Results 1 to 3 of 3

Math Help - Congruences

  1. #1
    Newbie
    Joined
    Sep 2007
    From
    Australia
    Posts
    5

    Congruences

    This is the problem which i'm currently stuck on, and I've been told to use principles based on mod, congruence equations and Fermat's theorum:

    Show that 27 * 23^n + 17 * 10^2n is divisible by 11 for all positive integers n
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,735
    Thanks
    642
    Hello, edwardteo.368!

    Show that 27\!\cdot\!23^n + 17\!\cdot\!10^{2n} is divisible by 11 for all positive integers n

    Let: N \:=\:27\!\cdot\!23^n + 17\!\cdot\!10^{2n}

    We have: . N \;\equiv\;5\!\cdot\!1^n + 6\!\cdot\!(\text{-}1)^{2n}\;\pmod{11}

    . . . . . . . . . . \equiv\;5\!\cdot\!1 + 6\!\cdot\!\left[(\text{-}1)^2\right]^n\;\pmod{11}

    . . . . . . . . . . \equiv\;5 + 6\cdot1^n\;\pmod{11}

    . . . . . . . . . . \equiv\;5 + 6\;\pmod{11}

    . . . . . . . . . . \equiv \;11\;\pmod{11}

    . . . . . . . . . . \equiv\;0\;\pmod{11}

    Therefore, N is a multiple of 11.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2007
    From
    Australia
    Posts
    5
    thnx for the help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Congruences
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 11th 2010, 12:52 PM
  2. Congruences
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 7th 2009, 02:26 PM
  3. Congruences
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 18th 2009, 05:12 AM
  4. More congruences
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: March 17th 2009, 10:40 PM
  5. Congruences
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 29th 2008, 09:49 AM

Search Tags


/mathhelpforum @mathhelpforum