Chapter 8 - Shorthand Division

So far, we have covered three out of the four basic arithmetical computations.

In adding, we learned to use complements for the tougher combinations—those that would add over ten if we ever added over ten—and to record tens in such a way that the answer forms naturally in our mind, just as, on the modern abacus, the answer forms naturally on the board.

In subtracting, we learned never to subtract a larger digit from a smaller, and to avoid that crude and precarious method of "borrowing," so that again the answer forms itself easily and naturally in the mind or on the paper.

In multiplying, we have torn apart the multiplication table so that we use only half of it at a time. This enables us to discard the idea of "carrying," and furthermore produces the answer from left to right. When we have to record tens in preceding digits in our answer, we adopt a simple and effective method that—again—gives us a natural development of the answer.

Now, what about dividing?

There is no single secret for speed division comparable to the secrets of complements or no-carry multiplication. But by leaning on both complements and no-carry multiplication, we can build a streamlined technique for division that, in its total effect, can save you as much time and effort in this field as the single secrets can in theirs.

In order to get our ground firmly established, let us look at a sample problem in division and work it in the traditional long-division way:

This is a fairly simple problem. It has no remainder. Everything comes out even. Yet a great deal of pencil work was involved. It looks complicated.

Just for comparison, although the figures will be meaningless to you at the moment, let us show what the same problem would look like in the shorthand method you will learn in this chapter:

Certain elements of these numbers should be familiar to you—the underline and the slashes. The shorthand method will rely on your confident handling of complement subtraction and no-carry multiplication.

The two hardest parts of traditional long division, you will undoubtedly agree, are (first) determining at a glance the next digit of the answer and (second) going through the complex pencil work of verifying that digit and finding the remainder into which you divide in order to determine the next digit of the answer.

This chapter will offer a simpler way of doing each of these processes. Before we go into them, however, consider a few basic facts about the process of division.

Continuous Approximation

Long division, by which we mean division by a number of several digits, is really a progressive estimate that gets more accurate as we finish more of it.

In this sense, division is radically different from adding, subtracting, or multiplying. It is the only one of the four processes that we were taught to do from left to right, in the natural way. Since this is true, division is already self-estimating, just as the new methods for doing the other three processes are.

The familiar process we call long division, incidentally, seems to be a special crutch developed only in England and America, which, because every single step is spelled out (and written down in detail), no rational person in school can fail to learn to handle. It is certainly accurate and easy enough, but it is also infernally slow and cumbersome.

For another comparison, look at the division we just examined, next to the same problem solved in the European short-hand method:

American                                              European

If you have never before confronted this European (in England it is called "Continental") method, you may feel some awe of European education when you learn that the difference is simply this: the multiplying and subtracting are done entirely in the head. They are never written down at all. The two lines of working figures you see under the problem are merely the results of each subtraction.

Difficult? To us, yes. To a French or German schoolboy it is something he is expected to learn; and learn it he does, or flunks out and spends the rest of his life hoeing potatoes. But to those of us trained in the "dot-every-i, put-down-every-digit" methods of American arithmetic, it is rather difficult to learn late in life.

What you will discover before this chapter is over, however, is that applying your new mastery of simplified left-to-right multiplication and subtraction will make it not only possible to divide in a way similar to the European method, but actually easier than it was in the standard long-division way.

Before we get into this subject, we will first explore a method for rapid answer-producing that removes the first major stumbling block to quick and easy dividing.

Automatic Division

The point at which most of us hesitate longest in working our way through any long division is deciding on the next digit of the answer.

Consider this example:

The first step is to divide 87 into 42 or, since this "won't go," into 426.

Almost all of us, no matter how good our number sense is otherwise, lack any sort of genuine feel for such an answer. We are not dividing by 8, but by 87. Think back, and you will probably find that you often try two or three "trial answers" in your mind before deciding on one to put down.
Here is a simple trick to overcome this difficulty—a trick that automatically delivers to you the next digit of your answer no matter how complicated the divider is. It makes dividing by 34,968 as simple (at this point) as dividing by 4.

