The Definitive Higher Math Guide on Integer Long Division (and Its Variants)

In this comprehensive guide, you'll find out how and why long division and its variants work, and how to combine their best features to your own advantage.

In mathematics, long division is a powerful procedure through which division can be carried out with pencil and paper. As powerful as it is, it is generally taught at the pre-college level without much consideration on its underlying theory or its alternatives…

In this guide though, we’ll take a look at both of those, and delve into the different types of division procedures under the different scenarios. We’ll also try to illustrate their role in promoting a holistic understanding about division and numbers in general — through candid analysis and a wide range of examples.

Table of Contents

(By the way, are you fairly acquainted with the content of this guide? If so, you might find this comprehensive summary worksheet both challenging and useful.)

In arithmetic, we often like to express a number $n$ as chunks of another (non-zero) number $d$. When that happens, we’re said to be dividing $n$ by $d$, where $n$ is known as the dividend, and $d$ the divisor.

In general, given a dividend $n$ and a divisor $d$, we can express $n$ as $dq + r$ for some integer $q$ and some number $r$, and while $n$ can have many $dq+r$ representations, there is only one such representation where $ 0\le r < |d|$.

In particular, in the case where the dividend and the divisor are integers, this result is known as the Division Theorem, but even if they are not, the following still apply in general:

$q$ — which is the unique integer in the above $dq+r$ representation — is known as the quotient.

$r$ — which is the unique number in the above $dq+r$ representation — is known as the remainder.

The act of finding those two numbers is called an Euclidean division, or more colloquially, division with remainder.

In Case You're Wondering...

The Division Theorem for integers is generally established by relying on the Well-Ordering Principle. For the general case, the Archimedean Property for real numbersmight be of help.

So what’s being implied here is that the dividend and the divisor are generally fixed numbers, while the interim quotients/remainders can change depending on the stage of the division we’re in. Furthermore:

The equation $n=dq+r$ is known as the algebraic representation of the division.

The equation $\displaystyle \frac{n}{d} = q + \frac{r}{d}$ is known as the fractional representation of the division.

(to be sure, there’s also the mixed fraction $\displaystyle q \, \frac{r}{d}$, though such notations are generally more prone to confusion when we move beyond the integers.)

And of course, there’s also the graphical representations of the division as well — which we’ll get into after some preliminary theory on division procedure.

Of course, the Division Theorem is a neat piece of logic guaranteeing that in an Euclidean division scenario, the quotient and the remainder exist and are unique. However, it doesn’t offer us much clue as to how we should go about finding those numbers in practice…

But then, that’s precisely where the primary topic of our interest — Euclidean division procedure — comes into play.

In a nutshell, these are the finite, recursive procedures that seek to find the quotient and the remainder through iterated subtractions. They tend to operate by deducting the dividend repeatedly — usually in a series of leaps — until it becomes smaller than the divisor itself.

And when that happens, the process comes to an end — with the last interim quotient and remainder being the answers to the original Euclidean division.

Or more specifically:

Given a dividend $n$ and a (non-zero) divisor $d$, we would begin the procedure by deducting $n$ (or its absolute value) by $d$ — an integer number of times (say $q_1$). This would lead to an interim remainder $r_1$, and the following interim algebraic representation of the division: \[ n = d q_1 + r_1 \]

At this point, if $0 \le r_1 < |d|$, then the division procedure is finished. If not, we would continue with the same process by deducting $r_1$ (or its absolute value) by $d$ — another integer number of times (say $q_2$). This would lead to the next interim remainder $r_2$, and the next algebraic representation of the division: \[ n = d q_1 + d q_2 + r_2 \]

So as we go through the $i^{\text{th}}$ iteration of the division, we would obtain:

The $i^{\text{th}}$-stage quotient $q_i$

The interim quotient $q_1 + \cdots + q_i$

The interim remainder $r_i$

And if we choose the stage quotients judiciously so that the $|r_i|$ decreases by, say, at least $1$ at each stage, then after a finite number of iterations (say $m$), we’d be guaranteed to generate an interim remainder $r_m$ such that $0 \le r_m < |d|$. And when that happens, our final algebraic representation would become: \begin{align*} n & = dq_1 + \cdots + dq_m + r_m \\ & = d (q_1 + \cdots +q_m) + r_m \end{align*} At which point, the division procedure would come to an end, yielding $q_1 + \cdots + q_m$ and $r_m$ as the quotient and the remainder of $n \div d$, respectively.

Here, notice that whether $n$ and $d$ are integers or not, the interim quotients must be always kept as integers — and are the only ones among the $n$, $d$, $q$s, $r$s that must satisfy that requirement.

For this guide though, we’ll restrict ourselves to the cases where both $n$ and $d$ are integers, and look at the division methods and approaches that come out of those. These include the long/short division methods, the chunking methods, the freeform method — among other less graphical division approaches.

Since the concept of division has its basis on the natural numbers, it makes sense that we first start from there and illustrate some of the primary methods being involved. Once there, we’ll slowly expand our toolset with other alternate methods and approaches — so that other scenarios of integer divisions can be included as well.

If you have a tape that’s 1529 cm long and you want to cut it down into pieces of 7cm chunks, how many of those pieces can you make? And after all is said and done, how much of the tape would remain?

Well, as you might guess, this is nothing more than the following division problem: $\require{enclose}$ \[ 7 \enclose{longdiv}{1529} \] which can be solved by using the traditional long division method, or — if we prefer — the abbreviated short division method as well.

As you might already know, the long division procedure consists in setting up the division tableaus so that at each stage, we would be cranking out a digit of the quotient — by deducting the dividend/interim remainder by the largest number of times possible.

In this case for example, since $1529$ can be deducted by $7$ $100$ times, but not $1000$ times, we know that we should begin the process by looking into the digit-multiples of $100$.

In fact, a bit of inspection would show that $1529$ can be deducted by $7$ $200$ times, but not $300$ times. This would lead us to choose $200$ as our first stage quotient, which would in turn produce the following tableau: \begin{array}{r} 200 \\[-0.35em] 7 \enclose{longdiv}{1529}\kern{-0.15ex} \\[-0.3em] \underline{1400} \\[-0.25em] 129 \end{array} And here is the beauty about this approach: since we’ve already established that $1529$ can be deducted by $7$ $200$ times, but not $300$ times, this means that we definitely won’t be further deducting $1529$ by $100$ times or more, and that all that remains is to focus at the digit-multiples of $10$ instead.

In fact, a little bit of inspection would show that $129$ — what remains now of $1529$ — can be deducted by $7$ $10$ times, but not $20$ times. This would lead us to choose $10$ as the next stage quotient, which in turn would lead to the following tableau: \begin{array}{r} 10 \\[-0.35em] 200 \\[-0.35em] 7 \enclose{longdiv}{1529}\kern{-0.15ex} \\[-0.3em] \underline{1400} \\[-0.25em] 129 \\[-0.3em] \underline{\phantom{1}70} \\[-0.25em] 59 \end{array} As you can see here, our notation is a bit unusual in that the numbers are being stacked up at the extremes, but those are there for a few reasons — which will become clearer in the next few sections.

But that aside, notice that since we’ve already established that what remained of $1529$ could be deducted by $7$ $10$ times, but not $20$ times, this means that we’ll definitely not be further deducting it by $10$ times or more — and that a deduction by a digit-multiple of $1$ should be our next step.

In fact, it doesn’t take a long time to realize that $59$ — what remains now of $1529$ — can be deducted by $7$ $8$ times, but not $9$ times. This would lead us to choose $8$ as our next stage quotient, which in turn would lead us to our next tableau: \begin{array}{r} 8 \\[-0.35em] 10 \\[-0.35em] 200 \\[-0.35em] 7 \enclose{longdiv}{1529}\kern{-0.15ex} \\[-0.3em] \underline{1400} \\[-0.25em] 129 \\[-0.3em] \underline{\phantom{1}70} \\[-0.25em] 59 \\[-0.3em] \underline{56} \\[-0.25em] 3 \end{array} Here, notice that since we’ve already established that what remained of $1529$ could be deducted $8$ times, but not $9$ times, it already followed by then that what would remain of $1529$ must be smaller than the divisor itself. In fact, this is reflected in our last tableau as well.

