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Math Help - Divisibility Problem

  1. #1
    Junior Member
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    Divisibility Problem

    Prove that there exist two distinct powers of 3 whose difference is a multiple of 2014. (The exponents need to be non-negative integers. To help clarify the question, if we had asked for 24 instead of 2014, then 3^4 and 3^2 are two powers of 3 whose difference 3^4 - 3^2 = 72 is a multiple of 24.)

    My attempt at the solution:
    2014|$3^x-3^y$
    $3^x \equiv 3^y (mod 2014)$
    I don't know where to go from here. How can I solve this for x and y?
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  2. #2
    MHF Contributor
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    Re: Divisibility Problem

    this looks to be an application of the pigeon hole principle.

    Powers of 3 Whose Difference Is Divisible by 1997

    is a similar problem and solution, scroll down a bit for the solution
    Thanks from nubshat
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