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# Introduction

## Forewords

The eastern abacus (simplified Chinese: 算盘; traditional Chinese: 算盤; pinyin: suànpán, Japanese:  そろばん soroban,  simply the abacus in this textbook), as an abacus of fixed beads sliding on rods, originated in China at an uncertain date, but by the late 16th century its use had entirely displaced counting rods as a computing tool in its home country. From China its use spread to other neighboring countries, especially Japan, Korea and Vietnam, remaining as the main calculation instrument until modern times. The way in which it was used, the “Traditional Method”,  remained stable for at least four centuries until the end of the 19th century, when an evolution began towards what we will call the “Modern Method”, that makes use of a “Modern Abacus”. This textbook is intended as an introduction to the traditional method, and is aimed at people who already know how to use a modern abacus using the modern method.

Modern abacus is of the 4+1 type, i.e. it has four beads on the lower deck and one on the upper deck.

 0 1 2 3 4 5 6 7 8 9

This is all that is needed to be able to perform decimal arithmetic with the abacus. However, traditional abacuses had additional beads, the most frequent being the 5+2 type (although the 5+1 type were also popular in Japan) and occasionally the 5+3 type.

With three upper beads we can store up to 20 on a single rod, which is convenient for traditional division and multiplication techniques. With two upper beads we can achieve the same by using the suspended bead technique (懸珠, Xuán zhū in Chinese[1], kenshu in Japanese), a kind of simulated or virtual upper third bead for the rare occasions when this third bead is required (see figure from 15 to 20).

 0 1 2 3 4 5 5 6 7 8 9 10
 10 11 12 13 14 15 15 16 17 18 19 20

With a lower fifth bead, we have two different ways to represent the numbers 5, 10 and 15. This means that we have options from which we can choose the one that is most convenient for us. In the case of addition and subtraction, the possibility of choosing between two representations for 5 and 10 will allow us to simplify the calculations somewhat.

Traditional techniques can be used on any type of abacus, with only the obvious exception of the use of the lower fifth bead on a 4+1 abacus, the difference between having or not additional upper beads is more a matter of comfort and reliability than of efficiency or capabilities.

Traditional method was used for at least four centuries, covering Ming and Qing dynasties in China and Edo period in Japan. Beginning with the Meiji Restoration in Japan, students of the abacus changed in the sense that they already knew written mathematics before they began to study the abacus, whereas students of earlier times did not know anything about mathematics previously. For most, the abacus was the only form of math they were going to know. This caused a slow adaptation of the teaching and the methods of the abacus to the new times and circumstances, giving rise, after several decades, to what we now call the Modern Method, in fact, a simplified method.

In the English language, the following two works by Takashi Kojima are frequently cited in reference to the modern method:

• The Japanese Abacus: its Use and Theory (1954)[2]
• Advanced Abacus: Theory and Practice (1963)[3]

Several editions of these books can still be found, including e-books formats, and the first one can be consulted at archive.org. In this wikibook, the reader is assumed to be familiar with the content of at least the first of these works.

Today, the modern method may seem optimal in many ways and we may think that some "oddities" of the traditional method, especially the way of organizing the division on the abacus, lack practical sense; but if the traditional method remained stable for centuries despite millions of users, including great figures of mathematics like Seki Takakazu who was a great promoter of the use of the soroban abacus in Japan, it can only be because it was also considered optimal by its users. Only the optimality criterion of the ancients differed from the one we may have today.

Unfortunately, no one in the past bothered to write why things were done that way, they just wrote about how to do things, and we can only speculate on the reasons underlying some of these ancient techniques.

## Main differences between traditional and modern methods

These are the three most important points that differentiate traditional techniques from modern ones:

• The use of the fifth lower bead in addition and subtraction to simplify both operations a bit, which extends to everything that can be done with the abacus since everything ultimately depends on addition and subtraction.
• The use of a division method using a division table that eliminates the mental effort required to determine the quotient figure. This method (kijohou, guīchúfǎ 帰除法) first described in the Mathematical Enlightenment (Suànxué Qǐméng, 算學啟蒙) by Zhū Shìjié 朱士傑 (1299)[4] using counting rods superseded the old division method based on the multiplication table and whose origin dates back to at least the 3rd to 5th centuries AD, to the book The Mathematical Classic of Master Sun (Sūnzǐ Suànjīng 孫子算經)[5][6]. This old method, being the basis of the short and long methods of written division, has in turn replaced the traditional method of dividing in modern times. That is, modern times have taken us back to the old!
• Traditional and modern methods also differ in the way the division operation was organized on the abacus. The traditional division arrangement is somewhat more compact than the modern one and also more problematic as it requires (or benefits from) the use of additional, higher beads. This arrangement of the division in turn conditions the way in which multiplication and roots are organized.

## The principle of least effort

As mentioned above, no author in the past has written about why things were done this way, only about how to do things; so we can only guess to try to understand why. But the reader will see throughout this book that the traditional techniques suppose, by comparison to the modern ones, a reduction of the mental effort necessary to use the abacus. This is especially clear in the case of division that uses a division table, but also in the rest of the techniques that will be described since they effectively involve a reduction in the number and/or the extent of "gestures" required to complete an operation. We call gesture here to:

• hand displacements
• changes of direction
• skipping rods (i.e. changing hand position from a starting rod to other  non-adjacent rod)

and each of these gestures,

• as a physical process, takes a time to complete,
• as governed by our brains, requires our attention, consuming (mental or biochemical) energy,
• as done by humans (not machines), has a chance to be done in the wrong way, introducing mistakes.

so that we can expect, by reducing the number and extension of these gestures, a somewhat faster, more relaxed and reliable calculation.

In view of the above, one is tempted to think that by adopting this principle of minimum effort, traditional techniques evolved in the sense of making life with the abacus easier, which could explain its validity throughout the centuries, but it is nothing more than a conjecture without documentary support.

If we think of the modern method, polarized towards simplicity, speed and effectiveness, we could say that it is the sprinter method while the traditional method is the Marathon runner method.

The reader, after following this textbook, will be able to draw their own conclusions about it.

## Abacus procedure tables, some terms and notation

As usual, in this book we will use tables to describe procedures on the abacus, for example:

Example of abacus procedure table
Abacus Comment
ABCDEFGHIJKLM
896 412 This time the divisor goes to the left and the dividend to the right
896 512 Column E: rule 4/8>5+0, change 4 in E into 5, add 0 to F
896 512 cannot subtract E×B=5×9=45 from FG,
-1 revise down E: subtract 1 from E,
896 492
etc. etc.

Where, on the left, either the digit by digit evolution of the state of the abacus or the current addition or subtraction operation is shown along with comments, on the right, about what is being done. The columns of the abacus are labeled with letters at the top (blank spaces represent unused / cleared rods).

This representation, which is perfect for the modern abacus, needs a couple of refinements to adapt it to the traditional abacus.

• A column of a traditional abacus can contain a number greater than 9 and it is not possible to write its two digits in our table without disturbing its vertical alignment. To get around this, we will use underline notation for values between 10 and 19 and the first digit (one) will be represented by an underline on the preceding column (see chapter Dealing with overflow for a reason). For example, the situation represented below occurs shortly after starting the traditional division of 998001 by 999
A B C D E F G H I K J L M 9 18 9 0 0 1 0 0 0 0 9 9 9
and is represented in procedure table as
Using underline notation
Abacus Comment
ABCDEFGHIJKLM
988001    999 Column B value is 18
• As seen above, numbers 5, 10 and 15 have two possible representations: using or not the 5th lower bead. When it is pertinent to distinguish between the two, we will use the following codes:
• F: to denote a lower five (five lower beads activated) as opposed to:
• 5: upper five (one upper bead activated).
• T: ten on a rod (one upper bead and five lower beads activated). On 5+2 type abacus, it is also a lower ten as opposed to t an upper ten (two upper beads activated).
• Q: lower fifteen on a rod (two upper beads and five lower beads activated) as opposed to q upper fifteen (suspended upper bead on the 5+2, three upper beads activated on the 5+3).

## External resources

### Soroban Trainer

If you are interested in traditional techniques but do not have a traditional abacus yet, you can use the JavaScript application

 Soroban Trainer

## References

1. Chen, Yifu (2013). L’étude des Différents Modes de Déplacement des Boules du Boulier et de l’Invention de la Méthode de Multiplication Kongpan Qianchengfa et son Lien avec le Calcul Mental (PhD thesis) (in French). Université Paris-Diderot (Paris 7). p. 40. {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
2. Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
3. Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0003-7
4. Zhū Shìjié 朱士傑 (1993) [1299]. Suànxué Qǐméng (算學啟蒙) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
5. Ang Tian Se; Lam Lay Yong (2004), Fleeting Footsteps; Tracing the Conception of Arithmetic and Algebra in Ancient China (PDF), World Scientific Publishing Company, ISBN 981-238-696-3
6. Sunzi 孫子 (3rd to 5th centuries AD). 孫子算經 (in Chinese). {{cite book}}: Check date values in: |year= (help); Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)

## Introduction

With any abacus type, addition is simulated by gathering the sets of counters representing the two addends, while subtraction is simulated by removing from the set of counters representing the minuend a set of counters representing the subtrahend. Addition and subtraction are the only two possible operations on any type of abacus. Everything else has to be decomposed into a sequence of addition and subtraction.

There is hardly any difference between addition and subtraction with a modern abacus or a traditional one, if the reader already knows how to perform these two operations fluently with a modern abacus, he will also do well with a traditional one. The only two additional points to consider are:

• use of the lower fifth bead to simplify the operations.
• alternating rightward and leftward operation to save hand displacements.

of which the first is by far the most important.

The lower fifth bead can be used in addition and subtraction operations just like its companions. Its use is demonstrated in some ancient books such as: Computational Methods with the Beads in a Tray (Pánzhū Suànfǎ 盤珠算法) by Xú Xīnlǔ 徐心魯 (1573)[1], but over time it ceased to appear in the manuals, perhaps as a non-fundamental technique it was no longer explained in the concise books of the past but surely it continued to be taught verbally, as a trick to abbreviate the operations. We dedicate the following chapter: Use of the 5th lower bead to this subject.

## Reverse operation

Some old books on the abacus, for instance, Mathematical Track (Shùxué Tōngguǐ 數學通軌) by Kē Shàngqiān (柯尚遷) (1578)[2], demonstrate the addition using an alternating direction of operation with the obvious intention of saving hand movements. If the reader has already studied the modern abacus he knows for sure why it is preferable to operate from left to right, and this is not only a question of the use of the abacus. In the 19th century, the well-known Canadian-American astronomer Simon Newcomb, a renowned human computer, recommended the practice of adding and subtracting from left to right using pencil and paper in the introduction to his tables of logarithms[3].

Therefore, the alternation of direction of operation should be considered a secondary matter. If it is mentioned here, it is because regardless of its limited usefulness it is a very interesting exercise that can be difficult at first, resulting in a small challenge that can lead the reader to interesting reflections on the order of movement of the fingers; in particular, on whether carries and borrows should be done before or after.

Chapter Extending the 123456789 exercise proposes its daily use as a way to perfect our understanding of beading.

## Learning the abacus in the past

It may be of interest to know that in the past people learned the abacus without having prior knowledge of mathematics, in particular without knowing anything like an addition or subtraction table; instead they memorized a series of mnemonic rules, verses or rhymes, short phrases in Chinese that indicated which beads had to be moved to result in the addition or subtraction of one digit to/from another digit[4][5][6]. We have an example in English of what these types of rules were like thanks to the booklet: The Fundamental Operations in Bead Arithmetic, How to Use the Chinese Abacus by Kwa Tak Ming[7], Printed in Hong Kong (unknown publisher and date), a work aimed to English-speaking Filipinos according to the author. Here are rules/rhymes/verses that appear on it and whose interpretation is left to the reader:

One; lower five, cancel four One; cancel five, return four
Two; lower five, cancel three Two; cancel five, return three
Three; lower five, cancel two Three; cancel five, return two
Four ; lower five, cancel one Four; cancel five, return one
One ; cancel nine, forward ten. (i.e. carry one to the left column) One ; cancel ten (i.e. borrow one from the left column), return nine
Two; cancel eight, forward ten Two; cancel ten, return eight
Three ; cancel seven, forward ten Three; cancel ten, return seven
Four ; cancel six, forward ten Four ; cancel ten, return six
Five; cancel five, forward ten Five; cancel ten, return five
Six; cancel four, forward ten Six; cancel ten, return four
Seven ; cancel three, forward ten Seven; cancel ten, return three
Eight ; cancel two, forward ten Eight; cancel ten, return two
Nine ; cancel one, forward ten Nine ; cancel ten, return one
Six ; raise one, cancel five, forward ten Six; cancel ten, return five, cancel one
Seven ; raise two, cancel five, forward ten Seven ; cancel ten, return five, cancel two
Eight; raise three, cancel five, forward ten Eight; cancel ten, return five, cancel three
Nine; raise four, cancel five, forward ten Nine; cancel ten, return five, cancel four

Clearly, the table above does not contain the trivial rules ; eg. "to add two, activate two lower beads" or "to subtract 6, deactivate both an upper and a lower bead". In the event that we cannot proceed with such rules because we do not have the necessary beads at our disposal, then, we use the non-trivial rules listed in the table.

Once the students learned to add and subtract with these types of rules, they began to memorize the multiplication and division tables also in the form of verses or rymas. In total, learning the basics of the abacus required memorizing about 150 rules that had to be recited or sung while applied.

We will have a chance to see more rules by studying the traditional division in this book.

## Chapters

### Use of the 5th lower bead

The specialized use of the 5th lower bead and the non-unique representation of numbers 5, 10 and 15 to simplify operations.

### Extending the 123456789 exercise

A plethora of addition and subtraction exercises that can be done without an exercise sheet.