But either way, the division procedure is now coming to its end, yielding $200 + 10 + 8 = 218$ as the quotient, $3$ as the remainder, and the following representations of the division as well:

Now, in the above example, there are two interesting peculiarities. The first one — which is easier to spot — is the fact that each stage quotient takes the form of a leading digit followed by zeros (or no zero), with the place of the leading digit moving to the right at each stage — without ever overlapping.

So the natural question becomes: “Does this apply to all long division tableaus at every stage?” Fortunately, the answer is yes, and that’s because by the nature of the algorithm, we always choose the highest digit-multiple at the highest available place at each stage.

And because of that, we usually prefer to adopt the convention of notating a stage quotient by its leading digitonly — and doing so from the left to the right as if we were writing the quotient on-the-fly.

As for the second peculiarity, it has to do with the fact that since the stage quotients are always of the form described above, we often end up with many numbers below the division sign whose “significant digits” are all on the left — followed by a string of zeros.

And while those zeros might be harmless in many cases, they can also compound very quickly as the dividend gets larger. Because of that, we often like to omit these zeros altogether — while keeping the “significant digits” of these numbers intact.

So if we were to redo our tape problem with the above conventions in mind (i.e., quotient merging, omission of zeros), the first iteration of the division tableau would have become: \begin{array}{r} 2\phantom{00} \\[-0.35em] 7 \enclose{longdiv}{1529}\kern{-0.15ex} \\[-0.3em] \underline{14\phantom{00}} \\[-0.25em] 1\phantom{29} \end{array} Here, notice that the placements of the partially-notated numbers above and below the division sign are now becoming crucial — as they are now notated with their “significant digits” only.

In fact, it is assumed by default that the decimal points of these numbers are aligned with that of the dividend, so the $2$ above the division sign actually stands for $200$, and the $14$ below actually stands for $1400$.

In Case You're Wondering...

Yes. There’s yet another omission convention which we haven’t covered yet. But then, we’ll leave it to you to figure out why it is useful, why it won’t undermine the calculation if correctly applied, and why it can be misleading if not well understood.

And with that, here comes the second tableau: \begin{array}{r} 21\phantom{0} \\[-0.35em] 7 \enclose{longdiv}{1529}\kern{-0.15ex} \\[-0.3em] \underline{14\phantom{00}} \\[-0.25em] 12\phantom{9} \\[-0.3em] \underline{\phantom{1}7\phantom{0}} \\[-0.25em] 5\phantom{9} \end{array} Here, notice that we’re merging the stage quotients on the fly already, which is a neat thing. As usual, the $21$ above actually stands for $210$, and the $7$ below actually stands for $70$.

And then, there’s the final version of the tableau as well: \begin{array}{r} 218 \\[-0.35em] 7 \enclose{longdiv}{1529}\kern{-0.15ex} \\[-0.3em] \underline{14\phantom{00}} \\[-0.25em] 12\phantom{9} \\[-0.3em] \underline{\phantom{1}7\phantom{0}} \\[-0.25em] 59 \\[-0.3em] \underline{56} \\[-0.25em] 3 \end{array} As you can see, this convention-based procedure for long division can be both a blessing and a curse. It’s a blessing because it eliminates a lot of unnecessary zeros and rewriting, and it’s a curse because it can give the illusion that one is simply manipulating small integers — when in fact the opposite is usually the case.

So whether you’re on the learning or the teaching side of things, make sure to understand what each of the partially-notated numbers in a tableau stands for. This alone will prevent a lot of misunderstanding, and dispel the perception of long division as a series of half-sensical algorithmic rituals.

While the traditional use of long division already has many shorthands built in, there’s yet another highly-abbreviated form of long division called short division — where the interim remainders are notated alongside of the dividend digits as superscripts.

And because the calculations leading to the interim remainders are entirely omitted in this case, this makes short division a notationally superior tool for handling small Euclidean divisions and divisions in general.

For example, if we were to redo the above tape division problem using short division instead, then our first tableau would have become: \begin{array}{r} 2\phantom{^{1}29} \\[-0.30em] 7 \enclose{longdiv}{15^{1}29}\kern{-0.15ex} \end{array} Here, notice that the $2$ above stands for $200$ as usual, and that the superscript $1$ is only there to inform us that the interim remainder at this point is $129$.

As you can guess, this clever notation puts many of the issues with the traditional use of long division to rest, in that not only is there little confusion about the value of the interim remainder at any given stage, but that the partially-notated numbers below the division sign are pretty much all gone as well.

And with that, here comes the second tableau: \begin{array}{r} 2\phantom{^1}1\phantom{^{5}9} \\[-0.30em] 7 \enclose{longdiv}{15^{1}2^{5}9}\kern{-0.15ex} \end{array} As in the case with long division, the stage quotients here are being merged on the fly as well, leaving us with an additional superscript $5$ — which is only there to indicate an interim remainder of $59$.

At this point, it should be clear what the final tableau looks like, but for the sake of completeness, here it is: \begin{array}{r} 2\phantom{^1}1\phantom{^{5}}8\phantom{^3} \\[-0.30em] 7 \enclose{longdiv}{15^{1}2^{5}9^{3}}\kern{-0.15ex} \end{array} In particular, notice that the last superscript, $3$, is now becoming the remainder of the division — just like the top number $218$ is now becoming the quotient of the division as well.

As another example, here’s the short division tableau for the division $2689 \div 13$: \begin{array}{r} 2 \phantom{^0} 0 \phantom{^8} 6\phantom{^11} \\[-0.30em] 13 \enclose{longdiv}{2 6^{0} 8^{8} 9^{11}}\kern{-0.00ex} \end{array} In particular, notice that:

A superscript can very well be zero.

A superscript can very well go into multiple digits.

A superscript can even be equal to the digit beneath it.

But whichever the case, the fact that one of these happens does not necessarily suggest inefficiencies. If anything, these superscripts can provide important cues as to which digits are finished — and which digits to focus on next.

As we’ve seen above, the long division method (shorthand-based or otherwise) is the one that likes to “keep all the tabs”, while the short division method would prefer to keep track of nothing — save the quotient and the interim remainders below the division sign.

As a result, for small numbers, doing divisions the “short” way can drastically minimize the chance of misunderstanding and notational errors. In fact, every short division tableau will look simpler than its long division counterpart — including the ones where only one single iteration is involved!

(if anything, one can also omit notating the quotient digits altogether when the remainder is the only thing that is sought after. This makes short division particularly useful for, say, the calculation of least positive residue in modular arithmetic.)

However, what usually goes unnoticed is that since there are even more omissions of numbers in short division, this can result in shifting the burden of calculation and retention from pencil and paper to one’s mental arithmetic faculty.

So as the dividend and the divisor get larger, there would come a point in time where “keeping the tabs in the head” can turn into more of a liability than an asset. After all, just because a method is notationally simple doesn’t mean that it’ll be the best for all cases!

If anything, both long and short division have a subtle blindspot that’s rarely addressed — and that has to do with the fact that both aregreedy algorithms looking to optimize on individual digits only.

And while this might not seem much on the surface, it does favor a certain form of reasoning that can come back to hurt us. In what follows, we’ll look at some of the methods that are not bounded by such constraints — but which lose their systematic-ness as a result.

In long and short division, the procedures are set up so that we’d base our decisions on the quotient digits, and try to crank them out one at a time.

And while this has the immediate benefit that once a quotient digit is determined, it would remain correct throughout the entire procedure, but beyond that, there’s really little reason as to why it should be the case.

So what would an Euclidean division look like if we allow ourselves to be a little bit more liberal? Well, that’s where division procedures such as the chunking method can play a big role.

In a nutshell, the chunking method is the ultimate embodiment of a division procedure, in that it consists in taking a large chunk off a number repeatedly — without much regard to the digits of the quotient itself.

Did You Know?

