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Math Help - Remainder?

  1. #1
    Junior Member
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    Remainder?

    Q) Find the remainder of 37^47^57 divided by 16?

    Any help will be appreciated?

    Thanks,
    Ashish
    Last edited by mr fantastic; October 26th 2009 at 04:01 AM. Reason: Deleted excessive ?'s
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  2. #2
    Super Member

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    Hello, Ashsh!

    I think I've solved it . . .


    Find the remainder of 37^{47^{57}} divided by 16
    Since 37 \equiv 5 \text{ (mod 16)}

    . . we have: . 5^{47^{57}} \text{ (mod 16)}


    Now consider consecutive powers-of-5 (mod 16):

    . . \begin{array}{|c|c|}<br />
2^n & \text{(mod 16)} \\ \hline \\[-4mm]<br />
5^1 & 5 \\ 5^2 & 9 \\ 5^3 & 13 \\ 5^4 & 1  \\ 5^5 & 5 \\ \vdots & \vdots \end{array}

    The remainders step through a 4-step cycle: . 5-9-13-1- \hdots


    Now we must determine: . 47^{57}\text{ (mod 4)}

    We have: . 47^{57}\text{ (mod 4)} \;=\;(\text{-}1)^{57}\text{ (mod 4)} \;=\;-1\text{ (mod 4)}\;=\;3\text{ (mod 4)}

    Hence: . 5^{47^{57}} \;=\;5^3\text{ (mod 4)}

    And: . 5^3 \div 16 has remainder 13.


    Therefore: . 37^{47^{57}} \div 16\,\text{ has remainder }13.



    But someone check my reasoning and work ... please!
    .
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