Results 1 to 7 of 7
Like Tree5Thanks
  • 1 Post By DenisB
  • 1 Post By DenisB
  • 2 Post By Idea
  • 1 Post By johng

Thread: Remainder Problem

  1. #1
    Member
    Joined
    May 2010
    Posts
    168

    Remainder Problem

    I am stuck on this problem, I am trying to work out the largest possible remainder when 2015 is divided 1 or 2 or 3 ...up to 1000.
    In other words, when 2015 is divided by 2 , it has a remainder 1, when divided by 5 its remainder is 0, when divided by 1000, its remainder is 15.
    Which positive integer between 1 and 1000 gives the largest remainder and what is it?
    I have tried pattern searching, prime factorising but no luck..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,935
    Thanks
    360

    Re: Remainder Problem

    2015 / 672 = 2 remainder 671
    Thanks from rodders
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2010
    Posts
    168

    Re: Remainder Problem

    how did you decide on 672?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2013
    From
    Colombia
    Posts
    1,915
    Thanks
    642

    Re: Remainder Problem

    $2015 \div 1008= 1 \, \text{remainder} \, 1007$



    I'd be very surprised if you can find a greater remainder.

    Oops, up to 1000.
    Last edited by Archie; Mar 9th 2018 at 05:33 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,935
    Thanks
    360

    Re: Remainder Problem

    Quote Originally Posted by rodders View Post
    how did you decide on 672?
    ceiling(2015 / 2) = 1008 (Archie's)
    ceiling(2015 / 3) = 672 (mine)
    ceiling(2015 / 4) = 504 ...next

    YOKAY?
    Thanks from rodders
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    855
    Thanks
    402

    Re: Remainder Problem

    Let $r_n$ denote the remainder when $2015$ is divided by $n$

    Obviously $0\leq r_n\leq n-1$

    The equation

    $2015=n\left(\frac{2016}{n}-1\right)+(n-1)$

    shows that

    if $n$ is a divisor of $2016$ then $r_n=n-1$

    Now $672$ is the largest divisor of $2016$ that is less than $1000$

    and $r_{672}=671$

    The proof is not complete. We need to check a few things.
    Thanks from rodders and topsquark
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    1,144
    Thanks
    477

    Re: Remainder Problem

    Here's a slight generalization of the original problem. At least this generalization can not be verified by a simple computer program. Maybe "watson" could prove it.

    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Nov 16th 2014, 02:09 AM
  2. A remainder problem.
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: May 29th 2012, 06:47 AM
  3. Remainder Problem
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: Sep 30th 2010, 01:39 PM
  4. remainder problem
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: Jul 22nd 2009, 09:15 AM
  5. remainder problem
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Jun 3rd 2008, 05:33 AM

/mathhelpforum @mathhelpforum