In primary school math education (particularly in the U.K.), the chunking method is also known as the partial quotient method or the hangman method — for reasons similar to those described in the theory section above.

To illustrate, suppose that we’re dealing with the following division problem: \begin{array}{r} 4 \enclose{longdiv}{5785} \end{array} At first, we might see that we can deduct $5785$ by $4$ at least $1400$ times. This would lead us to choose $1400$ as our first stage quotient, which would then result in the following tableau: \begin{array}{r} 1400 \\[-0.35em] 4 \enclose{longdiv}{5785}\kern{-0.15ex} \\[-0.3em] \underline{5600} \\[-0.25em] 185 \end{array} Here, we then see that $185$ — what remains now of $5785$ — can be further deducted by $4$ $45$ times (since $45 \times 4 = 90 \times 2 = 180$). This would lead us to adopt $45$ as our next stage quotient, which would then take us to the following tableau: \begin{array}{r} 45\\[-0.35em] 1400 \\[-0.35em] 4 \enclose{longdiv}{5785}\kern{-0.15ex} \\[-0.3em] \underline{5600} \\[-0.25em] 185\\[-0.3em] \underline{180} \\[-0.25em] 5 \end{array} At this point, it should be clear that the bulk of the work is done already, though we still have $5$ more to go. That said, all we have to do is to put on the finishing touches, and all should be good: \begin{array}{r} 1 \\[-0.35em] 45\\[-0.35em] 1400 \\[-0.35em] 4 \enclose{longdiv}{5785}\kern{-0.15ex} \\[-0.3em] \underline{5600} \\[-0.25em] 185\\[-0.3em] \underline{180} \\[-0.25em] 5 \\[-0.3em] \underline{4} \\[-0.25em] 1 \end{array} Here, since the interim remainder is already smaller than the divisor itself, we can safely declare the process finished, with $1400 + 45 + 1 = 1446$ and $1$ being the final quotient and remainder, respectively. We can also throw in the various representations of the division — as follows:

As we can see, the chunking method is unique in that it doesn’t come with a clear-cut algorithm out of the box, but because of that, it can be better than the traditional methods in promoting an intuitive and liberal way of thinking about division itself.

If anything, it even forces us to acquire strong numeracy skills and a genuine understanding about division — before we can even begin to use and wield the method effectively.

So even if such a method can be terribly inefficient for a computer to implement, for humans with strong arithmetic skills (or those who are simply eager to learn more about number patterns), this could very well be the missing method they’ve been looking for.

So far, we’ve looked at how the chunking method undoes the rigidity of the traditional, digit-based methods through the allowance of non-standard stage quotients. But as it turns out, the chunking method itself is not without its own restrictions either. In particular:

It operates under the framework of under-subtraction, but not over-subtraction.

It presumes that the interim remainders are $0$ or more, and disallows the presence of negative interim remainders.

It presumes that the stage quotients are positive in nature, and disallows the use of negative stage quotients.

To be sure, it’s always possible to argue that these restrictions are well-intentioned, in that they are based on our notions and understanding about physical quantities. and that if we were to remove them, then it would be possible for us to divide a number into more chunks than it actually has — before handling these outstanding chunks as if they were money from a bank account.

In other words, such unrestricted divisions would be bidirectional in nature, and would change the conception of division from a series of iterated subtractions — into one that includes iterated additions as well.

At its core, division is really just as much about iterated additions than it is about iterated subtractions. Click to Tweet

But then, if you think about it, this is actually not a new concept to begin with. If anything, most divisions beyond the integers and decimals already have some form of bidirectionality built in — and it’s only not so at the lower level because more conceptual understanding and numerical skills would be involved.

But either way, it’s not like that doing so would require a tremendous amount of background and effort, so it makes sense that we give this unrestricted form of division, bidirectional chunking, a try — by first looking at the following example: \begin{array}{r} 4 \enclose{longdiv}{2375} \end{array} Here, notice that if we were to stick to the old paradigm, then we would have chosen $500$ as our first stage quotient, but let’s just say that we beg to differ and choose to use $600$ instead — so that we could observe what would happen next: \begin{array}{r} 600 \\[-0.35em] 4 \enclose{longdiv}{2375}\kern{-0.15ex} \\[-0.3em] \underline{2400} \\[-0.25em] -25 \end{array} As expected, the interim remainder, $-25$, is now in the negative territory. However, if we just retract the divisor $7$ times, then we would be back into the positive territory. So let’s just do that and see what would happen: \begin{array}{r} -7 \\[-0.35em] 600 \\[-0.35em] 4 \enclose{longdiv}{2375}\kern{-0.15ex} \\[-0.3em] \underline{2400} \\[-0.25em] -25 \\[-0.3em] \underline{+ \phantom{-} 28} \\[-0.25em] 3 \end{array} Here, notice that two things are happening at the same time — as we restore the interim remainder back to normal:

The new stage quotient, $-7$, is negative — because we’re essentially “retracting the subtracting”.

As a result, there is a number that needs to be added back to the interim remainder — and hence the presence of the plus sign underneath.

But either way, since the interim remainder is now both non-negative and smaller than the divisor itself, we can already declare the division process finished, with $600-7=593$ and $3$ as the final quotient and the remainder, respectively. We can also spell out the entire division process algebraically — as follows: \begin{align*} 2375 & = 4 (600) + (-25) \\ & = 4 (600) + 4 (-7) + 3 \\ & = 4 (593) + 3 \end{align*} Or if we prefer the fractional representations instead: \begin{align*} \frac{2375}{4} & = 600 + \frac{-25}{4} \\ & = 600\, – 7 + \frac{3}{4} \\ & = 593 + \frac{3}{4} \end{align*} Now, you must be wondering: “Why on Earth do we want to go through such a process just to solve a division?” Well, in some cases, you may be right, but we only like to invoke bidirectionality for natural reasons, and in this case, it has to do with the fact that it actually makes the magnitude (i.e., absolute value) of the interim remainder smaller.

Hmm. Still not convinced of its usefulness? Then let’s look straight at an example with negative dividend instead — this time with $-36735 \div 13$: \begin{array}{r} 4 \\[-0.35em] -30 \\[-0.35em] 200 \\[-0.35em] – 3000 \\[-0.35em] 13 \enclose{longdiv}{-36735}\kern{-0.15ex} \\[-0.3em] \underline{+\phantom{-}39000} \\[-0.25em] 2265 \\[-0.3em] \underline{2600} \\[-0.25em] – 335 \\[-0.3em] \underline{+ \phantom{-} 390} \\[-0.25em] 55 \\[-0.3em] \underline{52} \\[-0.25em] 3 \end{array} Here, notice that the increase in division complexity naturally calls for more numbers to be involved, but either way, this is something that neither the long or the short division can even begin to handle — as they lack flexibility in the way a quotient is calculated and notated.

Furthermore, the above tableau also suggests that in bidirectional chunking, the magnitude of the interim remainder at each stage is always smaller or equal to its long/short division counterpart, which means that there will be equal or less iterations — along with calculations of smaller magnitudes as well.

But then, the bidirectional chunking approach is not without its own shortfalls either. For example:

By letting the stage quotients and the interim remainders to be both non-negative and negative, keeping tabs on everything can become a bit non-intuitive and confusing.

By stacking the positive and the negative stage quotients on top of each other, visualizing the size of the final quotient can be a bit difficult — if not impossible.

And of course, that’s not to mention that since bidirectional chunking is fundamentally a chunking method, the speed at which it can be carried out is largely dependent on one’s mental arithmetic and number recognition skills.

In other words, choose the right stage quotients and you’ll be done in no time, but if you choose the “wrong” ones, you might soon find yourself in some big arithmetic troubles!

But the good news is, there is a way to bypass the unidirectional, single-digit limitations of the traditional methods, all of the while benefiting from the structural and notational advantages these methods can offer us in return.

In fact, in the following section, we’ll introduce a brand-new method combining the unrestricted-ness of the bidirectional approach, with the systematic-ness and the concision of the traditional methods. This would result in an approach to division that’s flexible enough to cater to all scenarios — but which also offers an arena of new shortcuts and perspectives unavailable to the individual methods mentioned above.

