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Proof help!

  1. Dec 12, 2013 #1
    1. The problem statement, all variables and given/known data

    If n is a natural number, then n^4 ends in either zero, one, five, or six.

    2. Relevant equations



    3. The attempt at a solution

    Should I attempt this by cases?
     
  2. jcsd
  3. Dec 12, 2013 #2

    Office_Shredder

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    What cases did you have in mind? Why don't you show us what you are thinking.
     
  4. Dec 12, 2013 #3
    Actually not cases, but by a direct proof so:

    let n^4=(n^2)^2
    Let n be an odd number
    Then n=2k+1 for some integer k
    then n^2= (2k+1)^2
    =4k^2 + 4k +1
    =2(2k^2+2k) + 1

    I don't think this is proving anything. I will try something else
     
  5. Dec 14, 2013 #4
    Let n = 10m + k

    Where m and k are naturals and k lies in the interval [0,9].

    Take the 4th power of this expression. Can you find the term responsible for the final digit of n^4? Why is it responsible for the final digit? What are its possible values?
     
    Last edited: Dec 14, 2013
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