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Math Help - Two algebra problems? Yes I am a little stupid and even if you can do one, that's ok!

  1. #1
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    Exclamation Two algebra problems? Yes I am a little stupid and even if you can do one, that's ok!

    1. Show that 21*18^(2x)+36*7^(3x) is divisible by 19 for all positive integers x

    2. Find all pairs of odd integers m and n which satisfy the following equation:
    m+128n=3mn
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  2. #2
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    Re: Two algebra problems? Yes I am a little stupid and even if you can do one, that's

    Hello, someoneanonymous!

    \text{1. Show that }\,N \:=\:21(18^{2x})+36(7^{3x})\,\text{ is divisible by 19 for all positive integers }x.

    We have: . 21(18^2)^x + 36(7^3)^x

    . . . . . . =\;21(324)^x + 36(343)^x

    . . . . . . \equiv\;21(1)^x + 36(1)^x \pmod{19}

    . . . . . . \equiv\;21 + 36 \pmod{19}

    . . . . . . \equiv\;57 \pmod{19}

    . . . . . . \equiv\;0 \pmod{19}


    Therefore, N is divisible by 19.

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