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Manager  Joined: 27 Apr 2008
Posts: 151
Finding the remainder when dividing negative numbers  [#permalink]

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Suppose I want to divide -11 by 5. What is the the remainder and quotient?
Math Expert V
Joined: 02 Sep 2009
Posts: 65290
Re: Finding the remainder when dividing negative numbers  [#permalink]

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3
7
Wiki definition of the remainder:
If $$a$$ and $$d$$ are natural numbers, with $$d$$ non-zero, it can be proven that there exist unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r} < d$$. The number $$q$$ is called the quotient, while $$r$$ is called the remainder.

GMAT Prep definition of the remainder:
If $$a$$ and $$d$$ are positive integers, there exists unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r}<d$$. $$q$$ is called a quotient and $$r$$ is called a remainder.

Moreover many GMAT books say factor is a "positive divisor", $$d>0$$.

I've never seen GMAT question asking the ramainder when dividend ($$a$$) is negative, but if we'll cancel this restriction ($$dividend=a<0$$), and only this restriction, meaning that we'll leave the other one ($$0\leq{r}<d$$), then division of $$-11$$ by $$5$$ will result:

$$0\leq{r}<d$$, $$a=qd + r$$ --> $$0\leq{r}<5$$, $$-11=(-3)*5+4$$. Hence $$remainder=r=4$$.

Hope it helps.
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Intern  Joined: 22 Dec 2009
Posts: 10
Re: Finding the remainder when dividing negative numbers  [#permalink]

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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.
Manager  Joined: 27 Apr 2008
Posts: 151
Re: Finding the remainder when dividing negative numbers  [#permalink]

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handsomebrute wrote:
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?
Manager  Joined: 22 Dec 2009
Posts: 225
Re: Finding the remainder when dividing negative numbers  [#permalink]

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handsomebrute wrote:
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.

I doubt this...

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

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

Cheers!
JT
Manager  Joined: 27 Apr 2008
Posts: 151
Re: Finding the remainder when dividing negative numbers  [#permalink]

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jeeteshsingh wrote:
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...
Manager  Joined: 22 Dec 2009
Posts: 225
Re: Finding the remainder when dividing negative numbers  [#permalink]

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mrblack wrote:
jeeteshsingh wrote:
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...
Manager  Joined: 22 Dec 2009
Posts: 225
Re: Finding the remainder when dividing negative numbers  [#permalink]

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3
OK... finally I got some insight on this..

There is way of calculating the remainder when dealing with Negative Numbers....

Say you have a Number 'N' which is to be divided by 'd'. On division you get a quotient 'q' and remainder 'r'

Now if N is +ve, then
$$N = q*d + r$$
where $$q*d \leq N \leq (q+1)*d$$ and $$r = N - q*d$$

e.g. $$11 = 5*2 + 1$$

Now if N is negative, we have a change.... in this equation:
$$q*d \leq N \leq (q+1)*d$$... mutliply by -ve throughout and see the change as:
$$-(q+1)*d \leq -N \leq -(q)*d$$

Therefore remainder now would be:
$$r = (-N) - (-(q+1)*d) = (q+1)*d -N$$

e.g. Let N = -11, d = 5...
Hence $$-(3)*5 \leq -11 \leq -(2)*5$$
Therefore $$q = 2$$
Hence $$r = (2+1)*5-11 = 4$$

You can verify this as $$-11 = 5 * (-3) + 4$$
Therefore quotient = -3 and remainder = 4!
Hope it makes sence and is clear....

Cheers!
JT
Manager  Joined: 27 Aug 2009
Posts: 57
Re: Finding the remainder when dividing negative numbers  [#permalink]

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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.
Manager  Joined: 22 Dec 2009
Posts: 225
Re: Finding the remainder when dividing negative numbers  [#permalink]

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1
dmetla wrote:
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!
Manager  Joined: 27 Apr 2008
Posts: 151
Re: Finding the remainder when dividing negative numbers  [#permalink]

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Bunuel wrote:
Wiki definition of the remainder:
If $$a$$ and $$d$$ are natural numbers, with $$d$$ non-zero, it can be proven that there exist unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r} < d$$. The number $$q$$ is called the quotient, while $$r$$ is called the remainder.

