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Quick problem to do with powers

  1. Apr 22, 2004 #1
    hi, can someone try and represent:

    3(3^(2n+4) - 2^(2n))

    as a multiple of five. Been workin on it but I'm not very good at this sorta thing.

    Cheers

    Pat
     
  2. jcsd
  3. Apr 22, 2004 #2

    Njorl

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    It is easy to prove it is so, but I haven't seen any quick method for expressing it as a factor of 5, ie 5*f(x). The last digit of the "3" exponential term alternates between 9 and 1. The last digit of the "2" exponential term alternates between 4 and 6. The subtraction always yields a 5 for the last digit, x9-y4 or x1-y6.

    Njorl
     
  4. Apr 22, 2004 #3

    uart

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    Sorry, no obvious factors of five there.

    How about 5 ( 48.6 * 3^(2n) - 0.6 * 2^(2n) )

    :p
     
  5. Apr 22, 2004 #4
    That's cheating, uart ;)
     
  6. Apr 22, 2004 #5

    uart

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    Hehe that's why I put the ":p" there, it means I was poking my tongue out. :biggrin:


    Anyway it's actually pretty easy. You can readily reduce the problem to proving that (9^n - 4^n) is a multiple of 5. Just express the 9^n as (5+4)^n and use the binomial expansion formula. The 4^n terms cancel out you're left with only terms that are multiples of 5
     
    Last edited: Apr 22, 2004
  7. Apr 22, 2004 #6
    cheers for your help, altho I need to use the 3 because the original proof was by induction to prove f(x) = 3^(2n+4) - 2^(2n) is always a multiple of five, and I've been told to do it by:

    f(k+1) - f(k) = f(k+1) - 4f(k) + 3f(k)

    Iv proved f(k+1) - 4f(k) is a multiple of 5 by rearranging the powers so need to do it with the 3f(k) now. There should be a way.

    Thanks for your help

    Pat
     
  8. Apr 22, 2004 #7

    uart

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    Yes there's a way, you've already done it. The whole basis of proof by induction is that you can proof f(k+1) by assumming f(k). Of course you must verifiy f(1) yourself, but that is very easy.

    So you actually just assume that f(k) is a multiple of 5 and then show that this implys that f(k+1) is also a multiple of 5. You've done all the required work already. :)
     
  9. Apr 23, 2004 #8

    uart

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    Just to clarify to steps.

    1. Prove that g(k) = f(k+1) - 4*f(k) is a multiple of 5 (without making ANY assumptions that f(k) is a multiple of 5).

    2. Note that f(k+1) = g(k) + 4*f(k)

    3. Using 1. and 2. show that f(k) a mult of 5 implies that f(k+1) is a mult of 5.

    4. Verify f(1) is a mult of 5 and you're done.
     
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