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Problem with mod operation

  1. Sep 4, 2012 #1
    i know that it shows Remaining for example

    5 mod 3=2
    1 mod 3=1

    but if i select negative number what does it do?


    -10 mod 27 =?

  2. jcsd
  3. Sep 4, 2012 #2
    In math, mod is defined as a relation, rather than an operator. So we would say

    5 [itex]\equiv[/itex] 2 (mod 3)


    1 [itex]\equiv[/itex] 1 (mod 3)

    where the [itex]\equiv[/itex] in this context is pronounced "is congruent to."

    In other words a mod-n statement returns "true" or "false" when applied to pairs of numbers. The general rule is that

    a [itex]\equiv[/itex] b (mod n) if the number n divides a - b.

    Now a lot of people come to mod from programming languages, where mod is not a relation, but is rather an operator, meaning that it returns a single value. That's the usage you've written, so we say

    5 mod 3=2

    and so forth.

    But even though 5 [itex]\equiv[/itex] 2 (mod 3), it's also true that 5 [itex]\equiv[/itex] 47 (mod 3), right? Both 47 and 5 give the same remainder when divided by 3. [That's equivalent to the definition I gave earlier; but you should actually convince yourself of that]

    So if someone asks us what is 5 mod 3, what should the answer be? The convention is that we take the unique number x such that 5 [itex]\equiv[/itex] x (mod 3) and x is greater than or equal to 0, but less than 3.

    With that background, what is the answer to -10 mod 27 = ?

    Well, let's find x such that -10 [itex]\equiv[/itex] = x (mod 27), and x is between 0 and 26 inclusive. A moment's thought will convince you that x = 17 is the right answer here. So

    -10 mod 27 = 17

    That's because

    a) -10 - 17 is divisible by 27; and

    b) 17 is the unique number with that property that's also between 0 and 26, inclusive.

    That's a long answer but it's everything you need to know to make sense of this kind of problem.
    Last edited: Sep 4, 2012
  4. Sep 4, 2012 #3


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    Science Advisor

    Maybe a quick way of answering is a=b mod c is equivalent to : c|(b-a) , or, the

    remainder of dividing a by c is b*. And complement it with Stevel27's answer.

    * This is a technical point, since we usually choose the remainder to be within

    a given range, but we can add multiples.
  5. Sep 4, 2012 #4
    Thanks Dear SteveL27 & Bacle2
    you are my best teacher that dedicate your time to telling me the right answer.

    Thanks again
  6. Sep 4, 2012 #5


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    Science Advisor

    Thanks, but I did only a minimal part.

    That's how I like it, Stevel27 does 99%+ of the work and we split the credit in half ;) .
  7. Sep 4, 2012 #6
    I don't mind. I got 99% of the cash reward :-)
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