## References

1. Xú Xīnlǔ (徐心魯) (1993) [1573]. Pánzhū Suànfǎ (盤珠算法) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
2. Kē Shàngqiān (柯尚遷) (1993) [1578]. Shùxué Tōngguǐ (數學通軌) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
3. Newcomb, Simon (c1882), Logarithmic and other mathematical tables with examples of their use and hints on the art of computation, New York: Henry Holt and Company {{citation}}: Check date values in: |year= (help)
4. Suzuki, Hisao (鈴木 久男) (1982). "Chuugoku ni okeru shuzan kagen-hou 中国における珠算加減法". Kokushikan University School of Political Science and Economics (in Japanese). 57 (3). ISSN 0586-9749 – via Kokushikan. {{cite journal}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
5. Chen, Yifu (2018), "The Education of Abacus Addition in China and Japan Prior to the Early 20th Century", Computations and Computing Devices in Mathematics Education Before the Advent of Electronic Calculators, Springer Publishing, ISBN 978-3-319-73396-8 {{citation}}: Unknown parameter |editor1first= ignored (|editor-first1= suggested) (help); Unknown parameter |editor1last= ignored (|editor-last1= suggested) (help); Unknown parameter |editor2first= ignored (|editor-first2= suggested) (help); Unknown parameter |editor2last= ignored (|editor-last2= suggested) (help)
6. Chen, Yifu (2013). L’étude des Différents Modes de Déplacement des Boules du Boulier et de l’Invention de la Méthode de Multiplication Kongpan Qianchengfa et son Lien avec le Calcul Mental (PhD thesis) (in French). Université Paris-Diderot (Paris 7). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
7. Kwa Tak Ming (1920?), The Fundamental Operations in Bead Arithmetic, How to Use the Chinese Abacus (PDF), San Francisco: Service Supply Co. {{citation}}: Check date values in: |year= (help)

 T T T T T T T T 1

## Introduction

It is a mystery why traditional Chinese and Japanese abacuses had five beads in their lower deck as only four are required from the point of view of decimal numbers representation. As no extant ancient document seems to explain it, this mystery will probably last forever and we are limited to conjectures to try to understand its origin. In this line, we could think that, when they first appeared, fixed beads abacuses were conceived in image and likeness of counting rods, from which they were called to inherit every algorithm. With counting rods, the use of five rods to represent number five was compulsory in order to avoid the ambiguity between one and five, at least initially, when neither a representation for zero nor a checkerboard a la Japanese sangi were used. Furnishing the abacus with five lower beads allowed a parallel or similar manipulations of beads and rods, bringing some kind of hardware and software compatibility to fixed beads abacuses, in fact, the first Chinese books on suanpan also dealt with counting rods, so that both instruments were learned at the same time. We could also invoke a certain desire for compatibility between the abacus and rod numerals that, in one way or another, have been in use until modern times. So, for instance, one would like to change all 5’s to be represented by the five lower beads before writing down a result using rods numerals, in order to avoid very silly and catastrophic transcription mistakes.

Counting rods, by the way the most versatile and powerful abacus ever, had a flaw: it is extremely slow to manipulate. It is not a surprise that ancient Chinese mathematicians invented the multiplication table to speed up multiplication and that they also discovered the use of this multiplication table to also speed up division. Nor is it a surprise that they also discovered that, by using the abacus fifth bead, addition and subtraction operations could be somewhat simplified. They really had to be very sensitive to slowness.

In what follows a small set of rules for the use of the fifth bead is presented along with their rationale and scope of use. These rules are not explicitly stated in any of the classical works, but can be inferred from the addition and subtraction demonstrations present in them, (especially in the Panzhu Suanfa) as is done in Chen's thesis

## Some terms and notation

In what follows we will use these concepts and notation in reference to the use (or not) of the lower fifth bead.

• F: to denote a lower five (five lower beads activated) as opposed to:
• 5: upper five (one upper bead activated).
• T: ten on a rod (one upper bead and five lower beads activated). On 5+2 type abacus, it is also a lower ten as opposed to t an upper ten (two upper beads activated).
• Q: lower fifteen on a rod (two upper beads and five lower beads activated) as opposed to q upper fifteen (suspended upper bead on the 5+2, three upper beads activated on the 5+3).
• carry: this represents number 1 when it is to be added to a column as a carry from the right (addition).

• a1 Never use the 5th bead in addition except in the two cases that follow.
• a2 4 + carry = F
• a3 9 + carry = T

That is to say, when adding 1 to a rod you act as usual, for instance:

 A A A + 1 = 4 5

and

 A B A B B + 1 = 0 9 1 0

but when adding one as result of a carry you use the fifth lower bead:

 A B A B B + 5 = 4 6 F 1

text

 A B A B B + 5 = 9 6 T 1

You can see the above addition rules mentioned in a slightly different way in *Chen, Yifu (2018), "The Education of Abacus Addition in China and Japan Prior to the Early 20th Century", Computations and Computing Devices in Mathematics Education Before the Advent of Electronic Calculators, Springer Publishing, ISBN 978-3-319-73396-8 {{citation}}: Unknown parameter |editor1first= ignored (|editor-first1= suggested) (help); Unknown parameter |editor1last= ignored (|editor-last1= suggested) (help); Unknown parameter |editor2first= ignored (|editor-first2= suggested) (help); Unknown parameter |editor2last= ignored (|editor-last2= suggested) (help).

### The rationale behind

Rule a1 goal is simply to always leave an unused lower bead at our disposal in case the current column has to accept a future carry from the right, while rules a2 and a3 specify the use of the 5th bead in such a situation. Then, we can expect to obtain:

• a reduced number of finger movements because we avoid to deal with both upper and lower beads
• to avoid skipping rods and to reduce the left-right hand displacement span
• to avoid any “carry run” to the left (think of 99999+1=999T0 instead of 99999+1=100000)

The above advantages are automatically realized by using rules a2 and a3, but rule a1 is of a different nature. Rule a1 is a provision for the future, it will simplify things if a future carry actually falls on the current column (which happens about 50% of the time on average), but it will simplify nothing otherwise. Rule a1 is so a kind of a bet (subtraction rules below are also of the same nature).

### The scope of use

Rules a1 to a3 are for columns that can receive a carry, which excludes the rightmost column in normal (rightward) operation.

In inverse (leftward) operation, no column will receive a future carry from the right, so that rule a1 is out of scope and does not operate, but rules a2 and a3 should always be used. (This is mentioned because an ancient technique, now defunct, used leftward operation in alternation with normal operation to avoid long hand displacements. Not of general use but an extremely interesting exercise anyway).

Exceptionally, if you do know that some column will never receive a carry, you are also free of rule a1. (This seems a strange situation, but we need to introduce it to cope with the central part of the Test Drive below).

## Rules for subtraction

• s1 Always use lower fives (F) instead of upper fives (5). For instance: 7-2 = F
 A A A A - 2 = not 7 F 5
• s2 Never leave a cleared rod (0) if you can borrow from the adjacent left rod (but not from a farther one!), leave a T instead, i.e. 27-7 = 1T
 A B A B B - 7 = 2 7 1 T

should be preferred to 27-7 = 20

 A B A B B - 7 = 2 7 2 0

.

Remark: These two rules do not apply on rods where you are borrowing from, i.e. 112-7 = 10F (not TF)

 A B C A B C A B C ABC - 7 = not 1 1 2 1 0 F 0 T F
and 62-7 = 5F (not FF).
 A B A B A B AB - 7 = not 6 2 5 F F F

### The rationale behind

Both rules tend to leave activated lower beads at our disposal for the case we need to borrow from them in the future (it is like always holding small change in our pocket just in case), saving us some movements and/or wider or more complex hand displacements, such as borrowing from non-adjacent columns or skipping rods.

Is not automatically obtained, it is only fulfilled when we actually need to borrow from the present rod. This is similar to the case of addition rule a1.

### The scope of use

Once more, the rightmost column is outside the scope of these rules as we will never borrow from it.

Also, In leftward or inverse operation we will never borrow from the current column, so these rules do not apply (which may be seen as an additional reason to prefer rightward operation in normal use).

## Test drive

It was common in ancient books on the abacus to demonstrate addition and subtraction using the well-known exercise that consists of adding the number 123456789 nine times to a cleared abacus until the number 1111111101 is reached, and then erase it again by subtracting the same number nine times (This exercise seems to have the Chinese name: Jiǔ pán qīng 九盤清, meaning something like clearing the nine trays). You can find the sequence of intermediate results of the Panzhu Suanfa in this 1982 article by Hisao Suzuki (鈴木 久男): Chuugoku ni okeru shuzan kagen-hou 中国における珠算加減法 (Abacus addition and subtraction methods in China)[3]. This is a Japanese text (spiced up with some classical Chinese) that deals with addition and subtraction methods as they appear in various Chinese books from the 16th century. In pages 12-17, the Panzhu Suanfa version of the 123456789 exercise is graphically displayed on the upper series of 1:5 diagrams. The short Chinese phrases below each bar specify how the current digit was obtained (Table 1 in the Appendix A below serves a similar purpose but in a different and more convenient way for us).

Using the addition rules explained above, we should get the following sequence of results each time we complete the addition of 123456789 (see Table 1 for more details):

      000000000, 123456789, 246913F78, 36T36T367, 4938271F6,
617283945, 74073T734, 864197F23, 9876F4312,    ...


at this point, adding 123456789 once more results in 1111111101, but this number appears in the Panzhu Suanfa as:

      TTTTTTTT1


which cannot be obtained by the use of the above rules only. A similar situation occurs when repeating this exercise but starting with 999999999 instead of a cleared abacus (see Table 2), reaching 1TTTTTTTT0. This is why we introduced the last comment on the scope of addition rules above. It might be that, by inspection or intuition, we realize that using the 5th bead here does not generate any carry, so that we can overcome the a1 rule and proceed to this, somewhat theatrical, result ...

From here, by subtraction we should get:

      TTTTTTTT1, 9876F4312, 864197523, 740740734, 61728394F,
493827156, 36T370367, 246913578, 123456789, 000000000


As it can be seen here, few F’s and T’s appear on the intermediate results, but a few more appear in the middle of calculation (Table 1), being immediately converted to 4’s and 9’s by borrowing, which is the purpose for which they were introduced. The F’s and T’s remaining on the intermediate results are only the unused ones.

Of course, the rules for addition can also be directly used in multiplication and the rules for subtraction in division, roots, etc.

Additionally, if using traditional division method (see chapter: [[../../Division/Modern and traditional division; close relatives/|Modern and traditional division; close relatives]]) on the 2:5 or 3:5 abacus, we can introduce an additional rule:

• k1 Always use lower five’s, ten’s, and fifteen’s (F, T, Q) when adding to the remainder after application of the division rules.

This is so because, although we are adding to a rod, the next thing we will do is start subtracting from it (if the divisor has more than one digit). It is a kind of extension of the first rule for subtraction (s1). For instance, initiating 87÷98:

87÷98
Abacus Comment
ABCDEFG
87   98 Dividend AB, divisor FG
8Q   98 A: Rule 8/9>8+8
-64
886  98 etc.

Just after application of the division rule 8/9>8+8 we should have:

A B C D E F G 8 Q 0 0 0 9 8

By the way, you may sometimes find somewhat conflicting the use of the second rule for subtraction (s2) in Chinese division. For instance, 1167/32 = 36,46875

1167/32 = 36,46875
Abacus Comment
ABCDEFG
32 1167 1/3->3+1 rule
32 3267 -3*2=-6 in f, use 2nd subtraction rule
-6
32 31T7

Now, which rule should be used here? 1/3->3+1 or 2/3->6+2 ? In fact, we can use any of them and revise up as needed, but it is faster to realize that the remainder is actually 3207 so that the second Chinese rule is the appropriate one, so, simply change columns EF to 62 and continue.

Abacus Comment
ABCDEFG
32 3627
...

Finally, if you are using the traditional Chinese multiplication method or similar on the suanpan, you may face overflow on some columns, so that an additional rule:

• m1 [14] + carry = Q

can also be considered.

It is clear that the use of the 5th bead may reduce the number of bead or finger movements required in some calculations (Think of 99999 + 1 = 999T0 vs. 99999 + 1 = 100000). An estimate based on the 123456789 exercise and some of its derivatives (see the next chapter) leads to a reduction of 10% on average (counting simultaneous movements of upper and lower beads separately). This is a modest reduction, but the advantage of the 5th bead goes beyond simply reducing the number of finger movements, as it also reduces the number and/or the extent of other hand gestures required in calculations (hand displacement, changes of direction, skipping rods,...). As already stated in the [[../../Introduction/|introduction]] to this book, each gesture:

• as a physical process, takes a time to complete
• as governed by our brains, requires our attention, consuming (mental or biochemical) energy
• as done by humans (not machines), has a chance to be done in the wrong way, introducing mistakes

So, under this optic, we can expect that the use of the 5th bead will result in a somewhat faster, more relaxed and reliable calculation by reducing the total number of required gestures. It is not easy to measure this triple advantage using a single parameter.

Skipping columns, as Yifu Chen comments in his two works mentioned above, seems to have traditionally been viewed as something to be avoided as a possible source of errors. Without this concept the subtraction rule (s2) cannot be understood since it does not always lead to a reduction in the number of finger movements, but it always reduces the range of hand movement and the need to skip rods. Have you ever felt insecure with divisors or roots that contain embedded zeros? They force us to skip columns.

In any case, the advantage of using the fifth bead, although not negligible, is only modest, and each one must decide whether it is worth using it or not. After getting used to and becoming fluent in using the 5th bead, there is no better test of its efficiency than using a 4+1 abacus again and being sensitive to the amount of additional work required to complete tasks on it.