At this point, we’ve looked at 4 distinct procedures for handling Euclidean divisions for integers, and while each of these procedures are great in their respective domains, they can also be poor in other. Here’s a table illustrating our key finding thus far:

Long Division

Short Division

Chunking

Bidirectional Chunking

Quotient-Digit Correctness

Immediate

Immediate

Not always immediate

Not always immediate

Stage-Quotient Flexibility

Limited

Limited

Flexible

Very flexible

Quotient/Interim-Remainder Bidirectionality

No

No

No

Yes

Notational Efficiency

Moderate

High

Low to Moderate

Low

Presentation

Half-intuitive and half-concise

Half-intuitive and elegant

Intuitive but less concise

Intuitive but messy

Notational Standalone-ness

High

Low

Very high

Very high

Large-Number Scalability

Yes

No

Yes, but with more writing

Yes, but with more writing

Systematic-ness

High

High

Low

Very low

In light of this, we thought to ourselves: “hmm…is there a division method out there that is as close to being the best in all categories as possible?” And lo and behold, after a bit of experimentation, that’s when we stumbled upon the freeform method.

In a nutshell, this is a method that incorporates the philosophy of unlimited bidirectionality, but which does so without ever invoking any stage quotient at all. If anything, it does make use of some notational conventions heavily — so that things can be kept as concise as one needs to without sacrificing the scalability of the method as a whole.

And with that, let’s give our latest division method a try — this time with a 6-digit-by-2-digit example: \begin{array}{r} 17 \enclose{longdiv}{736871}\kern{-0.15ex} \end{array}

For the first attempt, if we begin by confining our focus to the first digit of the quotient, then we might notice that at the ten-thousand place, $5$ seems to work well as a quotient estimate, so let’s just put it there and see what’d happen: \begin{array}{r} 5\phantom{6871} \\[-0.35em] 17 \enclose{longdiv}{\underline{73}6871}\kern{-0.15ex} \\[-0.60em] -12\phantom{6871} \end{array} Here, notice that a few notational conventions are at play already:

On the quotient side, only the “significant digits” are notated.

Below the dividend, there’s an underline to signal the digit places being targeted — and that there’ll be an interim remainder underneath.

In general, we can choose to omit notating the products altogether — if the numbers are not yet large enough to warrant their stay.

Either way, we see that the (partial) interim remainder, $-12$, is now in the negative territory. As such, we need to restore it back to normal by retracting the quotient by $1$ at the ten-thousand place. And when that is done, the following tableau arises: \begin{array}{r} 4\phantom{6871} \\[-0.35em] 17 \enclose{longdiv}{\underline{73}6871}\kern{-0.15ex} \\[-0.60em] 5 \phantom{6871} \end{array} (here, notice that we’re erasing the traces of the quotient adjustment, as this is integral to a cleaner presentation. More on this later. )

And since the (partial) interim remainder is now non-negative and smaller than the divisor itself, we can be sure that the first quotient digit is now set and that no further correction would be needed. As for the next two digits of the quotient, since $30$ seems like a good initial guess, we’ll simply stick to it and see what’d happen: \begin{array}{r} 430\phantom{71} \\[-0.35em] 17 \enclose{longdiv}{\underline{73}6871}\kern{-0.15ex} \\[-0.60em] \underline{568} \phantom{71} \\[-0.25em] 58 \phantom{71} \end{array} As it turns out, this time we actually undershot the quotient digits by quite a bit, so let’s try to boost it up by $3$ and put the (partial) interim remainder into its ideal range: \begin{array}{r} 433\phantom{71} \\[-0.35em] 17 \enclose{longdiv}{\underline{73}6871}\kern{-0.15ex} \\[-0.60em] \underline{568} \phantom{71} \\[-0.25em] 7 \phantom{71} \end{array} Again, here we choose to omit notating the product altogether, because the numbers are still not big enough to warrant such notations. Either way, the tableau now has an interim remainder of $771$, and since $17 \times 40 = 680$ and $17 \times 50 = 850$, this means that $45$ might just be a good fit for the last remaining quotient digits: \begin{array}{r} 43345 \\[-0.35em] 17 \enclose{longdiv}{\underline{73}6871}\kern{-0.15ex} \\[-0.60em] \underline{568} \phantom{71} \\[-0.25em] 771 \\[-0.30em] \underline{765} \\[-0.25em] 6 \end{array}

Quick Tip on Choosing Quotient Digits

Whenever applicable, choosing quotient digits ending in $\mathbf{5}$, $\mathbf{25}$ or $\mathbf{125}$ can be a good idea — as these numbers are easier to multiply in the decimal system. In this case for example, we have that $17 \times 45 = 16 \times 45 + 45 = 8 \times 90 + 45 = 765$.

As it turns out, we were lucky enough to land upon the right digits without any adjusting, so that even if we didn’t omit the product this time, the procedure did rightfully reach its end pretty smoothly. Not a bad try for a first illustration of an experimental method!

As alluded to earlier, the freeform method can be generally broken down into the following steps — at the procedural level:

Choose the quotient digits to focus on.

Make a good guess/adjustment to those digits — by “dialing them up or down”.

(Optional) Notate the product partially — with the “significant digits” only.

Draw a partial line highlighting only the digits involved in the iteration.

Notate the next interim remainder partially.

Rinse and Repeat*.

(*Can only move on to other quotient digits if the current interim remainder has the same sign as that of the dividend — more on this later.)

On the surface, the freeform method doesn’t seem to be much different from the method of long division itself, but upon further examination, we see that it is actually a rather distinct method of its own. Let’s take a look at its various procedural aspects and see how it compares with other methods mentioned above:

On-the-fly Quotient Calculation

Similar to long and short division, the freeform method chooses to calculate the quotient on-the-fly through the use of partially-notated interim quotients. This is good from the presentational standpoint — as it does not require any quotient stacking like the way we do in the chunking methods.

However, while a quotient digit in long and short division is set once correctly calculated, it actually achieves so primarily through a trial-and-error heuristics. In contrast, the freeform method doesn’t put a priority on being instantly correct — which opens up the possibility of a more efficient and flexible way of determining the quotient digits throughmulti-digit bidirectional adjustments.

Of course, the act of adjusting the quotient on-the-fly can also necessitate more mental calculations as well, but that’s usually not much of an issue in practice — as changes in the quotient digits are generally minute (a fact that applies to all other methods discussed above as well).

Quotient Flexibility

As alluded to earlier, both long and short division advocate for a rigid, single-digit-based procedure for determining the quotient digits — an approach which can lead to a fair amount of trial-and-errors and wasted effort. With the freeform method, however, nothing is wasted — and there’s no need to be wary of a product that’s slightly too big or too small either.

In fact, because the freeform method is essentially a “multi-digit quotient dialing game“, it can basically replicate the full capabilities of both the chunking method and the bidirectional chunking method. All that’s required is a mix of overshooting, undershooting, quotient boosting and quotient retracting.

Don't Abuse the Flexibility

Of course, just because the freeform method can simulate both of the chunking methods doesn’t mean that it should always be used as such. As a rule of thumb, opt for a systematic approach which is good at exploiting number patterns — rather than a free-for-all approach with a series of pluses and minuses.

Notational Standalone-ness

While the freeform method incorporates a mix of notational shorthands and omissions below the division sign, it can still be made to be as complete as long division if called for. This is to be contrasted with short division — whose notational system is not only not standalone, but whose usage is limited to a small set of scenarios as well.

Above the division sign, the freeform method forgoes the use of stage quotients in favor of a single interim quotient calculated on-the-fly — a choice which undoubtedly makes it less notationally standalone than the two chunking methods, but which also contributes enormously to the method’s intuitiveness and presentability as well.

For the Sake of Completeness...

