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Math Help - Find the largest integer n, where 2009^n divides (2008^2009^2010 + 2010^2009 ^2008)

  1. #1
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    Find the largest integer n, where 2009^n divides (2008^2009^2010 + 2010^2009 ^2008)

    Find the largest integer n, where 2009^n divides (2008^(2009^2010) + 2010^(2009 ^2008))

    So, the only observation I've made is that you can rewrite this in the form:

    (x-1)^(x^(x+1)) + (x+1)^(x^(x-1))

    I also know that I have to use the binomial theorem, but I'm not exactly sure how that relates to finding the largest n possible...

    Help?
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  2. #2
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    Hint - Maybe this will help you move fwd
    In this case x = 2009 is odd.

    Had x been even life would have been simpler !

    I have not tried this completely but would go down this route first - hope it works
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by elemental View Post
    Find the largest integer n, where 2009^n divides (2008^(2009^2010) + 2010^(2009 ^2008))

    So, the only observation I've made is that you can rewrite this in the form:

    (x-1)^(x^(x+1)) + (x+1)^(x^(x-1))

    I also know that I have to use the binomial theorem, but I'm not exactly sure how that relates to finding the largest n possible...

    Help?
    From how this question is posed, I have the feeling that this is from an official math olympiad contest of some kind (due to the fact that the numerical values for years appear throughout the problem and it is something that pops up in olympiad questions from time to time).

    If you can justify this isn't from an olympiad of some kind, please message me and I'll consider reopening this thread.

    Until then, Thread Closed.

    EDIT: The user has informed me that this comes from an olympiad competition that took place in 2008. As a result, feel free to make contributions to this thread.
    Last edited by Chris L T521; November 18th 2010 at 10:12 AM.
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  4. #4
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    By looking at the powers of n that occur in the binomial coefficients n^{n-1}\choose k, you should be able to see that

    (1+n)^{n^{n-1}} = 1 + n^n + (\text{multiples of }n^{n+1}).

    In the same way, if n is odd then

    (-1+n)^{n^{n+1}} = -1 + (\text{multiples of }n^{n+1}).

    Therefore (n-1)^{n^{n+1}} + (n+1)^{n^{n-1}} is a multiple of n^n, but not a multiple of n^{n+1}.

    So the highest power of 2009 that divides 2008^{2009^{2010}}\!\! + 2010^{2009^{2008}} is 2009^{2009}.
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