## Table 1: The 123456789 exercise step by step

ABCDEFGHI       ABCDEFGHI       ABCDEFGHI       ABCDEFGHI       ABCDEFGHI
---------       ---------       ---------       ---------       ---------
000000000       123456789       246913F78       36T36T367       4938271F6
100000000  A+1  223456789  A+1  346913F78  A+1  46T36T367  A+1  5938271F6  A+1
120000000  B+2  243456789  B+2  366913F78  B+2  48T36T367  B+2  6138271F6  B+2
123000000  C+3  246456789  C+3  369913F78  C+3  49336T367  C+3  6168271F6  C+3
123400000  D+4  246856789  D+4  36T313F78  D+4  49376T367  D+4  6172271F6  D+4
123450000  E+5  246906789  E+5  36T363F78  E+5  49381T367  E+5  6172771F6  E+5
123456000  F+6  246912789  F+6  36T369F78  F+6  493826367  F+6  6172831F6  F+6
123456700  G+7  246913489  G+7  36T36T278  G+7  493827067  G+7  6172838F6  G+7
123456780  H+8  246913F69  H+8  36T36T358  H+8  493827147  H+8  617283936  H+8
123456789  I+9  246913F78  I+9  36T36T367  I+9  4938271F6  I+9  617283945  I+9

ABCDEFGHI       ABCDEFGHI       ABCDEFGHI       ABCDEFGHI
---------       ---------       ---------       ---------
617283945       74073T734       864197F23       9876F4312
717283945  A+1  84073T734  A+1  964197F23  A+1  T876F4312  A+1
737283945  B+2  86073T734  B+2  984197F23  B+2  TT76F4312  B+2
740283945  C+3  86373T734  C+3  987197F23  C+3  TTT6F4312  C+3
740683945  D+4  86413T734  D+4  987597F23  D+4  TTTTF4312  D+4
740733945  E+5  86418T734  E+5  987647F23  E+5  TTTTT4312  E+5
740739945  F+6  864196734  F+6  9876F3F23  F+6  TTTTTT312  F+6
74073T645  G+7  864197434  G+7  9876F4223  G+7  TTTTTTT12  G+7
74073T725  H+8  864197F14  H+8  9876F4303  H+8  TTTTTTT92  H+8
74073T734  I+9  864197F23  I+9  9876F4312  I+9  TTTTTTTT1  I+9   

### Subtraction

ABCDEFGHI       ABCDEFGHI       ABCDEFGHI       ABCDEFGHI       ABCDEFGHI
---------       ---------       ---------       ---------       ---------
TTTTTTTT1       9876F4312       864197523       740740734       61728394F
9TTTTTTT1  A-1  8876F4312  A-1  764197523  A-1  640740734  A-1  F1728394F  A-1
98TTTTTT1  B-2  8676F4312  B-2  744197523  B-2  620740734  B-2  49728394F  B-2
987TTTTT1  C-3  8646F4312  C-3  741197523  C-3  617740734  C-3  49428394F  C-3
9876TTTT1  D-4  8642F4312  D-4  740797523  D-4  617340734  D-4  49388394F  D-4
9876FTTT1  E-5  8641T4312  E-5  740747523  E-5  617290734  E-5  49383394F  E-5
9876F4TT1  F-6  864198312  F-6  740741523  F-6  617284734  F-6  49382794F  F-6
9876F43T1  G-7  864197612  G-7  740740823  G-7  617283T34  G-7  49382724F  G-7
9876F4321  H-8  864197532  H-8  740740743  H-8  6172839F4  H-8  49382716F  H-8
9876F4312  I-9  864197523  I-9  740740734  I-9  61728394F  I-9  493827156  I-9

ABCDEFGHI       ABCDEFGHI       ABCDEFGHI       ABCDEFGHI
---------       ---------       ---------       ---------
493827156       36T370367       246913578       123456789
393827156  A-1  26T370367  A-1  146913578  A-1  023456789  A-1
373827156  B-2  24T370367  B-2  126913578  B-2  003456789  B-2
36T827156  C-3  247370367  C-3  123913578  C-3  000456789  C-3
36T427156  D-4  246970367  D-4  123F13578  D-4  000056789  D-4
36T377156  E-5  246920367  E-5  123463578  E-5  000006789  E-5
36T371156  F-6  246914367  F-6  123457578  F-6  000000789  F-6
36T370456  G-7  246913667  G-7  123456878  G-7  000000089  G-7
36T370376  H-8  246913587  H-8  123456798  H-8  000000009  H-8
36T370367  I-9  246913578  I-9  123456789  I-9  000000000  I-9    

## Table 2: The 123456789 exercise over a background

    0          1           2           3           4
000000000  0111111111  0222222222  0333333333  0444444444
123456789  02345678T0  0345678T11  045678T122  05678T1233
246913F78  0357T24689  046913F7T0  057T246911  0691357T22
36T36T367  0481481478  0592592F89  06T36T36T0  0814814811
4938271F6  0604938267  0715T49378  082715T489  09392715T0
617283945  0728394TF6  08394T6167  09F0617278  1061738389
74073T734  08F18F1845  09629629F6  1074073T67  118F18F178
864197F23  097F308634  1086419745  1197F2T8F6  1308641967
9876F4312  109876F423  1209876F34  1320987645  14320987F6
TTTTTTTT1  1222222212  1333333323  1444444434  1555FFFF45
9876F4312  1098765423  1209876534  132098764F  1432098756
864197523  097F308634  108641974F  1197F30856  1308641967
740740734  08F18F184F  0962962956  0T74074067  118F18F178
61728394F  072839F056  0839F06167  09F0617278  0T61728389
493827156  05T4938267  0716049378  0827160489  093827159T
36T370367  0481481478  0592592589  06T370369T  0814814811
246913578  0357T24689  046913579T  0F7T246911  0691358022
123456789  023456789T  0345678T11  04F678T122  0F678T1233
000000000  0111111111  0222222222  0333333333  0444444444

5          6           7           8           9
0555555555  0666666666  0777777777  0888888888  0999999999
0678T12344  078T1234F5  08T1234F66  0T1234F677  11234F6788
07T2469133  091357T244  0T246913F5  11357T2466  1246913F77
0925925922  1036T36T33  1148148144  12592592F5  136T36T366
1049382711  115T493822  12715T4933  1382715T44  14938271F5
11728394T0  128394T611  1394T61722  1F06172833  1617283944
1296296289  14073T73T0  1F18F18F11  1629629622  174073T733
14197F2T78  1530864189  164197F2T0  17F3086411  1864197F22
1543209867  1654320978  176F431T89  1876F431T0  19876F4311
16666666F6  1777777767  1888888878  1999999989  1TTTTTTTT0
1F43209867  16F4320978  176F432089  1876F4319T  19876F4311
14197F3078  1F30864189  164197529T  17F3086411  1864197522
1296296289  140740739T  1F18F18F11  1629629622  1740740733
117283949T  12839F0611  139F061722  14T6172833  1617283944
0T49382711  115T493822  1271604933  1382716044  149382715F
0925925922  0T36T37033  1148148144  125925925F  136T370366
07T2469133  0913580244  0T2469135F  11357T2466  1246913577
0678T12344  078T12345F  08T1234566  0T12345677  1123456788
0FFF55555F  0666666666  0777777777  0888888888  0999999999

## References

1. Chen, Yifu. "L'étude des différents modes de déplacement des boules du boulier et de l'invention de la méthode de multiplication Kongpan Qianchengfa et son lien avec le calcul mental". theses.fr. Retrieved 13 July 2021.
2. Xú Xīnlǔ (徐心魯) (1993) [1573]. Pánzhū Suànfǎ (盤珠算法) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
3. Suzuki, Hisao (1982). "Zhusuan addition and subtraction methods in China". Kokushikan University School of Political Science and Economics (in Japanese). 57 – via Kokushikan.

• Heffelfinger, Totton; Hinkka, Hannu (2011). "The 5 Earth Bead Advantage". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

## External resources

You can practice using the fifth bead online with Soroban Trainer (see chapter: [[../../Introduction#External resources|Introduction]]) using this file 123456789-5bead.sbk that you should download to your computer and then submit it to Soroban Trainer (It is a text file that you can inspect with any text editor and that you can safely download to your computer).

# Addition and subtraction/Extending the 123456789 exercise

 1 2 3 4 5 6 7 8 9

## Introduction

As we have seen in the previous chapter, the "123456789 exercise", consisting of adding that number nine times to a cleared abacus until reaching the number 1111111101 and then subtracting it nine times until the abacus is cleared again, has been used since ancient times to illustrate and practice addition and subtraction. It is a convenient exercise because:

• it is long enough to be a non-trivial exercise
• if we do not return to the initial value (zero) it is a sign that we have made a mistake
• we do not need a book or exercise sheet
• uses many of the elementary cases of addition or subtraction of a digit to/from another digit

but it also has a couple of drawbacks:

• it does not use all pairs of digits (ex. a 3 is never added to a 5)
• after repeating it several times, one begins to mechanically memorize the exercise, so that we are no longer practicing addition and subtraction

To avoid these two problems we can extend the exercise in several ways.

## Using a background

Instead of using a cleared abacus, we can fill 9 columns with a digit (111111111, 222222222, etc.), this multiplies by 10 the number of exercises at our disposal. Now we are sure to use all possible cases of addition and subtraction digit by digit while mechanical memorization becomes harder.

The following table contains the intermediate values for reference. Such values are traversed from top to bottom in addition and from bottom to top in subtraction.

Table 1: The 123456789 exercise with background
+1..9 0 1 2 3 4 +1..9
0 000000000 111111111 222222222 333333333 444444444 0
1 123456789 234567900 345679011 456790122 567901233 1
2 246913578 358024689 469135800 580246911 691358022 2
3 370370367 481481478 592592589 703703700 814814811 3
4 493827156 604938267 716049378 827160489 938271600 4
5 617283945 728395056 839506167 950617278 1061728389 5
6 740740734 851851845 962962956 1074074067 1185185178 6
7 864197523 975308634 1086419745 1197530856 1308641967 7
8 987654312 1098765423 1209876534 1320987645 1432098756 8
9 1111111101 1222222212 1333333323 1444444434 1555555545 9
Table 1: The 123456789 exercise with background (continuation)
+1..9 5 6 7 8 9 +1..9
0 555555555 666666666 777777777 888888888 999999999 0
1 679012344 790123455 901234566 1012345677 1123456788 1
2 802469133 913580244 1024691355 1135802466 1246913577 2
3 925925922 1037037033 1148148144 1259259255 1370370366 3
4 1049382711 1160493822 1271604933 1382716044 1493827155 4
5 1172839500 1283950611 1395061722 1506172833 1617283944 5
6 1296296289 1407407400 1518518511 1629629622 1740740733 6
7 1419753078 1530864189 1641975300 1753086411 1864197522 7
8 1543209867 1654320978 1765432089 1876543200 1987654311 8
9 1666666656 1777777767 1888888878 1999999989 2111111100 9

## The 987654321 exercise

Additionally, instead of using the number 123456789 we can think of using any permutations of these digits that we are able to comfortably remember; for example, 987654321, the only one we will consider here. This gives us 10 other independent exercises for addition and subtraction practice. The following table shows us the intermediate values of this new series of exercises using a background.

In total, we already have 20 different exercises.

Table 2: The 987654321 exercise with background starting with addition
+9..1 0 1 2 3 4 +9..1
0 000000000 111111111 222222222 333333333 444444444 0
1 987654321 1098765432 1209876543 1320987654 1432098765 1
2 1975308642 2086419753 2197530864 2308641975 2419753086 2
3 2962962963 3074074074 3185185185 3296296296 3407407407 3
4 3950617284 4061728395 4172839506 4283950617 4395061728 4
5 4938271605 5049382716 5160493827 5271604938 5382716049 5
6 5925925926 6037037037 6148148148 6259259259 6370370370 6
7 6913580247 7024691358 7135802469 7246913580 7358024691 7
8 7901234568 8012345679 8123456790 8234567901 8345679012 8
9 8888888889 9000000000 9111111111 9222222222 9333333333 9
Table 2: The 987654321 exercise with background starting with addition (continuation)
+9..1 5 6 7 8 9 +9..1
0 555555555 666666666 777777777 888888888 999999999 0
1 1543209876 1654320987 1765432098 1876543209 1987654320 1
2 2530864197 2641975308 2753086419 2864197530 2975308641 2
3 3518518518 3629629629 3740740740 3851851851 3962962962 3
4 4506172839 4617283950 4728395061 4839506172 4950617283 4
5 5493827160 5604938271 5716049382 5827160493 5938271604 5
6 6481481481 6592592592 6703703703 6814814814 6925925925 6
7 7469135802 7580246913 7691358024 7802469135 7913580246 7
8 8456790123 8567901234 8679012345 8790123456 8901234567 8
9 9444444444 9555555555 9666666666 9777777777 9888888888 9

## Starting with subtraction

If you still do not have enough with the 20 previous exercises, you can count on another 20 independent exercises just start by subtracting 123456789 or 987654321 from the background nine times, after which we will return the abacus to its original state by adding the number nine times. In doing so, sooner or later we will find negative numbers that we can handle on "the other side" of the abacus; that is, by borrowing 1 from a real or imaginary column located further to the left. Before ending the exercise, that borrowed 1 will be returned with a carry to its real or imaginary column, and we will be able to finish the exercise with the abacus in its original state.

