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Show this is divisible by 6.

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    show that n(n-1)(2n-1) is always divisible by 6.
    3. The attempt at a solution
    I see how to get the factor of 2 if n is even then n is divisible by 2.
    is n is odd then n-1 is even.
    Here is my idea to get the factor of 3.
    we know that n and n-1 are consecutive integers.
    and we know that if n is divisible by 3 then so is 2n and the same goes for
    n-1 but 2n-1 comes right before 2n, so we could some how make an argument as to why
    n or n-1 or 2n-1 is divisible by 3. I need to think about it more.
     
  2. jcsd
  3. Jan 29, 2013 #2

    rollingstein

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    Gold Member

    If n or (n-1) are divisible by 3 then your problem is solved.

    If not, n+1 surely is divisible by 3.

    Hence

    n+1 = 3k, where k is some integer

    Hence

    2n -1 = 2 (3k-1) - 1

    2n-1=3(2k-1)

    Hence (2n-1) has to be divisible by 3.

    QED
     
  4. Jan 29, 2013 #3

    SteamKing

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    Homework Helper

    It's not clear what is meant by 'always divisible by 6'.

    By inspection, n(n-1)(2n-1) is divisible by 6 IFF n >= 2
     
  5. Jan 29, 2013 #4
    I would break this up into cases, some of which you already addressed. Consider the possible choices for n. The three cases are

    [tex]n = 3k[/tex]
    [tex]n=3k+1[/tex]
    [tex]n=3k+2[/tex]

    where k is an integer. Consider

    [tex]\frac{n(n-1)(2n-1)}{6}[/tex]

    What happens when you substitute each case in for n into the fraction above?
     
    Last edited: Jan 29, 2013
  6. Jan 29, 2013 #5
    I think the issue should actually be that n must itself be an integer. However, there is nothing wrong with n = 0, n = 1, (in both cases, the product is 0, which is divisible by 6), or even if n is a negative integer.
     
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