/Divisibility

    Yes, but n^2-n+3 doesn't have any real roots, if that's what you're talking about.
     
  9. Apr 23, 2012 #8
    Re: Arithmetic/Divisibility

    No, we group everything on one side of the equation so we have coefficients in terms of k and solve THAT.
     
  10. Apr 23, 2012 #9
    Re: Arithmetic/Divisibility

    So we get: n^2-k(n-2)-n+3=0, but where are we supposed to get b^2-4ac. Is b=k(n-2)?
     
  11. Apr 23, 2012 #10
    Re: Arithmetic/Divisibility

    Not quite. We get n^2-(k+1)n+(2k+3)=0 after multiplying out and grouping like terms, and it should be obvious how to proceed from here.
     
  12. Apr 23, 2012 #11
    Re: Arithmetic/Divisibility

    So for Δ i got k^2-6k-11 and then i counted the discriminant again and i got 80. Is what i've done so far correct?
     
    Last edited: Apr 23, 2012
  13. Apr 23, 2012 #12

    Mark44

    Staff: Mentor

    Re: Arithmetic/Divisibility

    Reread what scurty wrote in post #2. He laid out pretty much what you need to do.
     
  14. Apr 23, 2012 #13
    Re: Arithmetic/Divisibility

    So there is no need to find the discriminant of the equation : n^2-(k+1)n+(2k+3)=0?
     
  15. Apr 23, 2012 #14

    Mark44

    Staff: Mentor

    Re: Arithmetic/Divisibility

    I don't see any need for finding the discriminant in this problem.
     
  16. Apr 23, 2012 #15
    Re: Arithmetic/Divisibility

    But the quadratic formula involves finding the discriminant, or do i just use random values of k?
     
  17. Apr 23, 2012 #16

    Mark44

    Staff: Mentor

    Re: Arithmetic/Divisibility

    Of course you find the discriminant in the quadratic formula. What I meant was that there's no reason to find it and stop there.

    Again, take a look at what scurty said in post #2. You are asking questions that suggest you haven't read it or didn't understand it.
     
  18. Apr 23, 2012 #17
    Re: Arithmetic/Divisibility

    Well, he said to produce a new equation, which I did and then he said to find n using the quadratic formula, which will involve k, and that's what I did. I'm confused because we have just started this lesson, and we haven't done any problem like this so far.
     
  19. Apr 23, 2012 #18

    Mark44

    Staff: Mentor

    Re: Arithmetic/Divisibility

    Did you solve the quadratic equation for n? If so, what did you get?
     
  20. Apr 23, 2012 #19
    Re: Arithmetic/Divisibility

    For n^2-(k+1)n+(2k+3)=0 I got a discriminant of k^2-6k-11 then i solved that equation and I got Δ=80 so i got k1=3+2√5 and k2=3-2√5.
     
  21. Apr 23, 2012 #20

    Mark44

    Staff: Mentor

    Re: Arithmetic/Divisibility

    Solve n^2-(k+1)n+(2k+3)=0 for n.
     
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