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Math Help - 2 Induction Problems

  1. #1
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    2 Induction Problems

    1. Prove, by induction, that for every positive integer n, 4^n +14 \equiv 0 (mod 6)

    Base case, n=1, 6 \mid 18-0

    Assume: 4^{n+1} +14 \equiv 0 (mod 6)

    Then I could write is as  4^n * 4 +14 \equiv 0 (mod 6)

    I have no idea how to proceed.
    Last edited by yzc717; September 15th 2009 at 09:27 PM.
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  2. #2
    Super Member Matt Westwood's Avatar
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    If we assume 4^n + 14 \equiv 0 (mod 6) that means 4^n \equiv 4 mod 6

    So times it by 4 for 4^{n+1} \equiv 4 \times 4 \equiv 4 (mod 6) and there you go.
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  3. #3
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    Quote Originally Posted by Matt Westwood View Post
    If we assume 4^n + 14 \equiv 0 (mod 6) that means 4^n \equiv 4 mod 6

    So times it by 4 for 4^{n+1} \equiv 4 \times 4 \equiv 4 (mod 6) and there you go.
    so all we cares about is the 4^n, it is always divisible by 6, the constant doesn't matter?
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  4. #4
    Super Member Matt Westwood's Avatar
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    No, 4^n always leaves a remainder 4 when divided by 6. As 14 leaves a remainder 2 when divided by 6, so add them together and the sum is then divisible by 6.
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