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Math Help - Induction Proof: is congruent to....

  1. #1
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    Induction Proof: is congruent to....

    Hi Math Forum,

    Any one have any solutions for this:

    Prove by induction: that for every positive integer n, 4^n +14 is congruent to 0 (mod 6).

    I been at it for days, any help would be greatly appreciated
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  2. #2
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    Quote Originally Posted by brainspasm View Post
    Prove by induction: that for every positive integer n, 4^n +14 is congruent to 0 (mod 6).
    You do realize that means 4^n+14 is divisible by six for each n?
    Clearly it is true for n=1. No?
    Suppose that it is true for n=N now prove it is true for n=N+1.
    Look at this 4^{N+1}+14=\left(4^{N+1}+4\cdot 14\right)- 4\cdot 14 +14=4\left(4^{N}+ 14\right)- 6\cdot 7 .
    Is that a multiple of six?
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  3. #3
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    hey brainspasm,
    it's no so much an induction problem but a problem related to equivalence classes, note:
    <br />
(4^n + 14) \text{mod} 6<br />
    <br />
4 \text{mod} 6 = 4 \wedge 14 \text{mod} 6 = 2 \wedge (4 + 2) \text{mod} 6 = 0<br />

    Why does: 4^n \text{mod} 6 = 4.4^n \text{mod} 6 ?

    And I think you're done.
    Last edited by bmp05; February 6th 2010 at 08:05 AM. Reason: latex funkiness
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