Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 27 Apr 2008
Posts: 189

Finding the remainder when dividing negative numbers [#permalink]
Show Tags
06 Jan 2010, 12:36
1
This post was BOOKMARKED
Suppose I want to divide 11 by 5. What is the the remainder and quotient?



Intern
Joined: 22 Dec 2009
Posts: 13

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



Manager
Joined: 27 Apr 2008
Posts: 189

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



Senior Manager
Joined: 22 Dec 2009
Posts: 350

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
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
_________________
Cheers! JT........... If u like my post..... payback in Kudos!!
Do not post questions with OAPlease underline your SC questions while postingTry posting the explanation along with your answer choice For CR refer Powerscore CR BibleFor SC refer Manhattan SC Guide
~~Better Burn Out... Than Fade Away~~



Manager
Joined: 27 Apr 2008
Posts: 189

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



Senior Manager
Joined: 22 Dec 2009
Posts: 350

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
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...
_________________
Cheers! JT........... If u like my post..... payback in Kudos!!
Do not post questions with OAPlease underline your SC questions while postingTry posting the explanation along with your answer choice For CR refer Powerscore CR BibleFor SC refer Manhattan SC Guide
~~Better Burn Out... Than Fade Away~~



Senior Manager
Joined: 22 Dec 2009
Posts: 350

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
06 Jan 2010, 14:58
2
This post received KUDOS
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
_________________
Cheers! JT........... If u like my post..... payback in Kudos!!
Do not post questions with OAPlease underline your SC questions while postingTry posting the explanation along with your answer choice For CR refer Powerscore CR BibleFor SC refer Manhattan SC Guide
~~Better Burn Out... Than Fade Away~~



Manager
Joined: 27 Aug 2009
Posts: 108

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



Senior Manager
Joined: 22 Dec 2009
Posts: 350

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
06 Jan 2010, 15:51
1
This post received KUDOS
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!
_________________
Cheers! JT........... If u like my post..... payback in Kudos!!
Do not post questions with OAPlease underline your SC questions while postingTry posting the explanation along with your answer choice For CR refer Powerscore CR BibleFor SC refer Manhattan SC Guide
~~Better Burn Out... Than Fade Away~~



Math Expert
Joined: 02 Sep 2009
Posts: 44352

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
06 Jan 2010, 17:09
2
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 27 Apr 2008
Posts: 189

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



Math Expert
Joined: 02 Sep 2009
Posts: 44352

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
09 Jan 2010, 00:53
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 01 Jan 2008
Posts: 221
Schools: Booth, Stern, Haas

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
07 Jul 2010, 08:35
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?



Manager
Joined: 06 Jun 2014
Posts: 56

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



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5053
GPA: 3.82

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
21 Sep 2016, 20:12
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
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 3 month Online Course" "Free Resources30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons  try it yourself"



Manager
Joined: 28 Jun 2016
Posts: 207
Location: Canada
Concentration: Operations, Entrepreneurship

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
21 Sep 2016, 21:14
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



Intern
Joined: 11 Nov 2016
Posts: 4

Re: Finding the remainder when dividing negative numbers [#permalink]
Show Tags
24 Sep 2017, 09:29
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?




Re: Finding the remainder when dividing negative numbers
[#permalink]
24 Sep 2017, 09:29






