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N^3 + 100

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the largest positive integer n such that n^3 + 100 is divisible by n + 10.

    2. Relevant equations



    3. The attempt at a solution

    The hint ive been given is to use (mod n + 10) to get rid of the n.

    but i dont quite see how it would work :S

    all my attempts have gotten nowhere, lol.

    a little prod in the right direction would be nice :)

    cheers
     
  2. jcsd
  3. Oct 12, 2008 #2

    gabbagabbahey

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    My method would be to just use any polynomial dividing technique to find the remainder of (n^3+100)/(n+10). If n^3+100 is divisible by n+10, then the remainder will have to be an integer.
     
  4. Oct 13, 2008 #3
    right, i think i may have a solution to this problem. can someone please check it for me :) thanks.


    say that n^3 + 100 is divisible by n + 10.

    then n^3 + 100 = 0 (mod n + 10)

    n = -10 (mod n + 10)

    so (-10)^3 + 100 = 0 (mod n + 10)

    ... 0 = 900 (mod n + 10)

    we want to maximise n, and because the above line essentially means that 900 is an integer multiple of (n + 10), the maximum n would be when 900 = n +10

    so n = 890...


    is this reasoning correct?

    thanks:)
     
  5. Oct 13, 2008 #4

    gabbagabbahey

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    Looks good to me:approve: I got the same thing using polynomial division.
     
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