This is a post to clarify a potentially confusing passage in the OG. In the GRE OG 2e, on p. 134, we find the following passage about quotients and remainders.

“If an integer *n* is divided by a nonzero integer *d* resulting in a quotient *q* with a remainder *r*, then *n* = *qd* + *r*, where 0 ≤ *r* < |*d*|…. For example, when 20 is divided by 7, the quotient is 2 and the remainder is 6; when 21 is divided by 7, the quotient is 3 and the remainder is 0; and when –17 is divided by 7, the quotient is –3 and the remainder is 4.”

Well, using their formula, we can verify:

a) 20 = (2)(7) + 6

b) 21 = (3)(7) + 0

c) –17 = (–3)(7) + 4

All true. So far so good. The problem comes when we want to use all of this to convert an improper fraction to a mixed numeral. The guiding formula is what I will call the “Division Equation”:

This equation is true 100% of the time. With example (a), if we just plug everything into this formula we magically change the improper fraction into a mixed numeral.

Voila! The correct mixed numeral magically results. With example (b), we don’t need to worry about mixed numerals, because the divisor goes into the dividend evenly, and thus the result of the division is an integer: 21/7 = 3. With example (c), things get tricky:

The Division Equation above still works, but for negatives, we can’t just dump the numbers in and magically get the correct negative mixed numeral result. This has to do with the way ETS has chosen to define remainders for negative numbers. Let’s explore this.

## Remainders for negative numbers

In the passage from p. 134 quoted above, every math book on planet Earth, and probably most written by aliens on other planets, would agree 100% with what ETS said about quotients and remainders for positive numbers. All that is absolutely undisputable. The vast majority of math books deliberately ignore the entire topic of what happens to quotients and remainders with negative numbers, because of the mathematical trade-offs that arise. Most math books have the good sense to avoid this troublesome topic, but ETS brashly charges in and firmly establishes a position. (To follow knowledge like a sinking star beyond the utmost bound? or, fools rush in where angels fear to tread?)

The *disadvantage* of defining remainders of negative numbers in this way is that makes finding negative mixed numerals more difficult: we can no longer use the Division Equation — we’ll get back to that point below. The *advantage* is: when you divide by a certain divisor, say 7, and talk about all the numbers that have a remainder of 4, those numbers are equally spaced on the number line in both the positive and negative direction. The numbers which, when divided by 7, have a remainder of 4 are:

Notice that –17 appears in that pattern, and follows the same pattern as the familiar positive numbers which, when divided by 7, have a remainder of 4. It’s very important: if ETS mentions a dividend which, when divided by 7, has a remainder of 4, don’t automatically assume the dividend is positive: it could be negative, and it would follow this sort of pattern.

## Thoughts on mixed numerals

What exactly is a mixed numeral? That is to say, what do we mean when we write, say:

What this really means is:

In other words, ** any mixed numeral implicitly contains addition between the integer and the fraction.** That is the

**BIG**idea of mixed numerals.

It’s relatively easy to see that, for positive numbers, the Fraction Equation will result in a whole number quotient plus a fraction less than one, which automatically fits the pattern for a mixed number.

What’s going on with a negative mixed number? Well, that’s a little different:

The negative sign in front of the whole mixed number applies to both terms and distributes, so you get subtraction instead of addition. If we are using ETS’ definition of remainder for negative numbers, then the Fraction Equation still works, but it doesn’t help us convert a negative improper fraction to a negative mixed number, because it results in *adding *the fraction instead of *subtracting* it. Again, this part is the disadvantage of the convention ETS is following on this particular topic. The Fraction Equation, so helpful for converting positive improper fractions to positive mixed numerals, is ostensibly useless in helping us analogously with negative numbers.

## Negative mixed numerals

The question arises, then: how do we convert a negative improper fraction to a negative mixed numeral? For example, suppose we have the fraction –17/7. Again, as the OG p. 134 tells us, “when –17 is divided by 7, the quotient is –3, and the remainder is 4.” That will be consistent with the Fraction Equation, but, as we have seen above, that in turn will not lead to a correct mixed numeral. Therefore, everything OG says about quotients and remainders for negative number is useless if what we want to find is a negative mixed numeral. What do we do?