Table 3: The 123456789 exercise with background starting with subtraction
-1..9 0 1 2 3 4 -1..9
0 000000000 111111111 222222222 333333333 444444444 0
1 9876543211 9987654322 98765433 209876544 320987655 1
2 9753086422 9864197533 9975308644 86419755 197530866 2
3 9629629633 9740740744 9851851855 9962962966 74074077 3
4 9506172844 9617283955 9728395066 9839506177 9950617288 4
5 9382716055 9493827166 9604938277 9716049388 9827160499 5
6 9259259266 9370370377 9481481488 9592592599 9703703710 6
7 9135802477 9246913588 9358024699 9469135810 9580246921 7
8 9012345688 9123456799 9234567910 9345679021 9456790132 8
9 8888888899 9000000010 9111111121 9222222232 9333333343 9
Table 3: The 123456789 exercise with background starting with subtraction (continuation)
-1..9 5 6 7 8 9 -1..9
0 555555555 666666666 777777777 888888888 999999999 0
1 432098766 543209877 654320988 765432099 876543210 1
2 308641977 419753088 530864199 641975310 753086421 2
3 185185188 296296299 407407410 518518521 629629632 3
4 61728399 172839510 283950621 395061732 506172843 4
5 9938271610 49382721 160493832 271604943 382716054 5
6 9814814821 9925925932 37037043 148148154 259259265 6
7 9691358032 9802469143 9913580254 24691365 135802476 7
8 9567901243 9679012354 9790123465 9901234576 12345687 8
9 9444444454 9555555565 9666666676 9777777787 9888888898 9

Table 4: The 987654321 exercise with background starting with subtraction
-9..1 0 1 2 3 4 -9..1
0 000000000 111111111 222222222 333333333 444444444 0
1 9012345679 9123456790 9234567901 9345679012 9456790123 1
2 8024691358 8135802469 8246913580 8358024691 8469135802 2
3 7037037037 7148148148 7259259259 7370370370 7481481481 3
4 6049382716 6160493827 6271604938 6382716049 6493827160 4
5 5061728395 5172839506 5283950617 5395061728 5506172839 5
6 4074074074 4185185185 4296296296 4407407407 4518518518 6
7 3086419753 3197530864 3308641975 3419753086 3530864197 7
8 2098765432 2209876543 2320987654 2432098765 2543209876 8
9 1111111111 1222222222 1333333333 1444444444 1555555555 9

Table 4: The 987654321 exercise with background starting with subtraction (continuation)
-9..1 5 6 7 8 9 -9..1
0 555555555 666666666 777777777 888888888 999999999 0
1 9567901234 9679012345 9790123456 9901234567 12345678 1
2 8580246913 8691358024 8802469135 8913580246 9024691357 2
3 7592592592 7703703703 7814814814 7925925925 8037037036 3
4 6604938271 6716049382 6827160493 6938271604 7049382715 4
5 5617283950 5728395061 5839506172 5950617283 6061728394 5
6 4629629629 4740740740 4851851851 4962962962 5074074073 6
7 3641975308 3753086419 3864197530 3975308641 4086419752 7
8 2654320987 2765432098 2876543209 2987654320 3098765431 8
9 1666666666 1777777777 1888888888 1999999999 2111111110 9

## Using the 5th lower bead

This is the most interesting proposal in the context of traditional methods. The forty exercises above can be performed using the lower 5th bead, as explained in detail in the previous chapter: Use of the 5th lower bead, which will allow you to master this traditional technique.

With this, we have a total of 80 exercises!

## Using alternate operation

And finally, why not? Even if only for the pleasure of overcoming a different difficulty, we can combine the previous exercises with an alternating direction of operation, from left to right and from right to left, as explained in the introductory chapter to [[../../Addition and subtraction#Reverse operation/|addition and subtraction]].

Example
Abacus Comment
ABCDEFGHIJ
Cleared abacus
+1
+2
+3
+4
+5
+6
+7
+8
+9
123456789 First step completed
+9
+8
+7
+6
+5
+4
+3
+2
+1
246913578 Second step completed
etc.

With this, you could go one step further in your understanding of bead mechanics.

## Conclusion

With the 160 exercises presented here, you no longer have an excuse, you can practice addition and subtraction for hours, without exercise sheets, while comfortably seated on your sofa with only your abacus resting on your knees.

This is a door to mastery!

# Division

## Introduction

Of the four fundamental arithmetic operations, division is probably the most difficult to learn and perform. Being basically a sequence of subtractions, there are a large number of algorithms or methods to carry it out and many of these methods have been used with the abacus[1][2]. Of these, two stand out for their efficiency and should be considered the main ones:

• The modern division method (MD), shojohou in Japanese, shāng chúfǎ in Chinese (商除法); the oldest of the two, its origin dates back to at least the 3rd to 5th centuries AD, as it is cited in the book: The Mathematical Classic of Master Sun (Sūnzǐ Suànjīng 孫子算經). If we call it modern it is because it is the one taught today because it is the most similar to the division with paper and pencil. This method of division is based on the use of the multiplication table. During the Edo period it was introduced to Japan by Momokawa Jihei[3], but it did not gain popularity[4] until the 20th century with the development of what we have been calling the Modern Method.
• The traditional division method (TD), kijohou in Japanese, guī chúfǎ in Chinese (帰除法), first described in the Mathematical Illustration (Suànxué Qǐméng, 算學啟蒙) by Zhū Shìjié 朱士傑 (1299)[5]. Its main peculiarity is that it uses a division table in addition to the multiplication table, which saves the mental effort of determining what figure of the quotient we have to try. In addition, we can design custom division tables for multi-digit dividers that save us the use of the multiplication table.

Both methods were first used in China with Counting rods.

In this Part of the book we deal primarily with the traditional method of division while assuming that the reader already has experience with the modern method of division.

## Chapters

### Modern and traditional division; close relatives

In this chapter we try to show how modern and traditional methods, apparently so different, are actually closely related, while trying to justify why this method was invented.

### Guide to traditional division (帰除法)

Here, we will see how to use the traditional method.

### Learning the division table

It contains some indications that may make it easier for you to memorize the division table.

### Dealing with overflow

How to cope with the traditional division arrangement (TDA) using different types of abacuses, especially the modern 4+1 and the traditional Japanese 5+1.

### Special division tables

Division tables can be coined for multi-digit divisors, allowing dividing by them without resorting to the multiplication table.

A basic set of examples to illustrate all of the above.

### Division by powers of two

Another traditional division method different from 帰除法 based in fractions; a form of division in situ.

## References

1. Suzuki, Hisao (鈴木 久男) (1980). "Chūgoku ni okeru josan-hō no kigen (1 ) 中国における除算法の起源（1)". Kokushikan University School of Political Science and Economics (in Japanese). 55 (2). ISSN 0586-9749 – via Kokushikan. {{cite journal}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
2. Suzuki, Hisao (鈴木 久男) (1981). "Chūgoku ni okeru josan-hō no kigen (2 ) 中国における除算法の起源（2)". Kokushikan University School of Political Science and Economics (in Japanese). 56 (1). ISSN 0586-9749 – via Kokushikan. {{cite journal}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
3. Momokawa, Jihei (百川治兵衛) (1645). Kamei Zan (亀井算) (in Japanese). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
4. Smith, David Eugene; Mikami, Yoshio (1914), A history of Japanese mathematics, Chicago: The Open court publishing company, p. 43-44
5. Zhū Shìjié 朱士傑 (1993) [1299]. Suànxué Qǐméng (算學啟蒙) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)

# Division/Modern and traditional division; close relatives

## Introduction

As explained in the previous chapter, there are two main methods of division used with the abacus: the modern division and the traditional division. The modern method of division (MD), shojohou in Japanese, shāng chúfǎ in Chinese (商除法) , is actually the oldest, dating back to approx. 200 CE and only makes use of the multiplication table. By comparison, the traditional method (TD), kijohou, guī chúfǎ (帰除法), is more recent but also very old, dating back to the times of counting rods, at least from the 13th century. This method makes use of both the multiplication table and a specific division table. TD has been the standard method studied with the abacus for at least 4 centuries[1][2], losing popularity in the 1930s. The reason for this is that modern abacus students already know how to divide with pencil and paper before embarking on the study of the abacus and, having a very tight study program or being very busy, it is not a question of spending time learning a new method of division and memorizing a new table, but of taking advantage of what is already known; MD is the closest thing to written long division that can be done on the abacus.

It would be difficult to say which of the two division methods is more efficient, Kojima[3] does not dare to say it, what does seem generally accepted is that the traditional method is more comfortable or relaxed since one does not have to think about anything, just follow the rules. From what follows, one can think that MD is somewhat more efficient (faster) than TD, one could say that while MD is for the sprinter, TD is for the Marathon runner; i.e. for those who have to spend many hours a day doing divisions…

At first glance it may seem that these two methods of division are very different from each other, we will show in what follows that the two methods are so similar and related that they can be considered close relatives (with MD being the older brother and TD the youngest) to the point that if you are already skilled in MD you are also skilled in TD!  … although you don't know it yet and you are still far from getting all the power of TD.

For this purpose, we will go back to older division methods in order to position MD and TD within the framework of the Chunking methods[4] (sometimes also called the partial quotients method or the hangman method) which will allow us to show the extreme similarity of both approaches. After this, we will delve into the hidden beauty of TD and understand why it simplifies life with the abacus. In what follows we assume that you already know all about the modern division with the abacus, how to revise up and down, etc. as explained, for instance, by Kojima.

## First methods

Let's take 1225÷35 = 35 as an example. There is no simpler way to proceed than by repeated subtraction and since 35 is greater than the first two digits of the dividend, we will start subtracting 35 from 122 using a column from the abacus as a counter.

1225÷35 = 35
Abacus Comment
ABCDEFGHI
35   1225 Start, counter in D,
35 1  875 subtract 35 from GH, add 1 to counter D,
35 2  525 subtract 35 from GH, add 1 to counter D,
35 3  175 subtract 35 from GH, add 1 to counter D,
35 31 140 subtract 35 from HI, add 1 to counter E,
35 32 105 subtract 35 from HI, add 1 to counter E,
35 33  70 subtract 35 from HI, add 1 to counter E,
35 34  35 subtract 35 from HI, add 1 to counter E,
35 35  00 subtract 35 from HI, add 1 to counter E.
35 35 No remainder. Done, quotient is 35!

That was easy but a little long. If we can easily double the divisor and retain it in memory, we can shorten the operation by subtracting one or two times the divisor chunks.

times chunks
1 35
2 70
1225÷35 = 35
Abacus Comment
ABCDEFGHI
35   1225 Start, counter in D,
35 2  525 subtract 70 from GH, add 2 to counter D,
35 3  175 subtract 35 from GH, add 1 to counter D,
35 32 105 subtract 70 from HI, add 2 to counter E,
35 34  35 subtract 70 from HI, add 2 to counter E,
35 35  00 subtract 35 from HI, add 1 to counter E.
35 35 No remainder. Done, quotient is 35!

Or even better if we can build a table like the one below by doubling the divisor three times[5]:

times chunks
1 35
2 70
4 140
8 280
1225÷35 = 35
Abacus Comment
ABCDEFGHI
35   1225 Start, counter in D,
35 2  525 subtract 70 from GH, add 2 to counter D,
35 3  175 subtract 35 from GH, add 1 to counter D,
35 34  35 subtract 140 from HI, add 4 to counter E,
35 35   0 subtract 35 from HI, add 1 to counter E.
35 35 No remainder. Done, quotient is 35!

which is somewhat shorter and, clearly, nothing could be faster than having a complete multiplication table of the divisor

Multiplication table by 35
times chunks
1 35
2 70
3 105
4 140
5 175
6 210
7 245
8 280
9 315

then

1225÷35 = 35
Abacus Comment
ABCDEFGHI
35   1225 Start, counter in D,
35 3  175 subtract 105 from GH, add 3 to counter D,
35 35  00 subtract 175 from HI, add 5 to counter E.
35 35 No remainder. Done, quotient is 35!

There is no doubt, this is an optimal division method, nothing can be faster and more comfortable ... once we have a chunk table like the one above. But calculating the chunk table is time consuming and requires paper and pencil to write it and this extra work would only be justified if we have a large number of divisions to do with the same common divisor.

In 1617 John Napier, the father of logarithms, presented his invention to alleviate this problem consisting of a series of rods, known as Napier's Bones, with the one-digit multiplication table written on them and that could be combined to get the multiplication table of any number. For example, in our case

 1 35 2 70 3 105 4 140 5 175 6 210 7 245 8 280 9 315

There is no doubt that such an invention spread to the East and was used in conjunction with the abacus, but this use must be considered as exceptional; not everyone had Napier bones close at hand. Another tool is needed and that tool is the multiplication table learned by heart.

It should be noted that the above procedures do not exhaust the possibilities of the chunking methods. If you read The Definitive Higher Math Guide on Integer Long Division[4] article, you will be amazed at the variety of division methods that can be performed. Both MD and TD used in the abacus belong to this category, as we are going to see.

## Modern Division (商除法)

One of the key points of learning abacus is to be aware that this instrument allows us to correct some things very quickly and without leaving traces and this is specially useful in the case of division. So if we have to divide 634263÷79283, instead of busting our brain trying to find the correct quotient figure, we simply choose an approximate provisional or interim figure by simplifying the original problem to 63÷7 and test it by trying to subtract the chunk (interim quotient digit)✕79283 from the dividend; one of the following will occur:

• The interim quotient digit is correct
• It is excessive and we must revise it down
• It is insufficient and we must revise it up

Let's see it applied to our previous example. Instead of directly trying to solve 1225÷35 we simplify and try to solve 12÷3 using the memorized multiplication by 3  table.

 3×1 3 3×2 6 3×3 9 3×4 12 3×5 15 3×6 18 3×7 21 3×8 24 3×9 27
1225÷35 = 35
Abacus Comment
ABCDEFGHI
35   1225 12÷3↦4 from the table above as 3×4=12
+4 enter interim quotient in E
35  41225 Now try to subtract the chunk 4✕35 from FGH,
-12 first 4✕3 from FG
35  40025 then 4✕5 from GH
-20 Cannot subtract!
-1 Revising down interim quotient digit
35  30025
+3 return the excess subtracted from FG
35  30325
-15 continue normally, subtract 3✕5 from GH
35  3 175 17÷3↦5 from the table above as 3×5=15
+5 enter interim quotient in F
35  35175 Try to subtract chunk 5✕35 from GHI
-15 first 5✕3 from GH
35  35025
-25 then 5✕5 from HI
35  35 No remainder, done! 1225÷35 = 35

Instead of directly trying to solve the original problem 1225÷35 or the approximation used in MD 12÷3, we simplify still more and try to solve 10÷3; that is, we use a cruder approach to the original problem by ignoring the second digit of the dividend, so we must prepare to revise the interim quotient more frequently. By this change of focus from 12÷3 to 10÷3 we are adopting the philosophy of TD; it is only a slight variation of the chunking technique used in MD. This is why we can consider both division mechanisms as close relatives, members of the chunking methods family of division algorithms… and this is also why it can be said that if you are already proficient in modern division you are also already proficient in traditional division! but let us follow...