The trick is this: Do not divide by the divider. Divide only by its first digit, raised by one. Do not divide into the number divided. Divide only into its first digit (if that digit is larger than the divider digit) or into its first two digits (if the first digit is smaller than the divider digit).

This technique, by the way, is also adapted with minor variations from modern soroban theory. It is considered as basic to speed and ease on the abacus as is the use of complements for adding and subtracting.

In the example above, you do not try dividing 87 into 4263. Instead, divide 9 into 42. This is much, much easier. You should "see" the answer 4 at a glance.

The reason this works is that 87 is somewhere between 80 and 90, but for simplicity we consider it to be 90. A little over half the time, this first digit will be correct. Less than half the time, it will need revision—but the revision will be automatic and quick, just as it is on the abacus.

Try this technique on these examples:

On these three problems, our automatic division works like this:

5 (instead of 47) into 26 (instead of 268) is 5. This is the correct first digit of the answer.

7 (instead of 65) into 51 (instead of 513) is 7. Right.

3 (instead of 28) into 13 (instead of 136) is 4. This also checks out.

Caution: Note with special care that using this trick to "see" each successive digit of your answer does not alter the position of each answer digit. In the first example, you put the answer digit 5 over the 8, not over the 6. You still follow your classical rule for placing your answer: start as many digits over in the number divided as there are digits in your divider— plus one if you start by dividing into two digits instead of one:

In the first case, we "see" 8 into 9 and put down the answer digit 1 two places to the right because there are two digits in the divider. In the second case, we "see" 8 into 17 and put down the answer digit 2 three places to the right because there are two digits in the divider and we started the division into two digits of the number divided.

Now get the idea firmly in hand by trying these:

Remember that we are not yet finishing these divisions. At the moment, we are concerned only with developing this rapid and foolproof way to produce automatically each digit of the answer without hesitation.

Check your reactions to the above three examples. Did you see the first as 6 into 21, and put down 3 over the 6? Did the second become 10 instead of 9—indicating that the answer digit 1 goes over the 8? When you got to the third, did you "see" 4 into 8 as 2, and put it over the 3? If any of your answer digits got misplaced, review the general rule once more:

If your first division is into a single digit (4 into 9), the first answer digit appears as many places to the right over the number divided as there are digits in the divider.

If your first division is into two digits (4 into 23), the first answer digit moves one more place to the right.

The principle of finding each digit of the answer by dividing with only the first digit of the divider, raised by one, works with problems of any length. Experiment only on the examples provided, however, until we come to automatic revision.

Go through the following problem on your pad. Find the two digits of the answer by dividing with the first digit only (plus one) of the divider:

Work this out completely in your traditional handling of long division, applying to it at the moment only the new automatic digit-finder.

The final answer is 65. The first digit is produced by dividing 9 (not 876) into 57 (not 5714) and putting the resulting 6 over the 4. When you multiply out and subtract, you divide into the remainder 4580 for the second digit. 876 might make you hesitate between 5 and 6 for the second answer digit, but 9 into 45 can only be 5. We have produced two digits of the answer by simple inspection. For now, we will ignore the fractional remainder.

Now we will go on to the special aid that makes this technique useful on any problem at all, not merely on carefully selected examples.

Automatic Revision

Consider this case:

Start with the trick of dividing 9 into 43, instead of 876 into 4380. 9 will go into 43 no more than 4 times. Yet if you multiply out the divider by 4 and subtract, you find a remainder of 876. This is the divider itself. The answer to this problem is 5, not 4. Then it comes out even.

What is wrong? Nothing, really. We said earlier that division is really a continuous approximation from left to right. The digit of the answer we first put down is an approximation that may need revising before we finish.

On the abacus, each trial digit is produced by dividing with the first digit of the divider but without raising it first. Revision is frequently necessary, just as it is in this system. But revision on the abacus is always to reduce the trial digit by one (sometimes two), adding in this revision factor to the trial remainder. In our system of using a digit that is one higher than the first digit of the divider, the only way we ever have to revise is upward. As you will see when the technique develops fully, this is easier and more automatic with pencil and paper.

