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Mathematical Induction

  1. Jul 28, 2006 #1
    Hey there evryone

    I need some help with this problem as I dont know which direction to go with it.

    Prove by mathematical induction that (13^n)-(6^n) is divisible by 7.



    The Base Step is obviously ok....

    Then assume (13^K)-(6^K) is true

    Then have to prove (13^(k+1))-(6^(k+1)) is true...... how do i do this

    thank heaps....
     
  2. jcsd
  3. Jul 28, 2006 #2

    Office_Shredder

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    Can you factor (13^(k+1))-(6^(k+1)) ?
     
  4. Jul 28, 2006 #3
    yeah thats what i have been trying to do but i dont know how to get

    (13^(k+1) - 6^(k+1))

    as a multiple of (13^k - 6^k).

    I dont know if this is the correct direction to take
     
  5. Jul 28, 2006 #4

    StatusX

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    [tex]13^{k+1}-6^{k+1}=13 \cdot (13^k-6^k)+7 \cdot 6^k[/tex]
     
  6. Jul 28, 2006 #5
    how do you do that?
    like what is the step u take to get there
     
  7. Jul 28, 2006 #6
    i get

    13.13^K - 6.6^K

    then what do i do??
     
  8. Jul 28, 2006 #7

    Office_Shredder

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    Instead of trying that..... can you find a way to just show (13^(k+1) - 6^(k+1)) is divisible by seven? What looks like it equal seven in that equation?
     
  9. Jul 28, 2006 #8

    StatusX

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    =13.13^K-(13-7).6^K=13(13^K-6^K)+7.6^K
     
  10. Jul 28, 2006 #9
    thank you very much.....that makes a heap more sense now
    :smile:
     
  11. Jul 28, 2006 #10

    Gokul43201

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    In all this, you've missed out the slightly easier solution Shredder was trying to point you to. There is a standard factorization for a^n - b^n. You use a specific case of this factorization when you write the sum of a geometric series : [itex]1+b+b^2+...+b^{n-1} = (1-b^n)/(1-b) [/itex]
     
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