Continuing with our example

1225÷35
Abacus Comment
ABCDEFGHI
35   1225 10÷3↦3 from multiplication table
+3 enter interim quotient in E
35  31225 Try to subtract chunk 3✕35 from FGH,
-09 first 3✕3 from FG
35  3 325
-15 then 3✕5 from GH
35  3 175 ok.
35  3 175 10÷3↦3
+3 enter interim quotient in F
35  33175 Try to subtract 3✕35 from GHI,
-09 first 3✕3 from GH
35  33 85
-15 then 3✕5 from HI
35  33 70 remainder greater than divisor (35)
+1-35 Revising up
35  34 35 remainder equal to divisor (35)
+1-35 Revising up again
35  35 No remainder, done! 1225÷35 = 35

Note that MD and TD, as explained so far, can be freely intermixed during the same division problem. This is an interesting and recommended exercise that allows you to compare both strategies side by side.

TS uses a simpler and  lower approach to the original problem than MD, so that we can foresee some pros and cons

• Pros
• Some may consider this approach simpler
• It will be necessary to revise down less frequently (revising down is usually more difficult and prone to mistakes than revising up)
• Cons
• We need to revise the interim quotient more frequently, which is an efficiency issue.

The previous two pros probably played a role in the development of the sophisticated technique we know as traditional division, but understanding why it was the preferred method for centuries, despite the above con, requires reflecting on the origin of the mental effort made during division and discovering the hidden beauty of TD.

## The source of mental effort

When we learn the multiplication table we memorize a sequence of phrases like:

“nine times nine , eighty-one”
“nine times eight, seventy-two”
...

The order in which these phrases are learned can vary, but the structure of the phrases is similar in all languages, at least it is in Chinese and Japanese. It consists of a label that contains the two factors to be multiplied followed by the product. As soon as we think of the label, it, acting as an invocation, calls to our consciousness the value of the product. Let us represent it in the following way (read ➡ as the invocation):

Language Label Product nine times nine ➡ eighty-one 九九 ➡ 八十一 くく ➡ はちじゅういち 9✕9 ➡ 81

How do we use this multiplication table during division? Let's think about our example above using shojohou or modern division method: 17÷3↦5, from the multiplication by three table we need the largest product that can be subtracted from 17. We need to scan in our memory (represented by ⤷) at least a few lines of said table and for each product rescued, see if it is less than 17 and choose the maximum of those less than 17. A complicated process that can be represented as:

 3✕1 ➡ 3 3✕2 ➡ 6 ⤷ 3✕3 ➡ 9 yes ⤷ 3✕4 ➡ 12 yes ⤷ 3✕5 ➡ 15 yes select this one! ⤷ 3✕6 ➡ 18 no 3✕7 ➡ 21 3✕8 ➡ 24 3✕9 ➡ 27

This process is time and energy consuming. Computer specialists might find a similarity between this process and searching a relational database table on a non-indexed column; the inefficiency of such a search is well known. Creating a new index or key for that table based on the column and the search criteria can improve things drastically. Can we do something similar in our case to make the division more comfortable?

## Indexing the multiplication table (division table)

To do something similar to indexing the multiplication table in terms of the products to facilitate the search, we should memorize new phrases that contain those products as labels; that is, phrases that begin with them; for instance:

Label Quotient
3/3 1
6/3 2
9/3 3
12/3 4
15/3 5
18/3 6
21/3 7
24/3 8
27/3 9

That is, we have to memorize a division table, which is a hard work. Also think that the table above is not optimal in the sense that much of the numbers between 1 and 29 are missing; perhaps we should memorize a table of the following style instead:

Label Quotient Remainder
1/3 0 1
2/3 0 2
3/3 1 0
4/3 1 1
5/3 1 2
27/3 9 0
28/3 9 1
29/3 9 2

where the third column contains the remainders of the euclidean division. You will probably agree that memorizing such a table is out of the reach of ordinary humans (think of the table for 9!).

## The hidden beauty of traditional division

If we dedicate a lifetime to dividing with the abacus using MD method we would end up facing all elementary divisions of the type ab÷c where where a, b and c are digits and ab < c0, about 360 in total. However, if we were to use TD, we would be faced with all elemental divisions of the type a0÷c or (10✕a)÷c, only 36 in total! ... and this makes the memorization of a division table viable. In fact, to divide by 3 it is enough to memorize:

Label Quotient Remainder
10/3 3 1
20/3 6 2

or, in a more compact symbolic form

Rule
1/3 > 3+1
2/3 > 6+2

that we can use directly to solve our example without any thinking by simply choosing the figure suggested by the rule as the interim quotient.

1225÷35 = 35
Abacus Comment
ABCDEFGHI
35   1225 Use rule 1/3 > 3+1
+3 enter interim quotient in E
35  31225 Try to subtract chunk 3✕35 from FGH,
-09 first 3✕3 from FG
35  3 325
-15 then 3✕5 from GH
35  3 175 ok.
35  3 175 Use rule 1/3 > 3+1
+3 enter interim quotient in F
35  33175 Try to subtract 3✕35 from GHI,
-09 first 3✕3 from GH
35  33 85
-15 then 3✕5 from HI
35  33 70 remainder greater than divisor (35)
+1-35 Revising up
35  34 35 remainder equal to divisor (35)
+1-35 Revising up again
35  35 No remainder, done! 1225÷35 = 35

but we have not yet made use of the remainder that appears in the rules after the plus sign; that and other issues will be covered in the next chapter.

## The division table

Let's conclude by offering the complete division table used in TD. All elements are obtained from a0÷c terms by euclidean division.

 1/9>1+1 2/9>2+2 3/9>3+3 4/9>4+4 5/9>5+5 6/9>6+6 7/9>7+7 8/9>8+8 9/9>9+9 1/8>1+2 2/8>2+4 3/8>3+6 4/8>5+0 5/8>6+2 6/8>7+4 7/8>8+6 8/8>9+8 1/7>1+3 2/7>2+6 3/7>4+2 4/7>5+5 5/7>7+1 6/7>8+4 7/7>9+7 1/6>1+4 2/6>3+2 3/6>5+0 4/6>6+4 5/6>8+2 6/6>9+6 1/5>2+0 2/5>4+0 3/5>6+0 4/5>8+0 5/5>9+5 1/4>2+2 2/4>5+0 3/4>7+2 4/4>9+4 1/3>3+1 2/3>6+2 3/3>9+3 1/2>5+0 2/2>9+2 1/1>9+1

## References

1. Xú Xīnlǔ (徐心魯) (1993) [1573]. Pánzhū Suànfǎ (盤珠算法) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
2. Shinoda, Shosaku (篠田正作) (1895). Jitsuyo Sanjutsu (実用算術) (in Japanese). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
3. Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
4. a b "The Definitive Higher Math Guide on Integer Long Division (and Its Variants)". Math Vault. Archived from the original on May 14, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help); Unknown parameter |name= ignored (help)
5. Wilson, Jeff. "Long Division Teaching Aid, "Double Division"". Double Division. Archived from the original on March 02, 2021. {{cite web}}: Check date values in: |archivedate= (help); Text "year 2005" ignored (help); Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

# Division/Guide to traditional division (帰除法)

## Introduction

Traditional division method (TD), kijohou, guī chúfǎ (帰除法), is one of the two main methods of division used with the abacus. This method makes use of both the multiplication table and a specific division table and has been the standard method studied with the abacus for at least 4 centuries, losing popularity in the 1930s. As a digit-by-digit or slow division algorithm, has been introduced in the previous chapter, where its special characteristic is revealed: it does not require thinking but only following some rules. This document is an introduction to its use on the abacus and it is assumed that the reader is already proficient in the modern division (MD) method.

## The division table

In the previous chapter Modern and traditional division; close relatives, the following division table has been introduced

 1/9>1+1 2/9>2+2 3/9>3+3 4/9>4+4 5/9>5+5 6/9>6+6 7/9>7+7 8/9>8+8 9/9>9+9 1/8>1+2 2/8>2+4 3/8>3+6 4/8>5+0 5/8>6+2 6/8>7+4 7/8>8+6 8/8>9+8 1/7>1+3 2/7>2+6 3/7>4+2 4/7>5+5 5/7>7+1 6/7>8+4 7/7>9+7 1/6>1+4 2/6>3+2 3/6>5+0 4/6>6+4 5/6>8+2 6/6>9+6 1/5>2+0 2/5>4+0 3/5>6+0 4/5>8+0 5/5>9+5 1/4>2+2 2/4>5+0 3/4>7+2 4/4>9+4 1/3>3+1 2/3>6+2 3/3>9+3 1/2>5+0 2/2>9+2 1/1>9+1

where in each cell the result of the Euclidean division ${\displaystyle (10\times a)/b=q,r}$  (${\displaystyle q}$ : quotient, ${\displaystyle r}$ : remainder, ${\displaystyle a,b}$  digits from 1 to 9) is expressed in the form ${\displaystyle a/b>q+r}$  for reasons that we will see below. This means that the following hold:

${\displaystyle 10a=q\cdot b+r}$

The table has three zones corresponding to the following: If the divisor has n figures and we compare it with the first n digits (from the left) of the dividend, with added trailing  zeros if necessary, three cases can occur:

1. the dividend is greater than or equal to the divisor (ex. ${\displaystyle 770/689}$ )
2. the dividend is less than the divisor and the first digit of the divisor is equal to the first digit of the dividend (ex. ${\displaystyle 670/689}$ )
3. the dividend is less than the divisor and the first digit of the divisor is greater than the first digit of the dividend (ex. ${\displaystyle 570/689}$ )

The blank cells below the diagonal of the division table above correspond to case 1. They could be filled in the style of the tables that can be seen elsewhere[1], but we leave them empty for simplicity. If during the division we fall into this zone, we will proceed, for now, simply by revising up the previous quotient digit as we will see in the examples that follow.

The diagonal elements (in gray) correspond to the case 2 and can only occur if the divisor has at least two digits.

Finally, the other non-diagonal elements correspond to the case 3, which can be considered the most important to study.

There is no doubt that memorizing the division table takes time and effort and that you want to know if the traditional method of division is right for you before investing so much time and effort. Fortunately, the division by nine, five, and two tables are remarkably simple and can be memorized almost instantly (see below), as well as diagonal elements for multi digit divisors. This means that we can learn this traditional technique using divisors that start with only 9, 5 or 2 without much effort and thus be able to decide whether it is worth spending time learning the whole table or not. In what follows we will use examples based on such divisors.

Easy-to-memorize division rules
Diagonal Divide by 9 Divide by 5 Divide by 2
1/1>9+1 1/9>1+1 1/5>2+0 1/2>5+0
2/2>9+2 2/9>2+2 2/5>4+0
3/3>9+3 3/9>3+3 3/5>6+0
4/4>9+4 4/9>4+4 4/5>8+0
5/5>9+5 5/9>5+5
6/6>9+6 6/9>6+6
7/7>9+7 7/9>7+7
8/8>9+8 8/9>8+8
9/9>9+9

## Why do the division rules include remainders?

Suppose we are going to divide 35 by 9, the 3/9>3+3 rule tells us that we must use 3 as an interim quotient and the next step will be to subtract the chunk 3✕9=27 from 35, leaving a remainder of 8. If we also memorize the remainders, we can save this multiplication step as follows: we cancel, clear or erase the first digit of the dividend, in this case 3, then we add the remainder (3) to the next figure (5) of the dividend. In this way, we obtain the same result but without using the multiplication table. With one-digit divisors we will never have to resort to the multiplication table, and in the case of divisors with several figures, proceeding in the same way, we will save one of the necessary multiplications. We will see it on the abacus below, but first we need a few words about how we are going to arrange the division on the abacus.

## Modern Division Arrangement (MDA)

While using MDA you can use the same rules you already know about the unit rod if you need them.

Let us see the 35÷9 case  from the above section, first without using the (rule) remainders

35÷9 without using the (rule) remainders
Abacus Comment
ABCDEFGH
9     35 Divisor in A, dividend in GH, rule: 3/9>3+3
+3 enter quotient 3 in E
9    335
-27 subtract chunk 3✕9=27 from GH
9    3 8 new remainder/dividend in H
... ...

And now using the remainders

35÷9 using the (rule) remainders
Abacus Comment
ABCDEFGH
9     35 Divisor in A, dividend in GH, rule: 3/9>3+3
+3 enter quotient 3 in E
9    335
-3 clear first dividend digit in G
9    3 5
9     +3 add remainder 3 to H
9    3 8 new remainder/dividend in H
... ...

That is:

When using MDA, the rule a/b>q+r must be read: “write q as interim quotient digit to the left, clear a and add r to the right”

## One digit divisors

The number 123456789 has traditionally been used to demonstrate the use of multiplication and division tables in ancient Chinese[3] and Japanese works[4][5]. Here we will use it with the “easy divisors” 9, 5 and 2.

### Example 123456789÷9=13717421

123456789÷9=13717421
Abacus Comment
ABCDEFGHIJ (Divisor not indicated)
123456789 Rule 1/9>1+1
+1 enter quotient 1 into A
-1 clear B
1 33456789 Rule 3/9>3+3
13 6456789 Rule 6/9>6+6
1361056789
+1-9 revising up
137 156789 Rule 1/9>1+1
1371 66789 Rule 6/9>6+6
1371612789
+1-9 revising up
13717 3789 Rule 3/9>3+3
1371731089
+1-9 revising up
137174 189 Rule 1/9>1+1
1371741 99
+1-9 revising up
1371742  9
+1-9 revising up
13717421 Done!