Since you are in effect dividing by a number larger than your real divider, you could not possibly try too large an answer digit. It is child's play to revise your answer upward in our system, but it would be quite difficult to revise it downward.

You have learned in no-carry multiplication how to increase the value of a digit by one without rewriting it. You simply underline it. The answer to the example above, when finished, would look like 4. You read it as 5.

All of this will be drawn together as we assemble the various parts of the complete division system. For the moment, remember only that you speed up your division by "seeing"

Try this part of the technique once more. Do not bother to complete these examples. Just practice "seeing" the first answer digit by dividing with the first digit only of the divider, raised by one:

Now we will combine everything we know about multiplication and subtraction, both of which are continuously involved, with this simplified digit-finding technique, to make the complete shorthand division method both easier and faster than the cumbersome method of long division.

Shorthand Division

We began the explanation of no-carry multiplication by taking apart a sample problem and seeing how the answer develops. Let us do the same thing with a sample division:

We have called the process of division "continuous approximation." The first approximation you got in the above problem was really 60, not 6: 7 goes into 441 something more than 60 times. You know this because there is obviously another answer digit to come.

We get the second answer digit by finding out first how much of the 441 is left after subtracting from it exactly 60 7's. In long division, we multiply the answer digit by the divider and put this product under the portion of the number divided that produced the digit.

That product here is 420. We do not ordinarily bother with the 0, any more than we bothered with the 0 in 60, since careful placement of each digit takes them into account.

By subtracting 42 from 44 and then "bringing down" the remaining 1 in the number divided, we find that there is 21 left over. Actually, we really subtracted 420 from 441. The "bringing down" completes that process. It would be helpful to inspect two expressions of this situation:

Dividing now the 7 into the remainder, we find that it will go exactly 3 times. In long division, we verify this by multiplying 7X3 and subtracting it from the remainder, getting a final remainder of 0.

Again, try to feel the identity of these two expressions of the current situation:

Now we will accomplish the same result with a fraction of the pencil work involved in long division.

The two most tedious parts of long division are (1) multiplying the answer digit by the divider and writing it down as you go, and (2) subtracting this product from the part of the number divided involved, in order to establish the remainder so far.

The European system, you recall, involves doing these two steps in your head. You write down only the final result of each subtraction. But this involves handling several digits at once in your head—contrary to the best approach to speed mathematics.
 
Since you know how to multiply from left to right, digit by digit, and also how to subtract from left to right, digit by digit—without carrying or borrowing—you can combine the two and accomplish the European result without ever handling more than one digit at a time.

We will use that same problem as the first model:

The following process, remember, is multiplication and subtraction done in one-two order, one digit at a time:

One: 7 x 6 is in the 40's, and 4 from 4 is 0:

We do not bother to write the 0. As you become accustomed to this system, you will not even bother to make any mark at all for this result. The lack of a digit there shows you that the result was 0.

Two: 7X6 ends in 2, and 2 from 4 is 2:

Traditional long division would now require you to rewrite the next digit of the number divided—1—beside the 2. You do not need to do this. You can bring it down mentally and see that the remainder is now 21, by reading the problem like this:

The next digit of the answer is 3, and you know this is right merely by inspection. Just to get the technique thoroughly established, however, we will verify it as you would in a more complicated problem:
 
One: 7 X 3 is in the 20's, and 2 from 2 is 0. Two: 7X3 ends in 1, and 1 from 1 is 0.
 
Get out your pad and pencil and actively follow each step in this demonstration: 

Compare the work you have now finished with the same problem spelled out in long division:

Get out your pad and pencil and actively follow each step in this demonstration

Although we have not mentioned it before, you naturally divide by any single digit without raising it in value by 1. If this divider were 84, we would divide by 9 because 84 is somewhere between 8 and 9. But 8 is obviously nothing but 8.