Yes, we did contemplate on a fully notationally-standalone version of the freeform method as well, but after seeing the following tableau, we have had a second thought: \begin{array}{r} 345 \, \\[-0.35em] 43\enclose{horizontalstrike}{35} \phantom{1} \kern{-0.7ex} \\[-0.35em] \enclose{horizontalstrike}{45} \phantom{871} \kern{-0.5ex} \\[-0.35em] 17 \enclose{longdiv}{736871}\kern{-0.15ex} \\[-0.60em] \underline{765} \phantom{871} \\[-0.25em] \underline{-29}\phantom{871} \\[-0.25em] 587 \phantom{1} \\[-0.30em] \underline{595} \phantom{1} \\[-0.25em] \underline{-8} \phantom{1} \\[-0.25em] \underline{91} \\[-0.25em] 6 \end{array} In particular, notice how the tableau is now made a bit more involved and disorganized — by not erasing the traces of the quotient adjustments (i.e., overshooting + quotient retracting, undershooting + quotient boosting).

Conventions on Products

As illustrated in the previous examples, the freeform method is similar to long division in that it only notates the “significant digits” of the products. And while such notations can look a bit misleading at first, it’s also a necessary evil to avoid the compounding of zeros as the dividend gets larger.

Of course, for numbers that are sufficiently small, the products can be also omitted altogether as well. This means that notation-wise, the freeform method can be made to be just as concise as short division itself — as the following tableau illustrates: \begin{array}{r} 1368 \\[-0.35em] 7 \enclose{longdiv}{\underline{9}578}\kern{-0.15ex} \\[-0.60em] \underline{2\phantom{5}} \phantom{78} \\[-0.25em] \underline{4\phantom{7}}\phantom{8} \\[-0.25em] \underline{5\phantom{8}} \\[-0.25em] 2 \end{array}

Conventions and Constraints on Interim Remainders

Similar to long and short division, the freeform method only notates the “significant digits” of the interim remainders — a choice which would go on to prevent the compounding of zeros as the dividend gets larger.

If anything, the dropdown of digits (from the dividend to the interim remainder) can be also made to be optional for small numbers as well — making it a highly versatile method for tackling any kind of division in general.

However, unlike long and short division, the interim remainders are now free to assume any value that is positive, negative, large or small. As a result, we’ve now got some new procedural issues to handle. Take a look at the following tableau for instance: \begin{array}{r} 12500\phantom{6} \\[-0.35em] 4 \enclose{longdiv}{489876}\kern{-0.15ex} \\[-0.60em] \underline{50000} \phantom{6} \\[-0.25em] -1013\phantom{6} \end{array} Here, notice that the interim quotient, the product and the new interim remainder are all very dense with digits — since that’s the hidden contract for working with five quotient digits to begin with. Obviously, the goal of the freeform method is not to repeat the same mistakes of the chunking methods, which is why the following rule of thumb is in order:

Rule of Thumb on Quotient

For each iteration, only choose a large number of quotient digits to work with — if you have a way of making the magnitude of the interim remainder very small. Failure to do so can incur a series of “oscillations” along the way — and that’s not something you want to be dealing with.

Another procedural issue — which is a little bit more serious in consequences — has to do with the aggressive combination of shorthands with the bidirectional capability. Let’s take a look at our previous tableau once more: \begin{array}{r} 12500\phantom{6} \\[-0.35em] 4 \enclose{longdiv}{489876}\kern{-0.15ex} \\[-0.60em] \underline{50000} \phantom{6} \\[-0.25em] -1013\phantom{6} \end{array} Here, notice that there’s something fishy about the partially-notated numbers (i.e., interim quotient, product, interim remainder), in that they couldn’t really be interpreted to stand for anything. In particular:

The quotient $12500$ cannot be interpreted as $125000$.

The product $50000$ cannot be interpreted as $500000$.

The interim remainder $-1013$ cannot be interpreted as $-10130$, nor $-10136$ (i.e., dropdown of digits here is a no-no).

To get a better sense of what’s going on, let’s consider a similar tableau — this time with some better-behaving numbers instead: \begin{array}{r} 12200\phantom{6} \\[-0.35em] 4 \enclose{longdiv}{489876}\kern{-0.15ex} \\[-0.60em] \underline{48800}\phantom{6} \\[-0.25em] 187\phantom{6} \end{array} Hmm. It seems that this time, all the partially-notated numbers can be interpreted correctly as usual. So what caused the previous tableau to “malfunction” in the first place? Well, overshooting that is!

More specifically, when a quotient estimate results in overshooting, the magnitude of the product will exceed that of the interim remainder. This essentially creates a scenario where the operation is reversed — and hence the paradox with the partially-notated numbers.

In other words, as long as overshooting, reversed operations and incorrectly-signed interim remainders are at play, the result obtained will only be applicable to the digits being targeted — and cannot be further interpreted by involving other digits outside of the scope.

So does that mean any attempt at combining overshooting with partially-notated numbers will be met with dismay? Of course not! But it does mean that we need to put the following rule in place — so that we can prevent the method itself from malfunctioning:

Constraint on Overshooting

When working with a limited set of quotient digits, any incorrectly-signed interim remainder as a result of overshooting must be restored — before moving onto the next set of quotient digits.

In brief, think of overshooting as a temporary iteration to get the right quotient digits only. In practice, this often means that:

For divisions with positive dividend, any interim remainder must be eventually flipped back to $0$ or above — before tackling the next set of quotient digits.

For divisions with negative dividend, any interim remainder must be eventually flipped back to $0$ or below — before tackling the next set of quotient digits.

And with that rule firmly in place, let’s take a look at an example with a negative dividend as well — this time with $-12345 \div 4$: \begin{array}{r} -31\phantom{45} \\[-0.35em] 4 \enclose{longdiv}{-12345}\kern{-0.15ex} \\[-0.60em] \underline{+\phantom{;}124}\phantom{45} \\[-0.25em] 1\phantom{45} \end{array} In this first tableau, we’ve retracted the quotient by $31$ at the hundred place to minimize the magnitude of the interim remainder. However, in doing so, we’ve also incurred some overshooting as well — which resulted in an interim remainder of the wrong sign.

So if we want to proceed further, we’d have to boost the quotient by $1$ at the hundred place — and restore the interim remainder back to its default sign (which is negative): \begin{array}{r} -30\phantom{45} \\[-0.35em] 4 \enclose{longdiv}{\underline{-123}45}\kern{-0.15ex} \\[-0.60em] -3\phantom{45} \end{array} And since the magnitude of the interim remainder is smaller than that of the divisor, this means that the first two digits of the quotient are now set and that we can move on to the next two digits instead. In which case, since the interim remainder stands for $-345$ and we have that $4 \times 80=320$ and that $4 \times 90=360$, this hints at $85$ as a viable choice for the next quotient estimate: \begin{array}{r} -3085 \\[-0.35em] 4 \enclose{longdiv}{\underline{-123}45}\kern{-0.15ex} \\[-0.60em] -345 \\[-0.3em] \underline{+\phantom{-}340} \\[-0.25em] -5 \end{array} At this point, since we’ve already covered all the quotient digits, we no longer have to be worried about the constraint on overshooting. In which case, all that’s left to do is to further retract the quotient by $2$ — so that we can put the interim remainder into its ideal range: \begin{array}{r} -3087 \\[-0.35em] 4 \enclose{longdiv}{\underline{-123}45}\kern{-0.15ex} \\[-0.60em] -345 \\[-0.3em] \underline{+\phantom{-}340} \\[-0.25em] -5 \\[-0.3em] \underline{+ \phantom{-}8} \\[-0.25em] 3 \end{array} And with that, we can now declare the process finished, with $-3087$ and $3$ as the quotient and the remainder of $-12345 \div 4$, respectively. We can also look at the algebraic representation $-12345 = 4 (-3087) + 3$, and verify that the answer does indeed check out as well.

In brief, while the freeform method is uniquely subject to the constraint on overshooting, once fully mastered, it can also allow us to tackle integer divisions with great intuitiveness and efficiency.

And since the method also has a fair amount of flexibility and scalability built in, these reasons together makes it an interesting alternative to both the unidirectional, single-digit-based methods — and the chunking-based methods as well.