Actually, what you do is remarkably simple: pretend the negative sign isn’t there for the moment, and just convert the positive improper fraction to positive mixed numeral!

Now, we are dividing +17 by 7: of course, the quotient is 2 and the remainder is 3, so the Fraction Equation conveniently gives us

So far, so good. As a result of working with the positive numbers, we know:

But what we wanted was –17 divided by 7, not +17 divided by 7. How do we get that? We get that simply by multiplying the previous equation by a negative sign on both sides.

Again, to convert negative improper fraction to a negative mixed numeral: just pretend the improper fraction is positive, do the conversion with positive numbers, and then simply make the output negative when you are done.

## Practice questions

## Practice question solutions

1) Here, we must follow the ETS convention. According to this convention, when a negative number is divided the remainder must be positive. What is the multiple of 6 that is less than –23 and closest to –23? That would be –24, so 6 goes into –23 negative four times, with a remainder of (–23) – (–24) = +1. This checks out with the ETS’s equation: n = –23, d = 6, q = –4, and r = 1, and it’s true that

–23 = (6)( –4) + 1

So the quotient is –4 and the remainder is +1. Answer = **E**.

2) Here, we will first change it to a positive improper fraction. When +23 is divided 6, the quotient is 3 and the remainder is 5, so:

Now, just multiply both sides by a negative sign.

Answer = **B**.

thanks for this detailed explanation

Glad that this helped, Sarker! 🙂

Hello Mike,

I am facing problems each time this types of questions appear; where given,

K is an integer & is less than some mentioned integer,

when K/5, remainder is 4,

when K/23, remainder is 7,

what’s the value of K?

It will be a great help if you reply. Thanks

Hi Maha,

When trying to find an unknown that is defined by its relation to two different remainders, it’s often easiest to simply list out potential values for each case, and look for the overlap. For example, we need to find the cases where we can divide k by 5 and have 4 left over. Put another way, k will be a multiple of 5 with an extra four added! This allows us to list some possible values by adding 4 to multiples of 5 (for example 5*1=5, 5+4=9, so 9 satisfies this requirement.)

Possible values of k/5 has a remainder of 4: 9, 14, 19, 24, 29 …

Possible values of k/23 where the remainder is 7: 30, 53, 76, 99…

Here, we can see that since 23 is such a large integer, it will limit our answer choices more than the multiples of 5. We also notice that the possible values of k/5 remainder 4 all end with 4 or 9 (which makes sense since all multiples of 5 should end in 0 or 5). So, we need to find the k/23 remainder 7 that ends with a 4 or 9.

We see that if k=99, both requirements are met! Go ahead and check it for yourself to make sure 🙂

My confusion lies in how you show two completely different answers for the same question.

At the beginning, you show

-17/7 = -3 4/7

I get this, you say 7 goes into 17 3 times (21) so -4/7

But later down, you show

-17/7 = 2 3/7

You say 7 goes into 17 two times (14) so 3/7.

I do see how you get both answers. But which one the correct way of doing it?? Is it based on how the question is asked? thank you in advance!

Dear Misha,

I’m happy to respond. 🙂 I don’t know whether you noticed the unequal sign in the first part. Negative seventeen over seven does NOT equal negative three and four sevenths. Changing that unequal sign to an equal sign between the same quantities would be the mathematical equivalent of profanity!

The big idea in this blog is that the way that remainders work and the way that mixed numerals work fit together very nicely for positive numbers, but things are extremely different for negative numbers.

When we divide -17 by 7, then according to GRE conventions, 7 goes into -17 with a quotient of -3 and a remainder of 4. That is the correct quotient and remainder, by GRE standards, AND we are 100% forbidden to take that -3 and that 4 and use it to construct a negative mixed numeral number. For positive numbers, we can use the quotient and remainder to build our positive mixed numeral number, but it is 100% mathematically illegal to treat negative numbers in the same way.