The first answer digit, by inspection, is 7. Now we multiply and subtract digit by digit:

One: 8 x 7 is in the 50's, and 5 from 5 is 0. Two: 8X7 ends in 6, and 6 from 9 is 3.

The remainder so far is 336. In order to produce the next answer digit, we mentally bring down the 3 and divide 8 into 33. We put down the answer digit 4. Now we verify and produce the remainder:

One: 8 x 4 is in the 30's, and 3 from 3 is 0. Two: 8X4 ends in 2, and 2 from 3 is 1.
 
 The remainder at this point is 16. In the illustration above, we have already mentally brought down the 6 and put down the next answer digit, 2. Is there any remainder?

One: 8 x 2 is in the 10's, and 1 from 1 is 0. Two: 8X2 ends in 6, and 6 from 6 is 0.

The problem comes out even. A little later on, we shall get into fractional and decimal remainders.

Now try one entirely on your own. It will be an easy one, to get the technique firmly bedded in your habits before going on to more complicated problems. Do this one on your pad:

 After you have finished, check your working against this step-by-step explanation:

First digit: 7. One: 6 x 7 is in the 40's, and 4 from 4 is 0. Two: 6x7 ends in 2, and 2 from 5 is 3. Remainder (by mentally bringing down the 6), 36.

Second digit: 6. One: 6 x 6 is in the 30's, and 3 from 3 is 0. Two: 6x6 ends in 6, and 6 from 6 is 0.

Automatic "Borrowing*

The demonstrations so far have been chosen for simplicity. They are simple both because you are dividing by single digits and because there is no canceling ("borrowing") involved in the subtraction.

Consider this case:

 This problem will involve canceling. Yet because you have learned to use canceling in the answer instead of "borrowing" in the larger number, you will find it no trick at all to adapt what you already know to the smooth and efficient working of this kind of division.

The first answer digit, by inspection, is 2. We get the remainder so far in our usual way:

One: 9 x 2 is in the 10's, and 1 from 2 is 1. Two: 9X2 ends in 8, and 8 from 2 is —

STOP! Larger from smaller. Do not subtract. Add the complement of 8 (2) to 2 and slash left:

This answer should look perfectly normal after your work with left-to-right subtraction. It is simply 4. The (slashed) 1 has been reduced by the slash by one in value, to 0.

Mentally bringing down the next 2, you "see" the answer of 9 into 42 as 4 and put this down as the second answer digit. Now for the remainder:

One: 9 x 4 is in the 30's, and 3 from 4 is 1.

Two: 9X4 ends in 6, and 6 from 2 is—larger from smaller. Do not subtract. The complement of 6 (4) plus 2 is 6, and slash:

See if you can finish this problem yourself. Mentally bring down the proper digit and see the answer. After you have worked it out, check against this explanation:

The next answer digit is 7—9 into 63.

One: 9 x 7 is in the 60's, and 6 from 6 is 0.

Two: 9X7 ends in 3, and 3 from 3 is 0.

Try one example that involves "borrowing" before going on. Use your pad and do this problem just as we did the one above:

Do this with your pad and pencil before checking against the working figures below. A full understanding of the steps in shorthand division is essential before we get into longer problems. Once you have gone through the routine several times, you will find that you can handle any division with confidence.

Check your work and your answer with this finished problem. Here is how it should look:

That is all there is to it. No single part of this process is complicated. It is all based on techniques you have already mastered in earlier parts of this book. But the combination of them is new. If you have any trouble assembling the parts into a smooth-working whole, then go back and re-check the weak parts.

Make very sure you have everything so far down pat, because we are now going to add the third and final element in shorthand division that makes it just a bit more complex. You have already learned to handle this step in multiplication, but the mental processes will have to stretch one more notch when you apply it to division.

When you have done these, compare your results with these models:

Stop now and make sure you are ready for the next step by doing these two problems on your pad:
 
Now let us pick up the final technique from multiplying that enables you to handle shorthand division with dividers of any length.

