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Math Help - Can't prove this equation

  1. #1
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    Can't prove this equation

    I need help on how to prove that 9^(n+3) + 4^n is divisible by 5. Please help, I have no idea of how to solve this
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Can't prove this equation

    I would use induction, observing that:

    (9^{n+4}+4^{n+1})-(9^{n+3}+4^n)=5\cdot9^{n+3}+3(9^{n+3}+4^n)
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  3. #3
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    Re: Can't prove this equation

    $9^n 9^3 + 4^n$
    $(5+4)^n 729 + 4^n$
    Use binomial theorem to expand $(5+4)^n$
    All terms except the last term $4^n$ is divisible by 5.
    Take the two left over terms.
    $4^n * 729 + 4^n$. This is divisible by 5.
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  4. #4
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    Re: Can't prove this equation

    Or you can observe that

    9^{n+3} + 4^n = 729(9^n) + 4^n

    \equiv 729(4^n) + 4^n (mod 5)

    \equiv 730(4^n) (mod 5)

    \equiv 0 (mod 5)
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  5. #5
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    Re: Can't prove this equation

    Quote Originally Posted by richard1234 View Post
    Or you can observe that

    9^{n+3} + 4^n = 729(9^n) + 4^n

    \equiv 729(4^n) + 4^n (mod 5)

    \equiv 730(4^n) (mod 5)

    \equiv 0 (mod 5)
    even simpler:

    9 = 4 (mod 5), whence 93 = 43 = 4 (mod 5) (since 42 = 16 = 1 (mod 5)).

    thus 9n+3 + 4n = (4n)4 + 4n = 5(4n) = 0 (mod 5).

    why do this? because why should i have to calculate the cube of 9, when i can calculate the cube of 4 instead (729 is a number i don't use everyday)?
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  6. #6
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    Re: Can't prove this equation

    Quote Originally Posted by Deveno View Post
    why do this? because why should i have to calculate the cube of 9, when i can calculate the cube of 4 instead (729 is a number i don't use everyday)?
    Yeah that solution's slightly simpler than mine. I just happen to have 9^3 memorized.
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