Pro-Tip

When using the freeform method, consider cranking out the quotient 2 digits at a time — by using numbers ending in $0$ or $5$ as initial guesses, and adjusting up or down as necessary. This will allow us to cut down the number of iterations greatly — without necessarily making each of the iterations too difficult to solve.

So far, the cases we’ve dealt with all involve divisors which are positive, but what if the divisor itself is negative? Well, that’s where the adjustment trick of this section can come in handy.

Imagine that you were given the task to divide $1296$ by $-13$, what would you do? For one, questions as such are usually not covered in schools, and even if you could tackle it using some of the methods mentioned above, the fact is that adjusting for the negative sign at each iteration can also be pretty annoying — if not confusing as well…

Fortunately, there is a way out this sticky situation — and that is to simply consider the case where the divisor is positive, do the division, and then reverse the sign of the resulting quotient afterwards.

In this case for example, instead of looking at the original division $1296 \div -13$, we could try to tackle the corresponding division $1296 \div 13$ instead. With the freeform method, the first tableau would become: \begin{array}{r} 100 \\[-0.35em] 13 \enclose{longdiv}{\underline{1296}}\kern{-0.15ex} \\[-0.60em] -4 \end{array} Here, we were fortunate enough to be able to tackle all the quotient digits upfront, but since the interim remainder is negative, we’ll have to retract the quotient by $1$ — and bring the interim remainder back into its ideal range: \begin{array}{r} 99 \\[-0.35em] 13 \enclose{longdiv}{\underline{1296}}\kern{-0.15ex} \\[-0.60em] 9 \end{array} And with that, the division procedure is now finished, yielding $1296 = 13 (99) +9$ as the final algebraic representation of $1296 \div 13$. But then, if you think about it, this is the same as if we’re saying that: \[1296 = -13 (-99) +9\] And because we still have that $0 \le 9 < |-13|$ after the “flipping”, the Division Theorem guarantees when $1296$ is divided by $-13$, we must have $-99$ and $9$ as the quotient and the remainder, respectively. So all we had to do is to reverse the quotient and keep the same remainder, and the problem would be solved!

More generally, given a dividend $n$ and a negative divisor $-d$, instead of doing $n \div -d$ upfront, we can always resort to $n \div d$ instead. This would lead to the quotient $q$, the remainder $r$, and the algebraic representation \[n=dq+r\] And because this would imply that $n =-d(-q) + r$ (with $0 \le r < |-d|$), it must follow that $-q$ and $r$ are the quotient and the remainder of the original division, respectively.

And if you think about it, this “flip-the-quotient-and-keep-the-same-remainder” trick actually makes a lot of sense. After all, division by a negative number is nothing more than chasing the same dividend on the same real line — albeit from the opposite direction.

In the last example, we alluded to the idea that a negative-divisor problem can be tackled by considering its positive-divisor counterpart instead, but what about the case where both the dividend and the divisor are negative?

Well, as it turns out, the strategy is actually the same! All we need to do is to consider the case where the divisor is positive, and flip the sign of the quotient as we did in the previous example.

For example, if we were to deal with the division $-4643 \div -11$, we would begin by considering the corresponding division $-4643 \div 11$ instead. With the freeform method, our tableau would become: \begin{array}{r}-423 \\[-0.35em] 11 \enclose{longdiv}{-4643}\kern{-0.15ex} \\[-0.60em] \underline{+\phantom{,}44}\phantom{43} \\[-0.25em] -24\phantom{3} \\[-0.30em] \underline{+\phantom{.}22}\phantom{3} \\[-0.25em] -23 \\[-0.30em] \underline{+ \phantom{.}33} \\[-0.25em] 10 \end{array} Here, we were lucky enough that we only need to invoke overshooting all the way near the end, but either way, the tableau is final, which means that $423$ and $10$ must be the quotient and the remainder of the original division (after we’ve done the sign flipping, of course).

So whether it’s $+ \div +$, $- \div +$, $+ \div -$ or $- \div -$, we now have all the tools and tricks to deal with those!

Don't Do This 'Obvious' Thing

Unfortunately, when it comes to Euclidean divisions, $-n \div -d$ is actually not the same as $n \div d$. On that note, it’d make a good mental exercise trying to figure out why that would be the case!

In the previous section, we alluded to some examples where a division is solved by first tackling a simplified version of the division instead. These fall into category of simplification tricks, which — despite not being full-fledged methods themselves — can still come in handy in many scenarios.

In fact, there is one such trick that involves solving a division by rescaling it first — before undoing the scaling factor to retrieve the quotient and the remainder of the original division. In most cases, the division is scaled down to make the problem simpler, although that’s not to say that there aren’t cases where scaling up can actually be a better option.

As the size of the dividend and divisor gets larger, divisions can become increasingly difficult to solve. However, if these numbers share a (non-trivial) factor in common, then it’s usually a good idea to use this factor to scale down the division and make it simpler.

For example, let’s say that we’re given the task to divide $59504$ by $88$. At first, the size of the numbers might look a bit daunting, but upon further inspection, we see that both numbers are actually divisible by $2$, $4$ — and $8$ even.

In fact, a bit of mental arithmetic would show that when $59504$ and $88$ are divided by $8$, they’d become $7438$ and $11$ as a result. This means that instead of solving the original division upright, we might just as well consider its scaled version $7438 \div 11$ for a change. With the freeform method, this would become: \begin{array}{r} 676 \\[-0.35em] 11 \enclose{longdiv}{\underline{74}38}\kern{-0.15ex} \\[-0.60em] \underline{8\phantom{3}} \phantom{8} \\[-0.25em] \underline{4\phantom{7}}\phantom{8} \\[-0.25em] \underline{6\phantom{8}} \\[-0.25em] 2 \end{array} And since the tableau is final, this would give us the following fractional representation of the division: \[ \frac{7438}{11} = 676 + \frac{2}{11}\] And with that, the fractional representation of the original division can be also figured out as well — as follows: \begin{align*} \frac{59504}{88} & = \frac{7438}{11} \\ & = 676 + \frac{2}{11} \\ & = 676 + \frac{16}{88} \end{align*} Here, notice that since the original division was scaled down by $8$, we’ve had to scale up the last fraction by $8$ as well (so that we can restore the original denominator), but once there, it can be readily seen that the quotient and the remainder of $59504 \div 88$ must be $686$ and $16$, respectively.

As we can see, the downward scaling approach relies in general on both the dividend and the divisor sharing a (non-trivial) common factor, and choosing a scaling factor that’s both large enough — and that actually makes the scaled division easier to solve.

And while for small numbers, this approach might be a little bit of an overkill, for large numbers, it can definitely be a lifesaver in terms of both efficiency and workload.

Now. It was nice that our previous example worked out short and sweet, but how do we know that we weren’t just lucky with our little “fraction trick” — and that the technique is indeed generalizable to other scenarios where scaling is involved?

Well, the truth is, we don’t really know yet (although we do have great suspicion that it is so). This is why it’s a good idea to look at scaling as a whole — and see if we can replicate our previous result without any fraction involved.

More specifically, we are looking at the cases where $kn \div kd$ are being turned into $n \div d$ (with $n$, $d$, $kn$, $kd$ being integers, and $d$, $k$ being non-zero). The scaled division is then solved, yielding the quotient $q$, the remainder $r$, and the following algebraic representation of the division as well: \begin{align*} n& =dq+r & \text{(where $0 \le r < |d|$)} \end{align*} In which case, if we multiply both sides of the equation by $k$, we would get that: \begin{align*} kn = (kd)q + kr \end{align*} But then, here is where we spot a potential problem, in that we cannot allow the scaling factor $k$ to be negative (why?). Once that restriction is set though, multiplying the inequality above by $k$ would yield that $0 \le kr < k|d| = |kd|$ — which is exactly the condition needed to make $kn = (kd)q + kr$ the final algebraic representation of $kn \div kd$.

So whether we proceed by fractional or algebraic representations, we always get $q$ and $kr$ as the quotient and the remainder of the original division. In fact, the same argument also applies when we’re scaling upward (i.e., $k$ being a unit fraction) — and it even applies to the general case of Euclidean divisions with real numbers as well!