No-Carry in Division

So far, you have been dividing by only one digit. In such divisions, you would ordinarily do most or all of these steps entirely in your head anyway and not worry about putting down the remainders as you went along. It is really short division, and we have started with this only to get the general technique firmly established.

When you divide by numbers of two or more digits—by 653, for instance—you will add to your simultaneous left-to-right multiplication and subtraction the efficient and handy no-carry system.

This is the point where the European shorthand method becomes really difficult for most of us. When dividing by a number of two or more digits, the European system requires you to multiply (including carrying) and subtract (including borrowing), all in your head. This can involve juggling as many as six digits all at once in your mind. With our left-to-right methods, however, we can do everything digit by digit.

If you understand thoroughly both the no-carry multiplication method and the division method covered so far, then you could probably work out the entire method yourself without further help. Since it is using no-carry multiplication within a new framework, however, we will go into the entire process step by step.

Recall, as we get into this, the technique of dividing by only the first digit of the divider—raised by one:

Our first answer digit we "see" by dividing 4 into 13, and putting 3 over the 5. Remember that the first answer digit starts as many places over the number divided as there are digits in the divider—plus one if you start dividing into two digits instead of one.

Now we develop the remainder:

One: 3 x 3 is in the zeros, and 0 from 1 is 1

Three: 8X3 ends in 4, and 4 from 5 is 1:

Two: 3X3 ends in 9. 8 x 3 is in the 20's. 9 + 2 is (complement of 9 from 2) 1, and record. Now you have two things to do: subtract 1 from 3, and record the ten. Since we are subtracting while we multiply, this ten obviously gets subtracted. How? Just by canceling; slash left:

There was a curve hidden in point two of that example, but it seemed best to slide it in quietly. It is a new application of recording by slashing, because the only digits we jot down are the results of the subtraction and the final effect of a recorded ten from the multiplication is obviously to reduce the preceding digit in the answer to the subtraction by 1.

So one of the side-rules of speed division grows from this: when your no-carry multiplication involves a complement (and therefore a recorded ten), slash the digit to the left in your working figures.

Our remainder so far is 2,128. 38 will go into 13,528 300 times, with 2,128 left over. If the size of these figures jars you, inspect the work so far and think back to the identity-expressions at the start of this section.

Note that we have two digits in the answer to the subtraction. We still bring down the next digit (mentally), so the next division is 38 into 212. We "see" it as 4 into 21, and put down 5. Now we develop the remainder:

One: 3 x 5 is in the 10's, and 1 from 2 is 1:

Two: 3X5 ends in 5. 8 x 5 is in the 40's. 5 plus 4 is 9, and 9 from 1 is—larger from smaller. Add the complement of 9 to 1, and put down 2. Cancel by slashing to the left:

 This brings up an interesting point. You have two occasions to slash to the left when doing shorthand division: when a complement is used in no-carry multiplication, and when a complement is used in subtraction. Both involve the use of a complement, and both result in a slash to the left. One special result of this will develop later.

Three: 8x5 ends in 0, and 0 from 2 is 2:

The remainder at this point is 228. This is the excess after subtracting 38 x 350 from 13528.

Before going on to the final digit of this answer, and determining the remainder (if any), recall our earlier comments about revised digits. Dividing by only the first digit of the divider is the quickest and easiest way to produce the next digit of the answer, but sometimes it will need revising. This is a price paid happily by operators of the high-speed abacus, because it saves more time in producing each digit than it costs in revision.

The final digit of this answer will demonstrate such a case.

See the next digit of the answer as 4 into 22—5. Now let us work out the remainder (if any) and see what happens.

One: 3 x 5 is in the 10's, and 1 from 2 is 1:

Two: 3X5 ends in 5. 8 x 5 is in the 40's. 5 plus 4 is 9, and 9 from 2 is—complement of 9 plus 2, and slash:

Three: 8X5 ends in 0, and 0 from 8 is 8:

The remainder is the same as the divider. So our final answer digit must be raised by 1, and the problem will come out even. As we pointed out before, you raise the digit by underlining it, so the final answer is 3 5 5 which you read as 356.

