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Homework Help: Modular arithmetic

  1. Aug 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Okay, so I'm going to find the smallest positive remainder of (21999+31998+51997) divided by seven.


    2. Relevant equations



    3. The attempt at a solution
    Well, I did like this:
    23 is congruent to 1 (mod 7). Therefore, 21999= (23)1999/3 is congruent to 1 (mod 7).
    33 is congruent to (-1) (mod 7). Therefore, 31998=(33)666 is congruent to (-1)666=1 (mod 7).
    53 is congruent to (-1) (mod 7). Therefore, 51997=(53)665*25 is congruent to (-1)665*4 = -4 (mod 7)

    So, all in all the remainder should be two. However it says in the key that it is six, and I can't see where I'm wrong. Got any suggestions?
     
  2. jcsd
  3. Aug 1, 2009 #2

    Mark44

    Staff: Mentor

    Here's what I get:
    23 is congruent to 1 (mod 7). Therefore, 21999= (23)666*2 is congruent to 2 (mod 7) (not 1 mod 7 as you had).
    33 is congruent to (-1) (mod 7). Therefore, 31998=(33)666 is congruent to (-1)666=1 (mod 7).
    53 is congruent to (-1) (mod 7). Therefore, 51997=(53)665*25 is congruent to (-1)665*4 = -4 (mod 7) = 3 mod 7.

    Add 'em up and you get 6 mod 7.
     
  4. Aug 1, 2009 #3
    Ok, thanks a lot!
    I just wonder, why is it wrong to do the way I did, why doesn't it work?
     
  5. Aug 1, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi triac! :smile:
    Because 1999/3 isn't a whole number. :wink:
     
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