The very different question is: when we divide -17 by 7, what mixed numeral do we get? For this, I suggested that, by far, the simplest approach is simply to pretend that everything positive, build the positive mixed numeral, and just turn it back to negative at the end. Thus, +7 goes into +17 twice, quotient = 2, with a remainder of 3. From this positive quotient and remainder, we can easily construct the positive mixed numeral:

+17/7 = 2 3/7

Now that we know that, just multiply both sides by negative one:

-17/7 = -2 3/7

That final statement is the ONLY statement we should be making that involves (-17/7) and an equal sign. Saying that (-17/7) equals anything else would be grossly mathematically illegal.

Does all this make sense?

Mike 🙂

Thank you so much Mike for taking the time and answering this questions so properly. You guys are awesome

Hi Mike,

I had the same confusion regarding the negative improper fraction as Misha:

First, you show that the approach we use for positive improper fractions to turn into a mixed number does not work for negative improper fractions on the example -17/7 is not = -3 4/7.

However, later you give two practice questions and again suggest the following solution for the first practice question:

Practice question solutions

1) Here, we must follow the ETS convention. According to this convention, when a negative number is divided the remainder must be positive. What is the multiple of 6 that is less than –23 and closest to –23? That would be –24, so 6 goes into –23 negative four times, with a remainder of (–23) – (–24) = +1. This checks out with the ETS’s equation: n = –23, d = 6, q = –4, and r = 1, and it’s true that

–23 = (6)( –4) + 1

So the quotient is –4 and the remainder is +1. Answer = E.

This is NOT the same as:

2) Here, we will first change it to a positive improper fraction. When +23 is divided 6, the quotient is 3 and the remainder is 5, so:

{23/6} = 3{5/6}

Now, just multiply both sides by a negative sign.

-{23/6} =- 3{5/6}

Answer = B.

It makes sense that we should always use this 2) approach above. But why you still show 1) approach which is incorrect and mathematically illegal, as you have explained earlier?

Thank you!

Seda

Hi Seda,

I can see your confusion here! It’s important to remember that the ETS convention says that remainders CANNOT be negative. This means that we can’t just convert a negative mixed numeral to a quotient and remainder. In the first question, we are asked to give the quotient and remainder. The divisor must allow for the remainder to be positive, which is why (E) is the best answer.

However, question 2 is not asking us about the remainder in a question. As Mike says, “everything OG says about quotients and remainders for negative number is useless if what we want to find is a negative mixed numeral.” There is no special rule or case for negative mixed numerals–a special case only exists for remainders of negative numbers! So, when you convert a negative improper fraction to a mixed numeral, you do it as if it were a positive number. But be careful: a negative mixed numeral will not give you the remainder of a negative improper fraction!

The purpose of this blog post is to show this contradiction. When you are dealing with negative numbers, the numbers you get for a quotient and remainder will NOT ‘translate’ into the number you get when you convert an negative improper fraction into a negative mixed numeral. We haven’t done anything illegal here–rather, we must approach (1) differently because of the conventions in ETS.

Hello Mike, can you please elucidate me on what the conventions are concerning divisions where both the numerator and denominator are negative? I’m guessing it works the same way as a positive fraction since the negative signs cancel each other, but I just want to confirm. Thanks!

Dear Queque,

Yes, the negative signs cancel. Negative divided by a negative is positive. I hope this helps.

Mike 🙂

Thank you!

Dear Queque,

You’re quite welcome, my friend. 🙂 Best of luck to you in the future!

Mike 🙂

Hi Mike,

Very informative.Spent around 4 hours on this post & most importantly got each and every word you explained..initially it was in a dilemma,but kept reading the post over again & again..things then started becoming much clearer.