You were promised that digit-revision would be automatic. It is. Any digit in your answer may need revising, even the first. But the time to do so is signaled to you automatically, so you do not need to watch especially for such events.

This problem demonstrates why:

We "see" the first answer digit as the result of 7 into 54, which can be only 7. Now (use your pencil and pad to help build the technique) we find the first remainder:

One: 6 x 7 is in the 40's, and 4 from 5 is 1:

Three: 3X7 ends in 1, and 1 from 7 is 6:

Two: 6x7 ends in 2. 3 x 7 is in the 20's. 2 plus 2 is 4, and 4 from 4 is 0. Where a zero digit appears in the middle of a subtraction answer, as it does here, it is wise to put it down to avoid possible confusion:

Perhaps you have noticed that your remainder in this case is larger than your divider. Something is wrong, and what is wrong is that your answer digit needs raising by one.

You do not have to be especially alert to this situation, however. If you didn't notice at this point, you could not help but notice as soon as you tried to get the next answer digit. You mentally bring down the 4 and divide 63 into 1064— seeing it as 7 into 106.

Such an answer digit would be over ten. There is no such digit. This, in case you missed the signal that developed when your first remainder was larger than the divider, is the STOP signal that warns you to revise your answer digit.
 
This is automatic division. To raise the answer digit, you merely underline it. To adjust the remainder, you merely subtract the divider and put down the new remainder before going on:

The underlined 7 is, of course, now 8. The 43 is the answer we get after subtracting 63 (the divider) from 106 (the too large remainder). Now we are ready to continue, with everything adjusted and correct so far.

Finish this problem yourself on your pad. It does come out even, though one other answer digit will need revision. Finishing this problem will involve just about every technique in shorthand division covered so far.

Do it now.

If your final answer did not come out to an even 768, which you read or rewrite as 869, check the appearance of your working figures against this model:

This section, in its joining together of many different techniques from earlier parts of the book into one effective but apparently complex whole, has been perhaps the most difficult chapter to understand in one reading.

Sit back for a moment and let some of what you have done sink into your mind. Don't be discouraged if it takes several readings to understand fully what has been going on. It involves quite a new way of looking at numbers, a way really simpler than the traditional ways because much of it has been adapted from the simplest and highest-speed arithmetical system known—the modern Japanese abacus—but until you get used to it it does take some special lip-biting.

Division is the most complex of all our basic operations in arithmetic. There is simply no help for this; it is the nature of the beast.

What we have done so far is to reduce the process to the simplest series of easy steps that can possibly work. You never hold more than a digit or two in your mind at any one point; you work from left to right; you never have to carry as you multiply; and you never have to "borrow" as you subtract.
 
The seeming complexity at this point is inherent in the function itself. If you had never learned long division, and if somebody sat down to explain it to you, it would seem much more complex. There are many separate processes to be done, and if full accuracy is required every process must be done in full.

Except for one really minor special case, you now know everything you need to know for this rapid way to divide. Ease and speed will come with practice, which the next chapter will help to provide.

The one special case involves slashing a number not once, but twice. We hinted at this possibility when we pointed out that you slash left when you use a complement in multiplying, and you also slash left when you use a complement in subtracting. It does not come up very often, but it does come up and it is very easy to handle.

Here is the sort of problem in which you will find this necessity:

Let's begin this problem step by step, to drive the method deeper into your mind.

8 (not 74) into 50 (not 505) gives 6 as the first answer digit. Put it down, and determine the remainder:

One: 7 x 6 is in the 40's, and 4 from 5 is 1.

Two: 7X6 ends in 2. 4 x 6 is in the 20's. 2 plus 2 is 4, and 4 from 0 is (complement) 6 and slash.

Three: 4x6 ends in 4, and 4 from 5 is 1.

