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I Remainders, need Help proving a simple notion

  1. Mar 10, 2016 #1
    Hello

    Today I looked at something that seems like it should have a simple solution, but I may have looked at for too long, and can't solve it. My problem is as follows.

    When you devide a whole number by a whole number n, then it is clear that the possible remainders are 0,1,2,...n-1. If you then look at 7 devided by 5 you get remainder 2, 14 by five gives 4, 21 by 5 gives 1, 28 by 5 gives 3, and 35 by 5 gives 0, then the pattern repeats. That is all possible remainders is achieved by the first five multiplums of seven. I know it must have something to do with 5 and 7 being coprime. Because 18 and 15 shows a different pattern.

    In general is it possible to show that if m is lager than m and n and m are coprime, then the first n multiplums of m devided by n, will always achieve all the possible remainders?

    Are there anyway to prove what will in general happen to the pattern if n and m are not Co prime. It seems to me that all remainders are achieved, but one need to remove the remainders, in which the number that devides n and m, devides.

    Hope my two questions Makes sense.
     
  2. jcsd
  3. Mar 10, 2016 #2

    mathwonk

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    you are asking why 1.7, 2.7, 3.7, 4.7, 5.7 all have different remainders after divison by 5. do you know modular arithmetic? this is asking why none of those 5 numbers are equal mod 5. Well iof they were, then their difference would be zero mod 5. I.e. there would be a number k between 1 and 4, such that k.7 = 0 mod 5. This would be a number k such that 7.k is divisible by 5, and 0 < k < 5. Do you see why this cannot happen? can you generalize?
     
  4. Mar 10, 2016 #3
    Hey thanks i think i got it. I will take a closer look at it tommorrow and try to post the generalization
     
  5. Mar 10, 2016 #4

    mathman

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    For your first question: If two numbers give the same remainder, then their difference is a multiple of the divisor, which makes the original numbers not coprime.
     
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