So,i have a question. In the number line,how did you find all the numbers (4,11,28,35) which when divided by 7,yields 4 as the reminder? Is there a formulae for it? If so please specify…

Also found that,the numbers which lie on the negative side of the number line,is 1-(number on the positive side of the numberline)…is it true for ALL the cases? or is it something that pertains to this question itself?

Very informative article again.thanx

Regards,

Manish A.

Dear Manish,

To find the numbers that, when divided by 7, yield a remainder of 4, all I did was find the multiples of 7 (including 0, which is a multiple of every number):

{0, 7, 14, 21, 28, 35, 42, 49, 56 ….}

and then I added 4 to each one

{4, 11, 18, 25, 32, 39, 46, 53, 60 …}

The pattern about the relationship of the positive numbers and negative numbers is simply a coincidence here, and would not be true with most other numbers picked.

Does all this make sense?

Mike 🙂

Hi Mike,

Thank you for the reply…could you please also tell me how you found the numbers that, when divided by 7, yield a remainder of 4 on the NEGATIVE SIDE of the numberline?

Dear Manish,

First of all, just think about the list of negative multiples of 7: these would be all the positive multiples with a negative sign in front of them:

{-7, -14, -21, -28, -35, -42, -49, -56, …}

Then, just add 4 to each number.

{-3, -10, -17, -24, -31, -38, -45, -52, …}

Does this make sense?

Mike 🙂

I Have a question. I somehow cannot come around the logic to solve it.

When integer x is divided by 12 it leaves a remainder 5, what is the remainder when the square of integer x is divided by 8? 🙁

Dear Joyeeta,

That’s a considerably harder question. First, we need to write x using the remainder formula — see this GMAT post:

https://magoosh.com/gmat/2012/gmat-quant-thoughts-on-remainders/

x = 12N + 5, for some integer N

Then, we have to square that, using FOIL.

x^2 = (12N + 5)^2 = 144N^2 + 120N + 25

Clearly, 8 goes evenly into the first two terms. When 25 is divided by 8, the remainder is 1. That’s the answer. Does all this make sense?

Mike 🙂

Mike,

I am confused a little bit. Please correct me if i am wrong. So what you are saying is we have 2 different approaches,

– One to convert the negative improper fraction to negative mixed fraction

– Other one is to determine the quotient and remainder of negative improper fraction.

Am I right?

RB,

If you need to find a remainder, keep it negative and use the formula at the top. If you have a negative improper fraction that you want to change to a negative mixed numeral, make everything positive, make the change, and then make the result negative.

Does all this make sense?

Mike 🙂

Perfect Mike !!! Thank You so Much.

You are quite welcome. Best of luck to you!

Mike 🙂

Hello Mike…

The OG passage which is solved here has a MISTAKE !!!

It is given that the quotient is -3

and using the formula dividen/divisior= q+r/divisor

Why u have taken quotient as -2 instead of -3 ???

Please reply 🙂

Nitish,

Thank you for pointing that out. That indeed was a mistake, and you were the first person to point it out to us. I corrected it in the text above. Thanks again.

Mike 🙂

thanks for this detailed explanation

You are quite welcome.

Mike 🙂

Hi Mike,

I am still confused, so both are right answers?

Both answers to the practice questions are correct. What is your question?

Mike 🙂

So I understand how both answers E and B are correct. But what I am confused is when I do the reverse (let’s say to check my answers during the test):

For E) -23 = (6)(-4) +1

For B) -23 not = (6)(-3) +5 = -18 +5

Now, I know if I take -[(6)(3) + 5]= -23

Thus, I find these confusing because there is not a same way to place these calculations.

Please explain. Am I missing something here?

Thanks Mike.

N,

You are not missing something here. The fact that these two are very different is the point of this article. Because of the way ETS defines remainders, they way remainders work with negative dividends does not line up with the way we handle negative mixed numerals. As a consequence of the convention ETS is following, you have to treat (a) remainders with negative dividends, and (b) negative mixed numerals as two completely different and, more or less, unrelated things. That’s the whole point of this article. Does that make sense?

Mike 🙂