Our example now looks like this:

Inspection shows us that the next answer digit (8 into 61) is 7. We put it down and work out the remainder: One: 7 x 7 is in the 40's, and 4 from 6 is 2:

Two: 7X7 ends in 9. 4 x 7 is in the 20's. 9 plus 2 (complement, slash) is 1, from 1 is 0:

Three: 4X7 ends in 8, and 8 from 4 is (complement) 6 and slash.

BUT—when we slash a zero, we must always slash the digit to the left of it as well. The 2 to the left of the 0 is already slashed, but we slash it again. We have no choice. If one slash reduces a digit in value by one, two slashes reduce it in value by two, leaving nothing of that double-slashed 2:

Why does it happen this way? The answer is simply that we are multiplying and subtracting at the same time, digit by digit. The use of a complement in either case calls for recording a ten in multiplying (which means canceling a ten here, since we put down only the result of the subtraction) or canceling a ten in subtraction. Now and then, both may affect the same digit—as they did here.

That slashed 0, remember, is now a 9.

Let's go on. See the next answer digit as 8 into 96— MORE THAN TEN.

 
The remainder 222 comes from the subtraction of 74 from 96—really, of course, 740 from 962.

This signals the need for raising the answer digit by 1. This is not difficult. Just underline it, subtract the divider from the remainder, and the problem now looks like this:
 
Now we will find the last digit of the answer. 8 into 22 is 3. Multiply and subtract at the same time on your pad and find out whether or not this problem comes out even.

Do it yourself.

Does the problem come out even or not?

Longer Dividers

Perhaps you are already wondering whether dividing by numbers of three or more digits make things much more complicated.

The answer is, not much.

Nothing in our technique changes one bit, except that we repeat step two of no-carry multiplication as many times as we need to in order to get a full subtraction. Dividing by four- or five-digit dividers is no harder than dividing by two-digit dividers. There are more details and it will take a little longer, but the process is not really different

You still divide by only the first digit of your divider, raised in value by one, to produce automatically each succeeding digit of your answer. Should that answer digit need revising, that fact will be signaled to you when the remainder is larger than the divider. If you miss that signal, you are notified again when the next answer digit seems to be ten or more.

Once in a very great while an answer digit will need revising twice. After you have raised it once and adjusted your remainder, the remainder will still be larger than the divider —and the next answer digit will still be ten or more. In such rare cases, just underline the answer digit again (raising it in value by 2) and subtract the divider once again from the remainder before continuing.

The following example is admittedly a wild extreme, so obvious on the face of it that even by rote you could hardly get into this sort of situation. Yet, even should you abandon all your number sense and follow every rule without looking at the problem itself, the rules would eventually bail you out. You would have to raise your answer digit no less than five times, yet it would ultimately be right:

 This, as we said, is absurd. Yet it demonstrates the absolute reliability of the operating rules even when your own common sense sees nothing wrong. You divide 2 into 9 and see an answer digit of 4, so your first remainder is 55. You raise the answer by one and subtract the divider, giving a remainder of 44. This goes on through four more revisions of the answer digit, the last one taking care of the remainder 11.

The problem was selected especially to demonstrate this ultimate possibility. If an answer digit seems to need revising more than once or twice, sit back and look at the problem as a whole. Chances are you have overlooked something very obvious. The rules are as important to fast mathematics as trees are to a forest—but we take note of the forest first, then use the trees. The folk saying is too obvious to need repetition here.

One more point might be mentioned. If you work entirely by rote, you might sometimes be confused by the placement of your remainder when working out a problem like this:

The first answer digit, by inspection, is 2—4 into 8. But beware when you start to develop the remainder:

One: 3 x 2 is in the O's, and 0 from—

STOP. 0 from 8? This cannot be so. The answer digit would need several revisions. No, in this case it is 0 from 0— the unshown 0 to the left of 8.

Why? Your own number sense should give you a strong inkling. To distill it into an operating rule, it is because you divided into the first one digit of the number divided, rather than into the first two digits. This in effect moves the product of this first multiplication one place to the left.