### Example 123456789÷5=24691357.8

123456789÷5=24691357.8
Abacus Comment
ABCDEFGHIJ (Divisor not indicated)
123456789 Rule 1/5>2+0
2 23456789 Rule 2/5>4+0
24 3456789 Rule 3/5>6+0
246 456789 Rule 4/5>8+0
2468 56789
+1-5 revising up
2469  6789
+1-5 revising up
24691 1789 Rule 1/5>2+0
246912 789
+1-5 revising up
246913 289 Rule 2/5>4+0
2469134 89
+1-5 revising up
2469135 39 Rule 3/5>6+0
24691356 9
+1-5 revising up
24691357 4 Rule 3/5>6+0
246913578 Done!

### Example 123456789÷2=61728394.5

123456789÷2=61728394.5
Abacus Comment
ABCDEFGHIJ (Divisor not indicated)
123456789 Rule 1/2>5+0
5 23456789
+1-2 revising up
6  3456789
+1-2 revising up
61 1456789 Rule 1/2>5+0
615 456789
+2-4 revising up twice
617  56789
+2-4 revising up twice
6172 16789 Rule 1/2>5+0
61725 6789
+3-6 revising up three times
61728  789
+3-6 revising up twice
617283 189 Rule 1/2>5+0
6172835 89
+4-8 revising up four times
6172839  9
+4-8 revising up four times
61728394 1 Rule 1/2>5+0
617283945 Done!

## Multi Digit divisors

Consider, for example, ${\displaystyle 359936/9728=37}$ , in this case it is convenient to think of the divisor as made up of a divider, the first digit, followed by a multiplier, the rest of the digits of the divisor, that is, ${\displaystyle 9728=dmmm}$ , where ${\displaystyle d}$  is the divider (9) and ${\displaystyle mmm}$  is the multiplier (728). The Chinese and Japanese names for this division method (帰除 Guīchú in Chinese, 帰除法 Kijohou in Japanese) refer to this: 帰, Guī, Ki is the header and 除, chú, jo is the multiplier[6].

In this case, the way to act is as follows:

1. First we consider only the divider ${\displaystyle d}$  and do exactly the same as in the case of the single digit divisor i.e. we follow the division rule: get the interim quotient ${\displaystyle q}$  and add the remainder (from the rule) to the adjacent column
2. Then we subtract the chunk ${\displaystyle q\times {\text{multiplier}}}$  from the remainder if we can; otherwise we have to revise down ${\displaystyle q}$  and restore ${\displaystyle d}$  to the remainder using the following rules:
Rules to revise down (two-digit divisors)
While dividing by Revise q to Add to remainder
1 q-1 +1
2 q-1 +2
3 q-1 +3
4 q-1 +4
5 q-1 +5
6 q-1 +6
7 q-1 +7
8 q-1 +8
9 q-1 +9

These rules are for two-digit divisors, for divisors with more digits things may be more complicated, as in MD (see example ${\displaystyle 23712/5928}$  below). Let us see the above case

### Example 359936÷9728=37

359936÷9728=37
Abacus Comment
ABCDEFGHIJKLM
9728   359936 Rule 3/9>3+3
9728  3 89936 enter 3 to G, clear H and add 3 to I
-2184 subtract chunk 3✕multiplier 3✕728=2184a from I-L
9728  3 68096 Rule 6/9>6+6
9728  3614096 enter 6 to H, clear I and add 6 to J
-4368 subtract chunk 6✕multiplier 6✕728=4368 from J-M
9728  36 9728 revising up
+1-9728
9728  37 Done!

Note: ^a This is an abbreviated notation meaning that 3✕7, 3✕2 and 3✕8 have to be subtracted from IJ, JK, and KL respectively.

### Example 235÷59=3.98…

235÷59=3.98…
Abacus Comment
ABCDEFGHIJ
59   235 Rule 2/5>4+0
59  4 35 enter 4 to E, clear F and add 0 to G
-36 cannot subtract chunk 4✕multiplier 4✕9=36 from GH!
-1+5 revise down following above rules
59  3 85
-27 subtract chunk 3✕multiplier 3✕9=27 from GH
59  3 58 Rule 5/5>9+5
59  3913 enter 9 to F, clear G and add 5 to H
-81 subtract chunk 9✕multiplier 9✕9=81 from HI
59  39 49 Rule 4/5>8+0
... etc.

### Example 23711÷5928=3,9998…

Caption text
Abacus Comment
ABCDEFGHIJKLMN
5928   23711 Rule 2/5>4+0
5928  4 3711 enter 4 to G, clear H and add 0 to I
-36 subtract 4✕9=36 from IJ
5928  4  111
-8 subtract 4✕2=8 from JK
5928  4   31
-32 cannot subtract 4✕8=32 from KL!
-1+592 revise down and restore the subtracted excess to IJK
5928  3 5951
-24 continue normally, subtract 3✕8=24 from KL
5928  3 5927 Rule 5/5>9+5
... etc.

As commented above, there are two basic ways of arranging general division problems. Let us see them side by side:

• Modern division arrangement (MDA), as explained by Kojima[2],
MDA 25÷5=5
Abacus Comment
ABCDEF
5   25 Dividend starting in E
5  5 After division quotient begins in D

• Traditional division arrangement (TDA), as used in ancient books since the times of counting rods[7] to the first part of the 20th century[8],
TDA 25÷5=5
Abacus Comment
ABCDEF
5   25 25÷5=5 Dividend starting in E
5   5 After division quotient begins in E

So far we have used MDA with the traditional division without any problem. TDA, however, is problematic with any division method, the traditional one included. This troublesome nature is due to a collision between the divisor and the dividend/remainder that occurs frequently (that is, both require the simultaneous use of the same column), and special techniques or abaci are needed to deal with this collision. Despite this, the TDA has been used for centuries in conjunction with the traditional method of division, at least since the 13th century, while the MDA has been shelved until modern times. It is clear that certain advantages can be recognized to TDA, but it is not so clear that they are enough to justify its historical use:

• It uses one rod less
• Result does not displace too much to the left as in MDA, which is of interest in the case of chained operations. This and the above points makes TDA more suitable to abacuses with a small number of rods, like the traditional 13-rod suanpan/soroban.
• It saves some finger movements; for instance, in the operation 6231÷93=67 using traditional (chinese) division, I count 14 finger movements with TDA versus 24 with MDA.
• Hand displacements are shorter.
• It is less prone to errors as less rods are skipped.

The way to avoid the mentioned collision is to accept that the first column of the dividend/remainder, after the application of Chinese division rules, can overflow and temporarily accept a value greater than 9 (up to 18), while providing some mechanism to deal with such an overflow. Interestingly enough, it seems that no ancient text explains how to do the latter, but we will do it in chapter: Dealing with overflow!.

In the case of a 5+2 or 5+3 abacus we can use the additional upper bead(s) to represent values from 10 to 20, using the suspended bead (懸珠 xuán zhū in Chinese, kenshu in Japanese) in the 5+2 case .

The third or suspended bead is expected to be used only in about 1% of cases, which justifies the adoption of the 5+2 model as standard instead of the 5+3. (If you are interested in using TDA on any abacus, head over to the Dealing with overflow chapter to see how)

When using TDA, the rule a/b>q+r must be read: “change a into q as interim quotient digit and add r to the right”

For examples of TD using TDA, refer to the Traditional division examples chapter.

## About the efficiency of TD

As you can see in the examples with single digit divisors, TD efficiency deteriorates as the divisor starts with lower figures in the sense that we have to revise up more frequently. We can say that the efficiency is zero when the divisor starts with 1; in fact, we don't even have division rules except 1/1>9+1 (which is statistically excessive, see chapter: Learning the division table). For this last case, the trick is to divide by 2 in situ (chapter: Division by powers of two) both divisor and dividend, which is very fast, and proceed to divide both results normally; now the divisor begins with a digit between 5 and 9. for example: ${\displaystyle 128/16}$

Caption text
Abacus Comment
ABCDEFGHI
16    128 Divide in situ by 2
8     64 Rule 6/8>7+4
8    7 8
+1-8 revising up
8    8 Done!

${\displaystyle 128/16=(128/2)/(16/2)=64/8=8}$

In other cases, our intuition and experience with MD could help us.

This lower efficiency of TD compared to MD is the price to pay to save us the mental work of deducting the interim quotient figure that we have to try.

## References

1. "割り算九九". Japanese Wikipedia. {{cite web}}: Unknown parameter |Language= ignored (|language= suggested) (help); Unknown parameter |accesdate= ignored (|access-date= suggested) (help); Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
2. a b Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
3. Xú Xīnlǔ (徐心魯) (1993) [1573]. Pánzhū Suànfǎ (盤珠算法) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
4. Yoshida, Mitsuyoshi (吉田光由) (1634). Jinkoki (塵劫記) (in Japanese). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
5. Shinoda, Shosaku (篠田正作) (1895). Jitsuyo Sanjutsu (実用算術) (in Japanese). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
6. Lisheng Feng (2020), "Traditional Chinese Calculation Method with Abacus", in Jueming Hua; Lisheng Feng (eds.), Thirty Great Inventions of China, Jointly published by Springer Publishing and Elephant Press Co., Ltd, ISBN 978-981-15-6525-0
7. Zhū Shìjié 朱士傑 (1993) [1299]. Suànxué Qǐméng (算學啟蒙) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
8. Kwa Tak Ming (1922), The Fundamental Operations in Bead Arithmetic, How to Use the Chinese Abacus (PDF), San Francisco: Service Supply Co.

# Division/Learning the division table

## Memorization of the division table.

The division table contains 45 rules, including the 9 diagonal elements for multi-digit divisors.

 1/9>1+1 2/9>2+2 3/9>3+3 4/9>4+4 5/9>5+5 6/9>6+6 7/9>7+7 8/9>8+8 9/9>9+9 1/8>1+2 2/8>2+4 3/8>3+6 4/8>5+0 5/8>6+2 6/8>7+4 7/8>8+6 8/8>9+8 1/7>1+3 2/7>2+6 3/7>4+2 4/7>5+5 5/7>7+1 6/7>8+4 7/7>9+7 1/6>1+4 2/6>3+2 3/6>5+0 4/6>6+4 5/6>8+2 6/6>9+6 1/5>2+0 2/5>4+0 3/5>6+0 4/5>8+0 5/5>9+5 1/4>2+2 2/4>5+0 3/4>7+2 4/4>9+4 1/3>3+1 2/3>6+2 3/3>9+3 1/2>5+0 2/2>9+2 1/1>9+1

The same number of independent elements that we find in the multiplication table (given the commutativity of this operation) whose memorization was one of the feats of our childhood in school. Memorizing the division table is therefore a similar task to learning the multiplication table.

These rules:

• From an operational point of view, these rules should be read or interpreted slightly differently depending on whether we use the traditional (TDA) or the modern (MDA) division arrangement.
• when using MDA, the rule a/b>q+r must be read: “write q as interim quotient digit to the left, clear a and add r to the right”
• When using TDA, the rule a/b>q+r must be read: “change a into q as interim quotient digit and add r to the right”
• From a theoretical point of view, each rule expresses the result of a Euclidean division: ${\textstyle (10\times a)/b=q,r}$  (${\displaystyle q}$ : quotient, ${\displaystyle r}$ : remainder, ${\displaystyle a,b}$  digits from 1 to 9) or, equivalently ${\textstyle 10a=q\cdot b+r}$

If we think about this last point, in fact there is no need to memorize the division rules since we can obtain them in situ, when we need them, by a simple mental process. But then we would be making a mental effort similar to that required with the modern method of division and we would be moving away from the philosophy of the traditional method. There is no doubt, the efficiency and goodness of the traditional method is only achieved by memorizing the rules and we should only resort to the aforementioned mental process during the learning phase, when some rule resists coming to memory.

Fortunately, a series of patterns that appear in the division table come to our aid making it easier for us to learn it, leaving only 14 hard rules out of a total of 45.

## Easy rules

In the chapter: Guide to traditional division (帰 除法) we already mentioned that the division rules by 9, 5 and 2, as well as the diagonal rules, have a particularly simple structure that allows almost immediate memorization.

Easy rules
Diagonal Divide by 9 Divide by 5 Divide by 2
1/1>9+1 1/9>1+1 1/5>2+0 1/2>5+0
2/2>9+2 2/9>2+2 2/5>4+0
3/3>9+3 3/9>3+3 3/5>6+0
4/4>9+4 4/9>4+4 4/5>8+0
5/5>9+5 5/9>5+5
6/6>9+6 6/9>6+6
7/7>9+7 7/9>7+7
8/8>9+8 8/9>8+8
9/9>9+9

For this reason, the examples presented in that chapter only made use of divisors starting with 2,5 and 9. If you practice several examples with such divisors, it will not be difficult for you to memorize these 22 rules (almost half of the total!); which is a drastic reduction in the work to be done and not the only one.

## Division by 8

Of the remaining rules, the division by 8 series is the longest but not the most difficult, since it has an internal structure:

 1/8>1+2 5/8>6+2 2/8>2+4 6/8>7+4 3/8>3+6 7/8>8+6 4/8>5+0

Leaving aside 4/8>5+0 (think of this as 8x5 = 40), the two sub-series 1, 2, 3 and 5, 6, 7 have the same remainders and the quotients are as simple as 1, 2, 3 and 6, 7, 8; so, without a doubt, this will not be the series that will be the most difficult for you to learn.

## Subdiagonal rules

Finally, as a last resort for learning, note the following series of terms adjacent to the diagonal of the table.

 4/5>8+0 5/6>8+2 6/7>8+4 7/8>8+6 8/9>8+8

There are really only two new rules here, but grasping the structure of the table above will also help you memorize the rules for divisors 5, 8, and 9.

