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Math Help - Congreunce modulo 49

  1. #1
    Super Member dhiab's Avatar
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    Congreunce modulo 49

    Prove that :
    6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]
    Last edited by dhiab; December 20th 2009 at 11:34 PM.
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  2. #2
    Senior Member Shanks's Avatar
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    By using the squared method:
    6^{33}\equiv 6^{32}6\equiv 36^{16}6\equiv (-13)^{16}6
    \equiv 13^{16}6\equiv 169^{8}6\equiv 22^{8}6\equiv 484^{4}6 \equiv (-6)^{4}6\equiv 36^{2}6\equiv 13^{2}6\equiv 22\cdot 6\equiv 132\equiv 34
    I leave it to you to find 8^{33}.
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  3. #3
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    Entering this into my computer gives
    6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]<br />

    which might be why you are having trouble
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  4. #4
    Senior Member Shanks's Avatar
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    8^{33}\equiv 36 (mod 49)
    thus we get 21 not 0.
    There may be other better method to prove this congruence equation.
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by badgerigar View Post
    Entering this into my computer gives
    6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right] <br />

    which might be why you are having trouble
    Hello thank you you have 21
    Last edited by dhiab; December 21st 2009 at 05:54 AM.
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  6. #6
    Super Member dhiab's Avatar
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    Hello The general method is :

    \begin{array}{l}<br />
8^0 \equiv 1\left[ {\bmod 49} \right] \\ <br />
8^1 \equiv 8\left[ {\bmod 49} \right] \\ <br />
8^2 \equiv 15\left[ {\bmod 49} \right] \\ <br />
8^3 \equiv 22\left[ {\bmod 49} \right] \\ <br />
8^4 \equiv 29\left[ {\bmod 49} \right] \\ <br />
8^5 \equiv 36\left[ {\bmod 49} \right] \\ <br />
8^6 \equiv 3\left[ {\bmod 49} \right] \\ <br />
8^7 \equiv 1\left[ {\bmod 49} \right] \\ <br />
\end{array}<br />
    The period is 7 then : n = 7k +p
    but 33= 74+5
    8^{33} \equiv 36\left[ {\bmod 49} \right]<br />
     <br />
same method : <br />
\begin{array}{l}<br />
6^0 \equiv 1\left[ {\bmod 49} \right] \\ <br />
6^1 \equiv 6\left[ {\bmod 49} \right] \\ <br />
6^2 \equiv 36\left[ {\bmod 49} \right] \\ <br />
6^3 \equiv 20\left[ {\bmod 49} \right] \\ <br />
6^4 \equiv 22\left[ {\bmod 49} \right] \\ <br />
6^5 \equiv 34\left[ {\bmod 49} \right] \\ <br />
6^6 \equiv 8\left[ {\bmod 49} \right] \\ <br />
6^7 \equiv 1\left[ {\bmod 49} \right] \\ <br />
\end{array}<br />
    The period is 7 then : n = 7k +d

    but 33= 74+5

    6^{33} \equiv 34\left[ {\bmod 49} \right]
    Conclusion :
    6^{33} + 8^{33} \equiv \left( {34 + 36} \right)\left[ {\bmod 49} \right]


    6^{33} + 8^{33} \equiv \left( {34 + 36} \right)\left[ {\bmod 49} \right]because\left( {34 + 36} \right) \equiv 21\left[ {\bmod 49} \right]
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