Finding the remainder when dividing negative numbers

Just like in multiplication, when one but not both numbers are negative, the answer will be negative. If both numbers are negative, the answer will be positive. In number theory, remainders are always positive. -11 divided by 5 equals -2 remainder 1.

But according to the formula y=xq + r (q = quotient, r = remainder):

-11 = 5q + r

If q = -2 and r = 1 as you said, then:

-11 = 5(-2) + (1)
-11 = -10 + 1
-11 = -9

Did I miss a step somewhere?

I doubt this...

- 11 = 5 x (-2) + 1 ????

Guess remainder should be -1. It would then satisfy the equation above!

Cheers!
JT

Can you have negative remainders? Never heard of such a thing before...

No offence mate.. but I havent heard of anything as positive remainder (on a lighter note)

Usually we simply call them as remainders as long as I remember...

Reminder can be either positive or negative
Example:
If a and d are integers, with d non-zero, then a remainder is an integer r such that a = qd + r for some integer q, and with 0 ≤ |r| < |d|.
When defined this way, there are two possible remainders. For example, the division of −42 by −5 can be expressed as either
−42 = 9×(−5) + 3
as is usual for mathematicians, or
−42 = 8×(−5) + (−2).
So the remainder is then either 3 or −2.

the negative remainder is obtained from the positive one just by subtracting 5, which is d. This holds in general. When dividing by d, if the positive remainder is r1, and the negative one is r2, then
r1 = r2 + d.

I think this is not correct because.....

in −42 = 8×(−5) + (−2)... you are dividing -42 with 8 and choosing -5 as quotient...but -5 x 8 gives -40 which is greater than -42 and hence is not as per the divisibility methodology....

If u have to divide 11 by 5, you would use 10 (5x2) and not 15 (5x3) as 10 is less 11..
Similarly, for -42 you should use -48 (i.e. 8x-6) as -48 is less than -42 but -40 is more than -42!

The eq −42 = 8×(−5) + (−2) is correct mathematically but fails to ascertain the quotient and remainder correctly...!

More over your first example (highlighted in red) is wrong.. as you are dividing -42 with -5 which means you dividing 42 with 5 and hence the situation of a negative number being divided is void!

Thanks for the lengthy response Bunuel. The wiki entry seems to imply that there could be a +ve or -ve remainder. Referring to my original question once more, it seems that -11/5 can be either:

-11 = (-3)*5 + 4
or
-11 = (-2)*5 + (-1)

I think either case is valid.

Wiki is not a best source for Math but even its definition: \(0\leq{r} < d\), so \(0\leq{r}\), remainder must be positive. But again I wouldn't worry about this issue for GMAT.

Hi, bunuel

how did you come up with -3, I mean why exactly -3 and not -2 or -17?

This is what I think and according to the reminder theory :

when a=b*c => reminder of a/x i same as reminder of (b*c)/x
moreover it same as reminder of (reminder of b/x * reminder ofc/x)/x

I will try to ilsitrate with the example. actualy this concept I saw on the forum somewhere posted by EvaJager I think.

-11 = -1 *11 => R of -1*11/5 is Rof (R of-1/5 * R of 11/5)/5

now R of -1/5 is -1 and R of 11/5 is 1 => R of -1*1/5 is just R of-1/5 which is -1. since we cant really have negative reminder we need to add the negative

reminder back to the divisor 5, or 5-1 = 4, so 4 is the reminder

As always, questions like CMT 3,4 are often on the exam in a more developed form. Take a look at below and you’ll see the type of question that’s really common these days. This one, too, is a 5051-evel problem that is equivalent to CMT 4(A). You must get used to these. You need to know how variable approaches and CMT are related.

(ex 1) (integer) If x and y are postive integers, what is the remainder when 100x+y is divided by 11?
1) x=22
2) y=1
==> If you change the original condition and the prolem, you all ways get the remainder of 1 if you divide 100x by 11 regardless of the value of x. Thus, you only need to know y. Therefore, the answer is . This is a typical 5051 level problem. (CMT 4(A))
Answer: A

Is it B or A? I am getting C.

St 1:

y can be anything

St 2:

For 101---- r=2
For 201 ----r=3

Can the quotient be negative?

as per OG

If x and y are positive integers, there exist unique integers q and r , called the quotient and remainder, respectively, such that y = xq + r and 0 ≤ r < x . For example, when 28 is divided by 8, the quotient is 3 and the remainder is 4 since 28 = (8)(3) + 4. Note that y is divisible by x if and only if the remainder r is 0; for example, 32 has a remainder of 0 when divided by 8 because 32 is divisible by 8. Also, note that when a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer. For example, 5 divided by 7 has the quotient 0 and the remainder 5 since 5 = (7)(0) + 5.

So quotient can't be negative?

I happened upon this old thread today, and since it might be very confusing to any test taker who also happens upon it, I thought it might be helpful to clear up some issues and open questions in earlier posts:


Almost none of this makes sense. As the question is written, the answer is C; you do not always get a remainder of 1 when you divide 100x by 11 (you can get any remainder at all, depending on the value of x). I imagine instead the question intends to ask about 100^x + y (and not about 100x + y). In that case, you do get a remainder of 1 when you divide 100^x by 11, no matter what x is, but that means the answer to the question is B (not A). Nor is it true that this kind of question is "really common these days", except in the one instance where you are dividing specifically by 5 or by 10 (i.e. when the question is about units digits).


In theory, yes, if you divide a negative number by a positive one, you'll have a negative quotient. For example, if you divide -20 by 5, the quotient is -4. But in any actual GMAT question about quotients and remainders, the question will declare in advance that all of your quantities are positive integers. So you don't need to worry about how to answer the question "what is the remainder when -19 is divided by 5?", for example (though, for interest only, the answer is 'one', because -19 is one greater than a multiple of 5, specifically -20).

Bunuel IanStewart What about the opposite? Can we divide a numerator by a negative divisor? How would you determine the remainder?

In theory, yes, though on the GMAT you'd absolutely never need to be concerned about that situation. When you divide, say, 19 by 5, you're really asking "19 is how much larger than the nearest smaller multiple of 5?", and since 15 is the nearest smaller multiple of 5, the remainder is 4. If you instead divide 19 by -5, you're asking the same thing: 19 is how much larger than the nearest smaller multiple of -5? And since 15 is a multiple of -5 (it's -3 times -5), the remainder is again 4. So whether you divide by 5 or by -5, you'll get the same remainder either way.

That's using the standard definition of a remainder, but you might notice with this definition, when you divide 19 by -5, you get a remainder of 4, but when you divide -19 by 5, you get a remainder of 1 (because -19 is 1 larger than -20, a multiple of 5). That might seem paradoxical, because 19 divided by -5 and -19 divided by 5 are supposed to be the same thing, so some people use a different definition of remainder (they'd choose the smallest distance to any multiple, rather than the distance to the nearest smaller multiple) so that both situations produce the same answer.

That said, dividing by negatives is of very little theoretical or practical interest in divisibility or remainder situations (I don't think I ever was concerned with it once, and I did grad school in Number Theory). Dividing negatives by positives, on the other hand, turns out to be important in the subject called 'modular arithmetic', which is just the math jargon term for 'remainder arithmetic', so that situation can sometimes be useful to understand, at least in more advanced math, and occasionally on GMAT questions if you happen to use modular arithmetic methods (which can be convenient sometimes, but which aren't ever necessary).