GMAT Prep definition of the remainder:
If $$a$$ and $$d$$ are positive integers, there exists unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r}<d$$. $$q$$ is called a quotient and $$r$$ is called a remainder.

Moreover many GMAT books say factor is a "positive divisor", $$d>0$$.

I've never seen GMAT question asking the ramainder when dividend ($$a$$) is negative, but if we'll cancel this restriction ($$dividend=a<0$$), and only this restriction, meaning that we'll leave the other one ($$0\leq{r}<d$$), then division of $$-11$$ by $$5$$ will result:

$$0\leq{r}<d$$, $$a=qd + r$$ --> $$0\leq{r}<5$$, $$-11=(-3)*5+4$$. Hence $$remainder=r=4$$.

Hope it helps.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 65290
Re: Finding the remainder when dividing negative numbers  [#permalink]

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mrblack wrote:
Bunuel wrote:
Wiki definition of the remainder:
If $$a$$ and $$d$$ are natural numbers, with $$d$$ non-zero, it can be proven that there exist unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r} < d$$. The number $$q$$ is called the quotient, while $$r$$ is called the remainder.

GMAT Prep definition of the remainder:
If $$a$$ and $$d$$ are positive integers, there exists unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r}<d$$. $$q$$ is called a quotient and $$r$$ is called a remainder.

Moreover many GMAT books say factor is a "positive divisor", $$d>0$$.

I've never seen GMAT question asking the ramainder when dividend ($$a$$) is negative, but if we'll cancel this restriction ($$dividend=a<0$$), and only this restriction, meaning that we'll leave the other one ($$0\leq{r}<d$$), then division of $$-11$$ by $$5$$ will result:

$$0\leq{r}<d$$, $$a=qd + r$$ --> $$0\leq{r}<5$$, $$-11=(-3)*5+4$$. Hence $$remainder=r=4$$.

Hope it helps.

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.
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Manager  Joined: 01 Jan 2008
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Re: Finding the remainder when dividing negative numbers  [#permalink]

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Bunuel wrote:
Wiki definition of the remainder:
If $$a$$ and $$d$$ are natural numbers, with $$d$$ non-zero, it can be proven that there exist unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r} < d$$. The number $$q$$ is called the quotient, while $$r$$ is called the remainder.

GMAT Prep definition of the remainder:
If $$a$$ and $$d$$ are positive integers, there exists unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r}<d$$. $$q$$ is called a quotient and $$r$$ is called a remainder.

Moreover many GMAT books say factor is a "positive divisor", $$d>0$$.

I've never seen GMAT question asking the ramainder when dividend ($$a$$) is negative, but if we'll cancel this restriction ($$dividend=a<0$$), and only this restriction, meaning that we'll leave the other one ($$0\leq{r}<d$$), then division of $$-11$$ by $$5$$ will result:

$$0\leq{r}<d$$, $$a=qd + r$$ --> $$0\leq{r}<5$$, $$-11=(-3)*5+4$$. Hence $$remainder=r=4$$.

Hope it helps.

Hi, bunuel

how did you come up with -3, I mean why exactly -3 and not -2 or -17?
Intern  Joined: 06 Jun 2014
Posts: 41
Re: Finding the remainder when dividing negative numbers  [#permalink]

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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
Math Revolution GMAT Instructor V
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Re: Finding the remainder when dividing negative numbers  [#permalink]

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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))
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Re: Finding the remainder when dividing negative numbers  [#permalink]

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MathRevolution wrote:
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))

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
Intern  Joined: 11 Nov 2016
Posts: 3
Re: Finding the remainder when dividing negative numbers  [#permalink]

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handsomebrute wrote:
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.

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?
GMAT Tutor P
Joined: 24 Jun 2008
Posts: 2320
Re: Finding the remainder when dividing negative numbers  [#permalink]

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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:

MathRevolution wrote:
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))

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).

Srikar1729 wrote:
Can the quotient be negative?

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).
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Re: Finding the remainder when dividing negative numbers  [#permalink]

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_________________ Re: Finding the remainder when dividing negative numbers   [#permalink] 28 Jun 2020, 02:13

# Finding the remainder when dividing negative numbers  