Don't Scale With Negative Factors

As a reminder, don’t attempt to use negative integers as scaling factors. This means that while scaling $-3560 \div 26$ into $-1780 \div 13$ might be a good idea, scaling $-666 \div \, – 99$ into $74 \div 11$ is not!

As mentioned earlier, one thing we can also do is to scale up a division by solving a larger division first — before undoing the effect of the scaling factor. But then, why would such an idea be even useful in the first place?

Well, as it turns out, it’s about the number patterns and its complexity really, in that in some cases, scaling up might help us carry out the division with smaller or less “significant digits”, while in other cases, it might allow us to take advantage of the decimal system itself.

For example, let’s say that we were given the task to divide $4783$ by $125$. At first, this division might look a bit daunting, but after a bit of inspection, we also see that $125$ can be scaled up to $250$, which results in one less “significant digit“, or to $500$, which results in two less “significant digits”.

In fact, we can even scale up $125$ to $1000$, which results in the smallest and the least amount of “significant digits” possible. So let’s choose to scale up our division by $8$ — and see where this decision can take us to: \begin{align*} \frac{4783}{125} & = \frac{38264}{1000} \\ & = 38 + \frac{264}{1000} \\ & = 38 + \frac{33}{125} \end{align*} And… it looks like that after some algebra, the answers are already in front of us — with $38$ being the quotient and $33$ being the remainder of the original division. And while we have had to multiply and divide a few numbers by $8$, the numbers being involved are generally small — which gives the up-scaling approach a slight advantage over the direct division methods.

But whether it’s direct or otherwise, the point of introducing these tricks and methods is just so that we can elect to pick the one that’s the most fitting for each scenario. And while some of the methods might seem easier to learn and execute for most purposes, that doesn’t mean that they will always translate into more numerical thinking skill and mathematical growth!

So far, the division methods and approaches we’ve covered up to this point are all based on the decimal numeral system — the standard system which allows for $10$ numerals in a digit. In what follows, we’ll look at how division works in two alternate numeral systems: the binary numeral system and the hexadecimal numeral system.

In the binary system, only two numerals are available for each digit: $0$ and $1$. As a result, a binary number is always represented as a string of $0$s and $1$s. Here’s an illustration of how the binary number $101011$ works in the decimal system for example (decimal numbers in $\color{red}\text{red}$): \[ \overbrace{1}^{\color{red}2^5}\;0\;\underbrace{1}_{\color{red}2^3}\;0\;\overbrace{1}^{\color{red}2^1}\underbrace{1}_{\color{red}2^0} = {\color{red}2^5+2^3+2^1+2^0} = {\color{red}43} \] And while having only two numerals available for each digit clearly forces a number to have substantially more digits, the fact is that it also forces the arithmetic operations to be as simple as possible as well. Here’s how the first three operations work out at the digit level for instance:

Addition

$0+0=0$

$0+1=1$

$1+0=1$

$1+1=10$

Subtraction

$0\, – 0 = 0$

$0\, – 1 = -1$, $10\, – 1 = 1$

$1\, – 0 = 1$

$1\, – 1 = 0$

Multiplication

$ 0 \times 0 = 0$

$0 \times 1 = 0$

$1 \times 0 = 0$

$1 \times 1 = 1$

For additions and subtractions involving multiple digits, the calculations can be carried out by simply working through each digit one by one from the left to the right. Here are a few examples of multi-digit addition and subtraction to illustrate the point: \[ \begin{array}{r} 10101 \\[-0.3em] \underline{+\phantom{.0}1101} \\[-0.25em] 100010 \end{array} \qquad\qquad\qquad \begin{array}{r} 10101 \\[-0.3em] \underline{-\phantom{.0}1101} \\[-0.25em] 1000 \end{array} \] As for multiplications involving multiple digits, these can be carried out by first breaking the multiplier into its “digits” — before distributing the multiplicand over to these digits. Here’s how we’d calculate the product of $1101 \times 1010$ in binary for example: \begin{align*} 1101 \times 1010 & = 1101 \times (1000 + 10) \\ & = 1101 \times 1000 + 1101 \times 10 \\ & = 1101000 + 11010 \\ & = 10000010 \end{align*} Or more schematically: \begin{array}{r} 1101 \\[-0.3em] \underline{\times \phantom{.}1010} \\[-0.25em] 1101\phantom{000} \\[-0.3em] \underline{+\phantom{,00}1101\phantom{0}} \\[-0.25em] 10000010\end{array} At this point, addition, subtraction and multiplication all seem pretty — for a lack of a better term — binary, but what about division? How would it work out in its stead? And would we be able to continue to use the same methods and approaches this time around?

Well, as it turns out, binary division is really nothing more than the combination of the first three arithmetic operations, and as usual, we’ll still be able to use all of the aforementioned division tricks and methods — as long as we’re willing to keep up with the $0$s and $1$s incurred during the process of course!

So let’s take a look at an example with division then — this time with $100101001 \div 101$. As first, since we see that $101 \times 11 = 1111 < 10010$, this suggests that $11$ might have been a good candidate for the first two quotient digits: \begin{array}{r} 11\phantom{1101} \\[-0.35em] 101 \enclose{longdiv}{100101001}\kern{-0.15ex} \\[-0.60em] \underline{\phantom{1}1111}\phantom{1101} \\[-0.25em] 11\phantom{1101} \end{array} Here, notice that since the (partially notated) interim remainder is smaller than the divisor already, we can be assured that the first two quotient digits are correct — and that no further adjustment would be needed. As for the next two quotient digits, it seems that $11$ might just again do the trick: \begin{array}{r} 1111\phantom{01} \\[-0.35em] 101 \enclose{longdiv}{100101001}\kern{-0.15ex} \\[-0.60em] \underline{\phantom{1}1111}\phantom{1101} \\[-0.25em] 1110\phantom{01}\\[-0.3em] \underline{1111\phantom{01}} \\[-0.25em] -1\phantom{01} \end{array} As it turns out, we were quite close but still overshot by a little bit. That said, all we need to do is to retract the quotient by $1$ at the $2^2$ place — and all should be good: \begin{array}{r} 1110\phantom{01} \\[-0.35em] 101 \enclose{longdiv}{100101001}\kern{-0.15ex} \\[-0.60em] \underline{\phantom{1}1111}\phantom{1101} \\[-0.25em] \underline{1110}\phantom{01}\\[-0.25em] 100\phantom{01} \end{array} At this point, since the (partially-notated) interim remainder is again in the ideal range, this means that the last two quotient digits must be the only ones left. In which case, since $11$ is the largest candidate and that $101 \times 11 = 1111$ (i.e., smaller than the interim remainder $10001$), it follows that $11$ has to be correct: \begin{array}{r} 111011 \\[-0.35em] 101 \enclose{longdiv}{100101001}\kern{-0.15ex} \\[-0.60em] \underline{\phantom{1}1111}\phantom{1101} \\[-0.25em] \underline{1110}\phantom{01}\\[-0.25em] 10001 \\[-0.3em] \underline{1111} \\[-0.25em] 10 \end{array} And with that, the division procedure is now finished, yielding $111011$ and $10$ as the quotient and the remainder of $100101001 \div 101$, respectively. As a sanity check, let’s verify the algebraic representation that comes out of it — and see if we can get back to the same dividend as usual: \begin{align*} 101 \times 111011 + 10 & = (11101100 + 111011) + 10 \\ & = 100100111 + 10 \\ & = 100101001 \end{align*} Good. It looks like we were able to recover the original dividend safe and sound. Talk about what one can do with some seemingly-unending strings of $1$s and $0$s!

Unlike the binary system, the hexadecimal system has a significant amount of expressive power — in that it actually allows for $16$ numerals in a single digit. These include the usual numerals from $0$ to $9$ — plus the letters $A$, $B$, $C$, $D$, $E$, $F$ which correspond to $10$, $11$, $12$, $13$, $14$, and $15$ in the decimal system, respectively.