## Hard rules

In summary, of the 45 rules included in the division table, 31 fall within one of the previous patterns (grayed)

 1/9>1+1 2/9>2+2 3/9>3+3 4/9>4+4 5/9>5+5 6/9>6+6 7/9>7+7 8/9>8+8 9/9>9+9 1/8>1+2 2/8>2+4 3/8>3+6 4/8>5+0 5/8>6+2 6/8>7+4 7/8>8+6 8/8>9+8 1/7>1+3 2/7>2+6 3/7>4+2 4/7>5+5 5/7>7+1 6/7>8+4 7/7>9+7 1/6>1+4 2/6>3+2 3/6>5+0 4/6>6+4 5/6>8+2 6/6>9+6 1/5>2+0 2/5>4+0 3/5>6+0 4/5>8+0 5/5>9+5 1/4>2+2 2/4>5+0 3/4>7+2 4/4>9+4 1/3>3+1 2/3>6+2 3/3>9+3 1/2>5+0 2/2>9+2 1/1>9+1

and we are left with only 14 "hard" rules to memorize with no other help. This is no longer a huge job. Cheer up and don't give up! with some effort and practice, the greatest of the arcane mysteries of Traditional Bead Arithmetic will be yours!

## The combined multiplication-division table

What follows is a simple historical note with little or no practical relevance.

The multiplication table in the English language contains all the 81 two-digit products in any order; that is, it includes both 8x9 = 72 and 9x8 = 72, which is unnecessary given the commutativity of the multiplication. On the contrary, in Chinese it only contained one of the terms of these pairs 8x9 = 72; always with the first factor less than or equal to the second[1][2]. On the other hand, the division rules were enunciated by giving first the divisor that is always greater than the dividend, with the exception of the rules that we have called diagonals in which it is equal. This allows a combined multiplication-division table to be conceived that covers the entire "space" of pairs of digits as operands:

 9✕9 81 9\8 8+8 9\7 7+7 9\6 6+6 9\5 5+5 9\4 4+4 9\3 3+3 9\2 2+2 9\1 1+1 8✕9 72 8✕8 64 8\7 8+6 8\6 7+4 8\5 6+2 8\4 5+0 8\3 3+6 8\2 2+4 8\1 1+2 7✕9 63 7✕8 56 7✕7 49 7\6 8+4 7\5 7+1 7\4 5+5 7\3 4+2 7\2 2+6 7\1 1+3 6✕9 54 6✕8 48 6✕7 42 6✕6 36 6\5 8+2 6\4 6+4 6\3 5+0 6\2 3+2 6\1 1+4 5✕9 45 5✕8 40 5✕7 35 5✕6 30 5✕5 25 5\4 8+0 5\3 6+0 5\2 4+0 5\1 2+0 4✕9 36 4✕8 32 4✕7 28 4✕6 24 4✕5 20 4✕4 16 4\3 7+2 4\2 5+0 4\1 2+2 3✕9 27 3✕8 24 3✕7 21 3✕6 18 3✕5 15 3✕4 12 3✕3  9 3\2 2+6 3\1 3+1 2✕9 18 2✕8 16 2✕7 14 2✕6 12 2✕5 10 2✕4  8 2✕3  6 2✕2  4 2\1 5+0 1✕9  9 1✕8  8 1✕7  7 1✕6  6 1✕5  5 1✕4  4 1✕3  3 1✕2  2 1✕1  1

Where we have altered the writing of our division rules to adapt them to the order of arguments used in Chinese. To highlight this fact we have replaced "/" by "\", so that the division rules as they appear in the above table must be interpreted in the form: Read a\b c+d:  as: a divide into b0 c times leaving d as remainder.

The combined table has 81 elements or rules, to which we must add the diagonal rules

 Diagonal 1/1>9+1 2/2>9+2 3/3>9+3 4/4>9+4 5/5>9+5 6/6>9+6 7/7>9+7 8/8>9+8 9/9>9+9

and the rules for revising down given in the previous chapter.

Rules to revise down (two-digit divisors)
While dividing by Revise q to Add to remainder
1 q-1 +1
2 q-1 +2
3 q-1 +3
4 q-1 +4
5 q-1 +5
6 q-1 +6
7 q-1 +7
8 q-1 +8
9 q-1 +9

that were studied separately. This adds up to a total of 99 rules to which we can add the approximately 50 addition and subtraction rules. The traditional learning of the abacus consisted fundamentally of the memorization and practice of these 150 rules.

## Statistical rules

What follows is a matter that arises from practice, not from any book in the past. The diagonal rules for divisors 1 and 2

 2/2>9+2 1/1>9+1

are excessive in the sense that we are often forced to revise up the divisor several times. In practice the following two statistical rules (to give them a name) behave better allowing a faster calculation.

 2/2>7+6 1/1>7+3

# Division/Dealing with overflow

## Introduction

Excluding the so-called "special methods", there are two basic ways of arranging general division problems. Not knowing a standard designation for them, we have called them in the chapter: Guide to traditional division:

• Modern division arrangement (MDA), as explained by Kojima[3],
MDA 25÷5=5
Abacus Comment
ABCDEF
5   25 Dividend starting in E
5  5 After division quotient begins in D
• Traditional division arrangement (TDA), as used in ancient books like the Jinkoki (塵劫記)[4], or the Panzhu Suanfa (盤珠算法)[5]
TDA 25÷5=5
Abacus Comment
ABCDEF
5   25 25÷5=5 Dividend starting in E
5   5 After division quotient begins in E

MDA seems perfect for any division method; not just the modern and traditional ones, but also any of the amazing variety of methods one can imagine after reading a page like: The Definitive Higher Math Guide on Integer Long Division[6], and just using the beads of a 4+1 (modern) abacus. On the contrary, TDA is problematic with any division method since a collision between divisor and dividend/remainder frequently occurs, that is, both require the simultaneous use of the same column and, as this is not possible in principle, for example, in the case of modern division we would be forced to postpone the entry of the interim quotient digit in the abacus until the corresponding column be cleared by subtraction. As a result, special techniques or abaci are needed to cope with this collision. Even so, TDA has been used for centuries in conjunction with the traditional method of division while MDA seems to have been deprecated until modern times and the adoption of the modern abacus, even though MDA is the first idea that would occur to us if we tried to adapt the old division method used with counting rods (paradoxically MD!) to a single row instead of the usual three. Why? that could remain a mystery forever. However, certain advantages to TDA must be recognized:

• It uses one rod less
• Result does not displace too much to the left as in MDA, which is of interest in the case of chained operations. This and the above points makes TDA more suitable to small rod number abacuses, like the traditional 13-rod suanpan/soroban.
• It saves some finger movements; for instance, in the operation 6231÷93=67 using traditional (chinese) division, one can count 14 finger movements with TDA versus 24 with MDA.
• Hand displacements are shorter.
• It is less prone to errors as less rods are skipped.

Are they enough to justify its historical usage?

Regarding the traditional division (Guī chúfǎ, Kijohou 帰除法) using TDA, the way to avoid the mentioned collision is to accept that the first column of the dividend/remainder, after the application of Chinese division rules, can overflow and temporarily accept a value greater than 9 (up to 18), while providing some mechanism to deal with such an overflow. This is not a problem with a traditional 5+2 or 5+3 abacus; As already explained, the additional upper beads can be used to store values as high as 20 in one column of the abacus. The problem arises when we think that 5+1 type abaci were popular in Japan during the Edo period and it seems that no ancient Japanese text explains how to deal with overflow. This is the question: What can be done on an 5+1 or 4+1 abacus?.

In a post to Soroban and Abacus Group, a member presented two examples of traditional division using an apostrophe (‘) to mark the columns or rods that temporarily received a value higher than 9 (overflow)[7].

Example from Soroban and Abacus Group
Abacus Comment
ABC   abcdef
898   888122 見八無頭作九八（Div. table）...
898   9'68122 九九八十一引（Mul. table）...
... ...

The apostrophe has the hindrance of breaking the vertical alignment of the columns of the abacus in the procedure tables, but let us think of this apostrophe as a typographical representation of a small 1 (¹), a bead that should be pushed, set or activated somewhere, be it on a real or imaginary column. Note that if we could open or insert a new column in the place of the apostrophe (as it is commonly done in any spreadsheet) all our problems would go away by using the new column to receive the bead, but by doing so we would be using MDA. After a short digression, three alternatives will be described below to stay on TDA.

### On geeses and flocks

We will use the classical exercise 998001÷999=999 as an example to illustrate the three mentioned alternatives. This exercise is called in Chinese: The lone geese return (孤雁歸隊 Gūyàn guīduì). If you enter this division on the abacus, for instance:

Abacus
ABCDEFGHIJK
999  998001

and if you have an almighty imagination, no doubt, you will identify the lone bead set on K with a lone geese that has just left her flock FGH (you can see the place that she occupied in the lower part of column H). To convince her to rejoin her flock you only have to complete the division and obtain 999!

## First way: Brute force

In principle, we could add the small “1” in any unused column, for example the rightmost one; but this could be annoying and inconvenient because both the hand and the attention would have to be jumping from one place to another on the abacus with the risk of ending up working in the wrong column. Here, without any further consideration, we will simply add the small "1" to the column of the just entered interim quotient digit. This may sound strange or brutal (and indeed it is), but if we can keep the value of the interim digit in memory we can operate as usual and any anomaly will disappear from the abacus in a moment. Let's see it with the  998001999=999 example on an 4+1 abacus:

998001÷999 = 999; Brute force method on 4+1 abacus
Abacus Comment
ABCDEFGHIJK
999  998001 Chinese rule: 9/9->9+9, remember quotient digit 9!
999 1088001 ("carry run" to the left! Don’t panic!)
-81 -9*9
999 1007001
-81 -9*9
999  998901 Chinese rule: 9/9->9+9, remember quotient digit 9!
999 1007901 ("carry run" to the left! Don’t panic!)
-81 -9*9
999  999801
-81 -9*9
999  998991 Chinese rule: 8/9->8+8, remember quotient digit 8!
999  999791
-72 -8*9
999  999071
-72 -8*9
999  998999 finally, revising up
999  999 done!

On a 5+1 abacus, things are easier. We can use the 5th bead to avoid carry runs.

998001÷999 = 999; Brute force method on 4+1 abacus (2nd quotient digit)
Abacus Comment
ABCDEFGHIJK
...
999  998901 Chinese rule: 9/9->9+9, remember quotient digit 9!
999  9T7901
-81 -9*9
999  999801
... ...etc.

As we can see, we can do things this way but it does not seem like a very attractive method as we need memorization and a lot of attention to avoid making mistakes. So one should not attempt this method except as an exercise in concentration.

## Second way: Suspended lower beads

If we use a 5+1, instead of pushing the bead all the way up, effectively adding the small “1” to the interim quotient digit as in the previous case, it seems more reasonable to push it only halfway, leaving a suspended lower bead as illustrated at the top of the image to the right. This suspended bead will represent the overflow while respecting the integrity of the quotient digit.

This seems like a perfect method to deal with the overflow, both in division and multiplication, everything remains under our eyes and nothing has to be memorized. In fact, when using suspended lower beads there is no need for additional upper beads, and the 5+1 abacus becomes as powerful as the 5+2  or 5+3 instruments. This might help explain why the 5+1 abacus was so popular in the past and why the 5th lower bead survived for so long. Note in the bottom half of the figure that, with some complication, this method can also be extended to the 4+1 abacus. From here on,  We will use underlined digits to represent the overflow according to the figure, since the underline reminds us of what the suspended bead looks like and they don't mess up abacus procedure tables typed with monospaced fonts as the apostrophe does.

### 5+1 abacus

Let us repeat the above exercise with this technique. The divisor is no longer represented and some more details are also introduced to additionally illustrate how the fifth lower bead may be used in subtraction to somewhat simplify the operation (as usual, T is 10, 1 upper bead + 5 lower beads activated)

On an 5+1 abacus
Abacus Comment
ABCDEF
998001
988001 Chinese rule: 9:9 > 9+9
-8 Subtract 81 from BC
9T8001
-1
9T7001
-8 Subtract 81 from CD
999001
-1
998901
997901 Chinese rule: 9:9 > 9+9
-8 Subtract 81 from CD
999901
-1
999801
-8 Subtract 81 from DE
998T01
-1
998991
998791 Chinese rule: 8:9 > 8+8
-7 Subtract 72 from DE
998T91
-2
998T71
-7 Subtract 72 from EF
9989T1
-2
998999 Revising up
-9 (from right to left to save a hand displacement)
998990
-9
998900
-9
998000
+1
999000 Done!

See also division examples for illustrations of this division on 5+1, 5+2 and 5+3 type abacuses.

### 4+1 abacus

And now on a 4+1 abacus. We need to use the suspended group of four lower beads as a code for 9:

On an 4+1 abacus
Abacus Comment
ABCDEF
998001
988001 Chinese rule: 9:9 > 9+9
-81 Subtract 81 from BC
987001
-81 Subtract 81 from CD
998901
997901 Chinese rule: 9:9 > 9+9
-81 Subtract 81 from CD
999801
-81 Subtract 81 from DE
998991
998791 Chinese rule: 8:9 > 8+8
-72 Subtract 72 from DE
998071
-72 Subtract 72 from EF
998999 Revising up
999000 Done!

If you have tried this, you have probably noticed that the group of four suspended beads behaves the same as the  suspended upper bead used on the 5+2 abacus; i.e. with "inverse arithmetic", if you move the suspended bead toward the abacus bean you are subtracting instead of adding!.

## Third Way: Minimal memorization

It has been said above that using suspended lower beads seems a perfect method… but in fact it is somewhat annoying due to its inherent slowness. It is always difficult to suspend a bead, especially the small ones of modern abacus with little free space left on the rods, and this despite the silly trick of pinching the bead with two fingers and then retiring the hand as if taking a flower. It is true that with an 5+1 abacus there is no need of additional upper beads, but no doubt, if you have a lot of multiplications or divisions to do, you will prefer the speed that additional beads provide, since one very seldom need to suspend a bead on the 5+2, and never on the 5+3.

