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Finding the remainder when dividing negative numbers
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06 Jan 2010, 11:36
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
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06 Jan 2010, 16:09
Wiki definition of the remainder:If \(a\) and \(d\) are natural numbers, with \(d\) nonzero, 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
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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.



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Re: Finding the remainder when dividing negative numbers
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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): 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
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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
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06 Jan 2010, 13:25
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
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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
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06 Jan 2010, 13:58
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 \(r = (2+1)*511 = 4\)
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
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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 nonzero, 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
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06 Jan 2010, 14:51
dmetla wrote: Reminder can be either positive or negative Example: If a and d are integers, with d nonzero, 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. 8x6) 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
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06 Jan 2010, 18:51
Bunuel wrote: Wiki definition of the remainder: If \(a\) and \(d\) are natural numbers, with \(d\) nonzero, 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
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08 Jan 2010, 23:53
mrblack wrote: Bunuel wrote: Wiki definition of the remainder: If \(a\) and \(d\) are natural numbers, with \(d\) nonzero, 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
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07 Jul 2010, 07:35
Bunuel wrote: Wiki definition of the remainder: If \(a\) and \(d\) are natural numbers, with \(d\) nonzero, 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
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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 of1/5 * 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 of1/5 which is 1. since we cant really have negative reminder we need to add the negative reminder back to the divisor 5, or 51 = 4, so 4 is the reminder



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Re: Finding the remainder when dividing negative numbers
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21 Sep 2016, 19:12
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 5051evel 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
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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 5051evel 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



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Re: Finding the remainder when dividing negative numbers
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24 Sep 2017, 08:29
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 OGIf 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?



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Re: Finding the remainder when dividing negative numbers
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08 May 2019, 17:05
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 5051evel 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 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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