Results 1 to 3 of 3

Math Help - Mods.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Mods.

    FInd an integer m satisfying n \leq m \leq n+11 and m \equiv 7(mod \ 12) in each of the following questions.
    (i) n=100
    10=(-2) \ mod \ 12 where -n<r \leq 0 for x \equiv r (mod \ n).

    \Rightarrow 10^2=4 \ mod \ 12 so I need the number to be 100+3=103.

    (ii). n=10^6
    10=-2 \ mod \  12 where -n<r \leq 0 for x \equiv r (mod \ n).

    \Rightarrow \ 10^6=(-2)^6 \ mod \ 12 \Rightarrow 10^6=64 \ mod \ 12 \Rightarrow \ 10^6=4 \ mod \ 12.

    But I require a number m such that m=7 \ mod \ 12.

    Combining my above results reveals the number I need to be 10^6+3.

    (iii). n=-999
    -9= 3 \ mod \ 12 \Rightarrow \ -9(111)=3(111) \ mod \ 12 \Rightarrow \ -999=9 \ mod \ 12

    Comparing this with 7 \ mod \ 12 gives m=-999-2=-1001.

    My method seems to be working pretty well (i've checked them all on a calculator) but is there an easier way to do them?
    Last edited by Showcase_22; April 28th 2009 at 02:44 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,703
    Thanks
    1470
    Strictly speaking, if the problem is just "find a number, m, such that 0< m< n+ 11 and m= 7 (mod 12)" then "m= 7" is a perfectly valid answer for all n! Did you mean "such that n< m< n+ 11"?

    Any number m, satisfying m= 7 (mod 12) can be written m= 7+ 12k for some integer k. To find such a number such that n< m< n+ 11, divide n by 12 to see how large k must be. For example, with n= 10^6, the largest value of n that you have here, 12 divides into [tex]10^6= 1000000[/itex] 83333 times. Taking k= 83333, we have 7+ 12(83333)= 100003.

    For n= -999, divide -999 by 12: 12 divides into -999 -83 times. Taking k= -83, we have 7+ 12(-83)= -989 which satisfies -999< -989< -988.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Ah yes I did sorry!!

    One of the problems is that i'm not supposed to use a calculator, I did all the ones in my previous post using mental artihmetic.

    I'm just pretty worried that when I get tested on this I might have a question where instead of 10^6 or 10^2, I get a number like 10^{20} which is going to be a pain to do!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mods Help
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: April 27th 2010, 12:38 AM
  2. Mods help
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: April 21st 2010, 08:22 AM
  3. Mods and BN
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 17th 2010, 12:58 PM
  4. mods
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 23rd 2010, 04:08 AM
  5. Mods
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: March 24th 2007, 04:24 PM

Search Tags


/mathhelpforum @mathhelpforum