Problem with mod operation

In math, mod is defined as a relation, rather than an operator. So we would say

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

and

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.

 

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.

 

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 ;) .

 

I don't mind. I got 99% of the cash reward :-)