Results 1 to 5 of 5

Thread: 10^{2001}+1 is divisible by 11

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    10^{2001}+1 is divisible by 11

    $\displaystyle 10^{2001}+1$ is divisible by 11

    $\displaystyle \varphi(11)=10\Rightarrow 10^{10}\equiv 1 \ \mbox{(mod 11)}$

    $\displaystyle 10^{10*200+1}+1=(10^{10})^{200}*10+1=1^{200}*10+1\ Rightarrow$$\displaystyle 11\equiv 0 \ \mbox{(mod 11)}$

    Is this correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by dwsmith View Post
    $\displaystyle 10^{2001}+1$ is divisible by 11

    $\displaystyle \varphi(11)=10\Rightarrow 10^{10}\equiv 1 \ \mbox{(mod 11)}$

    $\displaystyle 10^{10*200+1}+1=(10^{10})^{200}*10+1=1^{200}*10+1\ Rightarrow$$\displaystyle 11\equiv 0 \ \mbox{(mod 11)}$

    Is this correct?
    Yes this is correct. Another way you could've gone is to apply the divisibility rule for 11.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Correct!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    $\displaystyle 10^{2001}+1$ is :

    $\displaystyle 1...(2001 \\ times \\ zero)+1$

    or


    $\displaystyle 1...(2001 \\ times \\ zero)...1$

    credit to Mr. chiph588@
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,115
    Thanks
    7
    This is how I would do it:

    $\displaystyle 10^{2001}+1=(11-1)^{2001}+1$

    On expanding the RHS with the binomial theorem, we see that it is indeed divisible by 11.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Feb 20th 2013, 09:32 AM
  2. Replies: 8
    Last Post: Jul 3rd 2011, 03:55 PM
  3. Replies: 1
    Last Post: May 7th 2010, 11:49 PM
  4. Replies: 5
    Last Post: Jan 1st 2010, 01:59 AM
  5. Mathcad 2001
    Posted in the Math Software Forum
    Replies: 17
    Last Post: Sep 14th 2008, 03:38 AM

/mathhelpforum @mathhelpforum