As a result, a hexadecimal number is always represented as a string of numbers — along with the letters from $A$ to $F$. Here’s an illustration of how the hexadecimal number $FACE50$ works in the decimal system for instance (decimal numbers in $\color{red}\text{red}$): \begin{align*} \overbrace{F}^{\color{red}15\times16^5}\underbrace{A}_{\color{red}10\times16^4}\overbrace{C}^{\color{red}12\times16^3}\underbrace{E}_{\color{red}14\times16^2}\overbrace{5}^{\color{red}5\times 16^1}\;0\; & = {\color{red} 15\times16^5 +10\times16^4+12\times16^3} \\[-30em] & \phantom{ = } {\color{red} \:+ \,14\times16^2 + 5\times16^1} \\[1em] & = {\color{red} 16436816 } \end{align*}

At the single-digit level, addition and subtraction work pretty much the same way as in the decimal system, except that the tables now have just gotten a bit larger. For that reason, it’s usually a good idea to have a solid grasp on the concept of complement — so that we can come up with the “opposite” of a hexadecimal numeral when called for:

\begin{array}{l c c c c c c c c c c c c c} \text{Number} & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & A & B & C & D & E & F \\ \text{Complement} & F & E & D & C & B & A & \cdots & 6 & 5 & 4 & 3 & 2 & 1 \end{array}

Once there, single-digit additions and subtractions — if involve carry or borrowing — can be calculated by re-expressing a number as the complement of the other. Here are some illustrations of the key cases of single-digit additions and subtractions for instance:

Addition (without carry)

$5 + 8 = D$

$4 + A = E$

$ C + 3 = F$

Addition (with carry)

$5 + C = 1 + 4 + C = 11$

$6 + F = 6 + A + 5 = 15$

$A + B = A + 6 + 5 = 15$

Subtraction

$F \, – 8 = 7 \: (\text{since }8 + 7 = F)$

$ 4 \, – B =\, – (B\, – 4) =\, – 7$

$14 \, – B = 9 \: (\text{since }B + 9 = B + 5 + 4 = 14)$

As for additions and subtractions involving multiple digits, these can be carried out by working digit-by-digit from the left to the right as usual. Here are a few tableaus to illustrate how multi-digit addition and subtraction work at the schematic level: \[ \begin{array}{r} FACE \\[-0.3em] \underline{+\phantom{F}BAD} \\[-0.25em] F\phantom{ACE} \\[-0.25em] 1\;0\; 5\phantom{CE} \\[-0.25em] 1\;0\;6\;6\phantom{E} \\[-0.25em] \boxed{1\;0\;6\:7B} \kern{-0.7ex}\end{array}\qquad\qquad\qquad \begin{array}{r} FACE \\[-0.3em] \underline{-\phantom{F}BAD} \\[-0.25em] F\phantom{ACE} \\[-0.25em] EF\phantom{CE} \\[-0.25em] EF\:2\phantom{E} \\[-0.25em] \boxed{EF\:2\:1} \kern{-0.7ex}\end{array} \]

Similar to the case in the decimal system, single-digit multiplications can be likened to iterated additions — except that the table is now almost three times larger. For that reason, mastering the multiplication table (such as the one below) can be useful:

And with that, we can now carry out multi-digit multiplications as usual — by breaking the multiplier into its digits and distributing the multiplicand over to those digits. Here’s a sample tableau of $BAD \times D09$ for instance — as we perform the multiplication from the left to the right: \begin{array}{r} BAD \\[-0.3em] \underline{\times \phantom{.}D\:0\:9} \\[-0.25em] 9\:7\,C\:9\phantom{AB} \\[-0.45em] 0 \phantom{D} \\[-0.45em] \underline{+\phantom{,9\:7\:}6\:9\:1\:5} \\[-0.3em] 9\:8\:3\:2\:1\:5 \end{array}

Calculation Through Conversion

For the record, all hexadecimal operations can also be carried out by converting the numbers into decimal form (or binary form, since each hexadecimal digit corresponds to exactly four binary digits). However, doing so will also forego the advantages of working within the hexadecimal system itself.

As with the case in binary, divisions in hexadecimal also involve a mix of additions, subtractions and multiplications, and while the “operational tables” now are obviously much larger, the division methods and tricks introduced in the previous sections are nevertheless still as applicable as before.

As an example, suppose that we’re given the division $CDEF \div AB$ to solve in hexadecimal. At first, it’s not totally clear what the first two digits of the quotient should be, but after a bit of reflection, we also see that:

$AB \times 10 = AB0$

$AB \times 11 = AB0 + AB =B5B$

$AB \times 12 = B5B + AB = C06$

$AB \times 13 = C06 + AB = CB1$

And since $13$ seems to be as close enough as it gets, this suggests that we might just as well inject it into our first tableau — and see what’d happen: \begin{array}{r} 1\;3\phantom{D} \\[-0.35em] AB \enclose{longdiv}{CDEF}\kern{-0.15ex} \\[-0.60em] \underline{CB\:1}\phantom{D} \\[-0.25em] 2\,D\phantom{F} \end{array} As it turns out, we did manage to get the (partially-notated) interim remainder $2D$ to be smaller than the divisor itself, which means that our first two quotient digits must be correct — and that we can move on to the last quotient digit instead. In which case, a bit of reflection would show that:

$AB \times 2 = AB + AB = 156$

$AB \times 3 = 156 + AB = 201$

$AB \times 4 = 201 +AB = 2AC$

And since $2AC$ seems to be the closest to the interim remainder $2DF$, this suggests that $4$ might be the best estimate for the last quotient digit. With it, the next tableau becomes: \begin{array}{r} 1\;3\: 4 \\[-0.35em] AB \enclose{longdiv}{CDEF}\kern{-0.15ex} \\[-0.60em] \underline{C\,B\:1}\phantom{D} \\[-0.25em] 2\,DF \\[-0.3em] \underline{2\,AC} \\[-0.25em] 3\;3 \end{array} As luck would have it, the interim remainder is again in the ideal range already, which means that the procedure is also finished, yielding the following algebraic representation as a result: \[ CDEF = AB \times 134 +33\] And most interestingly, the following fractional representation as well: \[ \frac{CDEF}{AB} = 134 + \frac{33}{AB}\] which is a living testament of how quickly a few letters can be turned into numbers — at a snap of fingers!

Pro-Tip

In general, choose the quotient digits and adjust up or down depending on the scenario. In particular:

For the lower-end digits, start with $0$ and scale up.

For the middle-end digits, start with $8$ (i.e., midpoint) and adjust up or down.

For the higher-end digits, start with $0$ and scale down instead.

Congratulation! If you’ve made it this far without skipping, you’re probably a true division geek! In fact, the main goal of this guide is to convey the idea that many methods and approaches for integer Euclidean division actually exist, and these include both the mechanical approaches such as long and short division, and the flexible approaches such as chunking and the bidirectional chunking method.

And while both types of methods are undoubtedly advantageous in their own ways, they also each have their own share of weaknesses at many different levels. As a result, a new method — the freeform method — was developed, in an attempt to mitigate these weaknesses and harness the best of both worlds.

In doing so, we’ve also introduced an adjustment trick which can be used to extend Euclidean divisions to the full range of integers — along with a technique of scaling which can be used to reduce the complexity of a problem. But then, the story doesn’t stop there…

In fact, by toying around with the different concepts and tools, we see that a division approach does not only derive its value from its ability to find the answers quickly — but by its ability to promote higher mathematical thinking and understanding as well. If anything, this is something that binary/hexadecimal division illustrates pretty well.

On that note, if you’re looking to further hone in the methods and approaches discussed in this guide, then you might find the following comprehensive summary worksheet both challenging and useful:

And if you’re looking forward to introduce some of the lesser-known division methods and approaches into your community, then of course, please do, but either way, the saga with long division and other division procedures continues!

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Since the word “computer” is referenced in this article and presumably, programming language implementations of division are relevant, you might consider citations: (The exact URLs can be obtained from internet search results):

Division and Modulus for Computer Scientists DAAN LEIJEN University of Utrecht Dept. of Computer Science

The Euclidean Definition of the Functions div and mod RAYMOND T. BOUTE University of Nijmegen