So when you divide the first digit of the divider into the first digit only of the number divided, start right out by ignoring what that first answer digit and the first digit of the divider would "be in." They wouldn't "be in" anything but the zeros, or you would have divided into the first two digits of the number divided.

The rest of this particular problem, by the way, shows a typical example of two-digit revisions. Let's go through it. So far, the figures look like this:

The next answer digit, by inspection, is 6—4 into 25. Put it down and work out the remainder:

One: 3 x 6 is in the 10's, and 1 from 2 is 1.

Three: 2X6 ends in 2, and 2 from 6 is 4:
 
Two: 3X6 ends in 8. 2 x 6 is in the 10's. 8 and 1 is 9, and 9 from 5 is—complement, slash:

Look at the remainder. Your growing number sense might show right away that it is exactly twice the divider. If not, you would at least notice that it is larger than the divider, so you underline the 6 to raise it to 7, and subtract the divider from the remainder: 3 from 6 is 3. 2 from 4 is 2.

Look at the remainder again. It is the same as the divider. Underline the underlined 6 once again, subtract the divider, and the problem comes out even. The answer is 2 6, which you read or rewrite as 28.   =

Get out your pad now, have an absolutely clean page on top, and start one problem with a five-digit divider. This one problem embodies just about every possible wrinkle in shortcut division.

This problem is bound to be a bit tedious, in any method of arithmetic. We can get a rapid estimate, as the next chapter will show, but for a complete answer there is no avoiding a number of steps.

Take a deep breath and plunge in.

4 into 18—our way of producing the first answer digit automatically—is 4. This 4 goes over the sixth digit of the number divided, since there are five digits in the divider and we started by dividing into two digits rather than one. Now we start on the remainder:

One: 3 x 4 is in the 10's, and 1 from 1 is 0.

Two: 3X4 ends in 2. 6 x 4 is in the 20's. 2 plus 2 is 4, and 4 from 8 is 4.
(We are still spelling out every step to make it clear. But as you practice the smooth handling of these steps, your eye and mind should begin to jump from one fact to the other almost without intermediate thought. Step two above will, with experience, become "2—A—4.")

Three: 6x4 ends in 4. 1 x 4 is in the O's. 4 from 9 is 5.

Four: 1 x 4 ends in 4. 8 x 4 is in the 30's. 4 and 3 are 7, and 7 from 2 is (complement) 5 and slash left.

Five: 8x4 ends in 2. 2 x 4 is in the O's. 2 from 6 is 4.

Six: 2x4 ends in 8. 8 from 8 is 0.

 Are you ready to go on? Or would it be a good idea to take another look at the remainder? The remainder is larger than the divider. So underline the 4—raising it to 5—and subtract 36182 from 44640, left to right, canceling in the answer:

If you followed these steps on your pad, here is what your work should look like:

If 18468 is smaller than 36182, we are ready to go on. Divide 4 into 8 and put down 2 as the second answer digit. Start developing the remainder:

One: 3 x 2 is in the 0's. Ignore it, because we divided into one digit rather than two.

Two: 3X2 ends in 6. 6 x 2 is in the 10's. 6 plus 1 is 7, and 7 from 8 is 1.

Three: 6x2 ends in 2. 1 x 2 is in the 0's. 2 from 3 is 1.

Four: 1X2 ends in 2. 8 x 2 is in the 10's. 2 plus 1 is 3, and 3 from 5 is 2.

Five: 8x2 ends in 6. 2 x 2 is in the 0's. 6 from 8 is 2.

Six: 2 x 2 ends in 4. 4 from 2 (complement) 8, and slash.

Check your jottings against the model at this point:

This should be enough step-by-step explanation. Go ahead and finish this problem. You will find that it does not come out even. There will be a remainder. Make sure to examine the remainder carefully. There is a reason why you should.

Do not look ahead to the finished problem until you have a remainder that satisfies you. Then, if you feel you understand all the techniques in this chapter, go ahead to build speed in division.
 
The remainder, as you undoubtedly discovered, held within it a revision of the last digit:

Final answer, 523; remainder, 3,638.

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