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Congruence Modulo n Proving

  1. Jan 15, 2010 #1
    Can anyone give me hints to how to prove this??

    Prove that for any positive integer n, n^5 and n have the same units digit in their base 10
    representations; that is, prove that n^5 = n (mod 10).

    Thanks!
     
  2. jcsd
  3. Jan 15, 2010 #2
    What does the Euler-Fermat theorem tells you, when applied to a congruence mod 10?
     
  4. Jan 15, 2010 #3
    I'm sorry, I'm still a bit lost. Can you please explain what the Euler-Fermat theorem is and how I can apply that to this problem?

    Thanks
     
  5. Jan 15, 2010 #4
    See the following link:

    http://planetmath.org/encyclopedia/EulerFermatTheorem.html" [Broken]

    And notice that you only have to prove that:

    [tex]n^{4}\equiv 1 \left(mod 10\right)[/tex]

    For n coprime with 10.
     
    Last edited by a moderator: May 4, 2017
  6. Jan 16, 2010 #5

    Mark44

    Staff: Mentor

    This is equivalent to proving that n5 - n [itex]\equiv[/itex] 0 (mod 10)

    You can show this by proving that n5 - n is even, and is divisible by 5.
    The first part is easy, since two of the factors of n5 - n are n and n + 1, one of which has to be even for any value of n.
    The second part, showing that n5 - n is divisible by 5 can be done by math induction, and isn't too tricky.
     
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