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Dvisibility of large numbers

  1. Apr 23, 2007 #1
    1. 6^17 + 17^6 is divisble by 3 or 7?



    2. Relevant equations



    3. 6^1 = 0(mod 3)
    (6^1)^17=0(mod 3) so 6^17 is divisible by 3
    how do u do this?
     
    Last edited: Apr 23, 2007
  2. jcsd
  3. Apr 23, 2007 #2

    matt grime

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    17 to any power is prime.... No.

    What is the stray T in your first line. Where did 61 come from? Why are you raising it to the power 1?
     
  4. Apr 23, 2007 #3
    sorry i changed it
     
  5. Apr 24, 2007 #4

    matt grime

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    Fermat's little theorem.
     
  6. Apr 24, 2007 #5

    HallsofIvy

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    For 3 at least, you don't need anyone as powerful as Fermat! Suppose 617+ 176 were divisible by 3. Then we would have 617+ 176= 3n for some integer n. Then 176= 3n- 617= 3n- 317217= 3(n- 316217) which is impossible: since 17 is prime, no power of 17 is divisible by 3.
     
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