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

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06 Jan 2010, 12:26

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.

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

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06 Jan 2010, 13:09

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):

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

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06 Jan 2010, 13:09

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, 13:33

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, 13:58

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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, 14:16

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.

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

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06 Jan 2010, 14:51

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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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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:

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

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06 Jan 2010, 18:51

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:

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:

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:

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, 07:35

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:

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

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12 Sep 2014, 02:32

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

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02 May 2015, 10:57

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]

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17 Sep 2016, 08:38

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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21 Sep 2016, 20:14

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)) 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

gmatclubot

Re: Finding the remainder when dividing negative numbers
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21 Sep 2016, 20:14

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