Results 1 to 6 of 6

Thread: Congreunce modulo 49

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    582
    Thanks
    3

    Congreunce modulo 49

    Prove that :
    $\displaystyle 6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]$
    Last edited by dhiab; Dec 20th 2009 at 10:34 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    By using the squared method:
    $\displaystyle 6^{33}\equiv 6^{32}6\equiv 36^{16}6\equiv (-13)^{16}6$
    $\displaystyle \equiv 13^{16}6\equiv 169^{8}6\equiv 22^{8}6\equiv 484^{4}6$$\displaystyle \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 $\displaystyle 8^{33}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Melbourne
    Posts
    428
    Entering this into my computer gives
    $\displaystyle 6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]
    $


    which might be why you are having trouble
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    $\displaystyle 8^{33}\equiv 36$ (mod 49)
    thus we get 21 not 0.
    There may be other better method to prove this congruence equation.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    582
    Thanks
    3
    Quote Originally Posted by badgerigar View Post
    Entering this into my computer gives
    $\displaystyle 6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]$$\displaystyle
    $

    which might be why you are having trouble
    Hello thank you you have 21
    Last edited by dhiab; Dec 21st 2009 at 04:54 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    582
    Thanks
    3
    Hello The general method is :

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

    but 33= 74+5

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


    $\displaystyle 6^{33} + 8^{33} \equiv \left( {34 + 36} \right)\left[ {\bmod 49} \right]because\left( {34 + 36} \right) \equiv 21\left[ {\bmod 49} \right]$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. modulo ?
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: Apr 16th 2010, 03:07 AM
  2. Modulo of squares = modulo of roots
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: Dec 1st 2009, 09:04 AM
  3. Modulo 16
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jul 3rd 2009, 07:01 AM
  4. Modulo 42
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jun 2nd 2009, 05:00 PM
  5. a modulo m
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Feb 19th 2008, 09:38 PM

Search Tags


/mathhelpforum @mathhelpforum