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# Finding the remainder when dividing negative numbers

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Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 12:36
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Suppose I want to divide -11 by 5. What is the the remainder and quotient?
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 13:26
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.
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 14:09
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?
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 14:09
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
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 14:25
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...
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 14:33
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...
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 14:58
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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
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 15:16
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.
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 15:51
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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!
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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 17:09
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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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Re: Finding the remainder when dividing negative numbers [#permalink]

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06 Jan 2010, 19:51
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.
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Re: Finding the remainder when dividing negative numbers [#permalink]

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09 Jan 2010, 00:53
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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Re: Finding the remainder when dividing negative numbers [#permalink]

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07 Jul 2010, 08:35
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?
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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]

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02 May 2015, 11:57
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
Re: Finding the remainder when dividing negative numbers   [#permalink] 02 May 2015, 11:57
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