Rather than physically moving/suspending the overflow bead, it is enough to think that the bead has been already suspended on the quotient rod, or pushed on an imaginary rod flying around your abacus, around you..., or simply remember that the “overflow status” has been set to ON and that it needs to be unset back to OFF as soon as possible. This last way is similar to the process of setting flags ON/OFF in old electronic calculators programming. Obviously, moving no bead is faster than moving any bead, so nothing can be faster than this alternative. Nevertheless, we should expect to need some practice to get used to this method and prepare to make some more mistakes due to memorization. However, memorizing a digit, as in the brute force method, is worse than simply memorizing an alert condition as required here.

No need for a new example. The previous ones can be followed under this new view simply by interpreting the underlines as something like OverflowFlag: ON.

## Conclusion

We have seen here three techniques to deal with overflow on 4+1 and 5+1 abacuses that pushes the small “1” up on the interim quotient column:

1. All way, effectively adding it as a carry to the quotient
2. Only half way, leaving a suspended lower bead
3. Nothing at all (but in our minds)

These methods bring us the possibility of using traditional techniques and arrangements on any abacus type by simply adapting the mechanics to the presence/absence of additional beads. This is an advantage if you finally end up convinced by traditional techniques.

It has been mentioned that no ancient Japanese text explains how to deal with overflow with a 5+1 abacus. Most likely the form used was one of the last two methods introduced here. Consider that the second method can be demonstrated to others in just seconds, and that once seen, it is neither forgotten nor requires further explanation; It is so obvious. So there is not much need to write long texts to convey that knowledge.

## One-digit divisors (short division)

The number 123456789 has also been used to demonstrate multiplication and division in many ancient books on the abacus. Some, like the Panzhu Suanfa[8], start with the traditional multiplication (see chapter: [[../../Multiplication/|Multiplication]]) of this number by a digit and use the division to return the abacus to its original state; others, like the Jinkoki[9], do it the other way around, starting with division and ending the exercise with multiplication. The latter is what we do here.

The number 123456789 is divisible by 3, 9 and 13717421, so divisions by 2, 3, 4, 5, 6, 8 and 9 have results with finite decimal expansion (2 and 5 are divisor of the decimal basis or radix 10 ). Only division by 7 leads to a result with an infinite number of decimal places, so here we will cut it off and give a remainder.

Unfortunately, this exercise does not use all the division rules, but it is a good start and allows you to practice without a worksheet.

### 123456789 divided by 9

123456789 divided by 9
Abacus Comment
ABCDEFGHIJKLM Divisor 9 at M
123456789   9 Column A: Apply 1/9>1+1
133456789   9 Change 1 in A into 1 and add 1 to B
136456789   9 Column B: Apply rule 3/9>3+3 Change 3 in B into 3 and add 3 to C
136T56789   9 Column C: Apply rule 6/9>6+6 Change 6 in C into 6 and add 6 to D
136056789   9 (Same as above)
137156789   9 Revise up
137166789   9 Column D: Apply rule 1/9>1+1 Change 1 in D into 1 and add 1 to E
137162789   9 Column E: Apply rule 6/9>6+6 Change 6 in E into 6 and add 6 to F
137173789   9 Revise up
137173089   9 Column F: Apply rule 3/9>3+3 Change 3 in F into 3 and add 3 to G
137174189   9 Revise up
137174199   9 Column G: Apply rule 1/9>1+1 Change 1 in G into 1 and add 1 to H
137174209   9 Revise up
137174210   9 Revise up. Done! 123456789/9=13717421

### 123456789 divided by 8

123456789 divided by 8
Abacus Comment
ABCDEFGHIJKLM Dividend in A-I, divisor 8 at M
123456789   8
143456789   8 Column A: rule 1/8>1+2, change 1 in A into 1, add 2 to B
153456789   8 Column B: rule 4/8>5+0, change 4 in B into 5, add 0 to C
153T56789   8 Column C: rule 3/8>3+6, change 3 in C into 3, add 6 to D
153056789   8 (Same as above)
154256789   8 Revise up C, add 1 to C, subtract 8 from D
154296789   8 Column D: rule 2/8>2+4, change 2 in D into 2, add 4 to E
154316789   8 Revise up D, add 1 to D, subtract 8 from E
154318789   8 Column E: rule 1/8>1+2, change 1 in E into 1, add 2 to F
154320789   8 Revise up E, add 1 to E, subtract 8 from F
154320849   8 Column G: rule 7/8>8+6, Change 7 in G into 8, add 6 to H
154320969   8 Revise up G, add 1 to G, subtract 8 from H
154320973   8 Column H: rule 6/8>7+4, change 6 in H into 7, add 4 to I
154320985   8 Revise up H, add 1 to H, subtract 8 from I
1543209862  8 Column I: rule 5/8>6+2, change 5 in I into 6, add 2 to J
15432098624 8 Column J: rule 2/8>2+4, change 2 in J into 2, add 4 to K
1543209862508 Column K: rule 4/8>5+0, change 4 in K into 5, add 0 to L.

Done! 123456789/9=15432098.625

### 123456789 divided by 7

123456789 divided by 7
Abacus Comment
ABCDEFGHIJKLM Dividend in A-I, divisor 8 at M
123456789   7
153456789   7 Column A: rule 1/7>1+3, change 1 in A into 1, add 3 to B
174456789   7 Column B: rule 5/7>7+1, change 5 in B into 7, add 1 to C
175956789   7 Column C: rule 4/7>5+5, change 4 in C into 5, add 5 to D
176256789   7 Revise up C, add 1 to C, subtract 7 from D
176256789   7 Column D: rule 2/7>2+6, change 2 in D into 2, add 6 to E
176346789   7 Revise up D, add 1 to D, subtract 7 from E
176351789   7 Column E: rule 4/7>5+5, change 4 in E into 5, add 5 to F
176364789   7 Revise up E, add 1 to E, subtract 7 from F
176365289   7 Column F: rule 4/7>5+5, change 4 in F into 5, add 5 to G
176366589   7 Revise up F, add 1 to F, subtract 7 from G
176366799   7 Column G: rule 5/7>7+1, change 5 in G into 7, add 1 to H
176366829   7 Revise up G, add 1 to G, subtract 7 from H
176366825   7 Column H: rule 2/7>2+6, change 2 in H into 2, add 6 to I
176366841   7 Revise up H twice, add 2 to H, subtract 14 from I.
Stop here! 123456789/9=17636684, remainder = 1

### 123456789 divided by 6

123456789 divided by 6
Abacus Comment
ABCDEFGHIJKLM Dividend in A-I, divisor 8 at M
123456789   6
163456789   6 Column A: rule 1/6>1+4, change 1 in A into 1, add 4 to B
203456789   6 Revise up A, add 1 to A, subtract 6 from B
205456789   6 Column C: rule 3/6>5+0, change 3 in C into 5, add 0 to D
205696789   6 Column D: rule 4/6>6+4, change 4 in D into 6, add 4 to E
205736789   6 Revise up D, add 1 to D, subtract 6 from E
205756789   6 Column E: rule 3/6>5+0, change 3 in E into 5, add 0 to F
205760789   6 Revise up E, add 1 to E, subtract 6 from F
205761189   6 Revise up F, add 1 to F, subtract 6 from G
205761129   6 Column G: rule 1/6>1+4, change 1 in G into 1, add 4 to H
205761309   6 Revise up G twice, add 2 to G, subtract 12 from H
205761313   6 Revise up H, add 1 to H, subtract 6 from I
205761315   6 Column I: rule 3/6>5+0, change 3 in I into 5, add 0 to J.
Done! 123456789/6=20576131.5

### 123456789 divided by 5

123456789 divided by 5
Abacus Comment
ABCDEFGHIJKLM Dividend in A-I, divisor 8 at M
123456789   5
223456789   5 Column A: Rule 1/5>2+0, change 1 in A into 2, add 0 to B
243456789   5 Column B: Rule 2/5>4+0, change 2 in B into 4, add 0 to C
246456789   5 Column C: Rule 3/5>6+0, change 3 in C into 6, add 0 to D
246856789   5 Column D: Rule 4/5>8+0, change 4 in D into 8, add 0 to E
246906789   5 Revise up D, add 1 to D, subtract 5 from E
246911789   5 Revise up E, add 1 to E, subtract 5 from F
246912789   5 Column F: Rule 1/5>2+0, change 1 in F into 2, add 0 to G
246913289   5 Revise up F, add 1 to F, subtract 5 from G
246913489   5 Column G: Rule 2/5>4+0, change 2 in G into 4, add 0 to H
246913539   5 Revise up G, Add 1 to G, subtract 5 from H
246913569   5 Column H: Rule 3/5>6+0, change 3 in H into 6, add 0 to I
246913574   5 Revise up H, add 1 to H, subtract 5 from I
246913578   5 Column I: Rule 4/5>8+0, change 4 in I into 8, add 0 to J.
Done! 123456789/5=24691357.8

### 123456789 divided by 4

123456789 divided by 4
Abacus Comment
ABCDEFGHIJKLM Dividend in A-I, divisor 8 at M
123456789   4
243456789   4 Column A: rule 1/4>2+2, change 1 in A into 2, add 2 to B
303456789   4 Revise up A, add 1 to A, subtract 4 from B
307656789   4 Column C: rule 3/4>7+2, change 3 in C into 7, add 2 to D
308256789   4 Revise up C, add 1 to C, subtract 4 from D
308556789   4 Column D: rule 2/4>5+0, change 2 in D into 5, add 0 to E
308616789   4 Revise up D, add 1 to D, subtract 4 from E
308628789   4 Column E: rule 1/4>2+2, change 1 in E into 2, add 2 to F
308640789   4 Revise up E twice, add 2 to E, subtract 8 from F
308641389   4 Revise up F, add 1 to F, subtract 4 from G
3086417T9   4 Column G: rule 3/4>7+2, change 3 in G into 7, add 2 to H
308641929   4 Revise up G twice, add 2 to G, subtract 8 from H
308641959   4 Column H: rule 2/4>5+0, change 2 in H into 5, add 0 to I
308641971   4 Revise up H twice, add 2 to H, subtract 8 from I
3086419722  4 Column I: rule 1/4>2+2, change 1 in I into 2, add 2 to J
3086419725  4 Column J: rule 2/4>5+0, change 2 in J into 5, add 0 to K.
Done! 123456789/4=30864197.25

### 123456789 divided by 3

123456789 divided by 3
Abacus Comment
ABCDEFGHIJKLM Dividend in A-I, divisor 8 at M
123456789   3
333456789   3 Column A: rule 1/3>3+1, change 1 in A into 3, add 1 to B
403456789   3 Revise up A, add 1 to A, subtract 3 from B
410456789   3 Revise up B, add 1 to B, subtract 3 from C
411156789   3 Revise up C, add 1 to C, subtract 3 from D
411366789   3 Column D: rule 1/3>3+1, change 1 in D into 3, add 1 to E
411506789   3 Revise up D twice, add 2 to D, subtract 6 from E
411520789   3 Revise up E twice, add 2 to E, subtract 6 from F
411522189   3 Revise up F twice, add 2 to F, subtract 6 from G
411522399   3 Column G: rule 1/3>3+1, change 1 in G into 3, add 1 to H
411522609   3 Revise up G three times, add 3 to G, subtract 9 from H
411522630   3 Revise up H three times, add 3 to H, subtract 9 from I.
Done! 123456789/3=41152263

### 123456789 divided by 2

123456789 divided by 2
Abacus Comment
ABCDEFGHIJKLM Dividend in A-I, divisor 8 at M
123456789   2
523456789   2 Column A: rule 1/2>5+0, change 1 in A into 5, add 0 to B
603456789   2 Revise up A, add 1 to A, subtract 2 from B
611456789   2 Revise up B, add 1 to B, subtract 2 from C
615456789   2 Column C: rule 1/2>5+0, change 1 in C into 5, add 0 to D
617056789   2 Revise up C twice, add 2 to C, subtract 4 from D
617216789   2 Revise up D twice, add 2 to D, subtract 4 from E
617256789   2 Column E: rule 1/2>5+0, change 1 in E into 5, add 0 to F
617280789   2 Revise up E three times, add 3 to E, subtract 6 from F
617283189   2 Revise up F three times, add 3 to F, subtract 6 from G
617283589   2 Column G: rule 1/2>5+0, change 1 in G into 5, add 0 to H
617283909   2 Revise up G four times, add 4 to G, subtract 8 from H
617283941   2 Revise up H four times, add 4 to H, subtract 8 from I
617283945   2 Column I: rule 1/2>5+0, change 1 in I into 5, add 0 to J.
Done! 123456789/2=61728394.5

## Multi-digit divisors (long division)

### Division of 998001 by 999

Division of 998001 by 999
Abacus Comment
ABCDEFGHIJKLM Dividend in A-F, divisor 8 in K-M
998001    999
988001    999 Chinese rule: 9/9>9+9
-8 Subtract 81 from BC
9T8001    999
-1
9T7001    999
-8 Subtract 81 from CD
999001    999
-1
998901    999
997901    999 Chinese rule: 9/9>9+9
-8 Subtract 81 from CD
999901    999
-1
999801    999
-8 Subtract 81 from DE
998T01    999
-1
998991    999
998791    999 Chinese rule: 8/9>8+8
-7 Subtract 72 from DE
998T91    999
-2
998T71    999
-7 Subtract 72 from EF
9989T1    999
-2
998999    999
-9 Revising up (from right to left to save a hand displacement)
998990    999
-9
998900    999
-9
998000    999
+1
999000    999 Done! 998001/999 = 999

### Division of 888122 by 989

Division of 888122 by 989
Abacus Comment
ABCDEFGHIJKLM Dividend 888122 in A-F, divisor 989 in K-M
888122    989
868122    989 Focus on A and use rule: 8/9>8+8 i.e. change 8 in A to 8 (nothing to do) and add 8 to B
804122    989 Subtract A×L=8×8=64 from BC
896922    989 Subtract A×M=8×9=72 from CD
895922    989 Focus on B and use rule: 9/9>9+9 i.e. change 9 in B to 9 (nothing to do) and add 9 to C
898722    989 Subtract B×L=9×8=72 from CD
897912    989 Subtract B×M=9×9=81 from DE
897612    989 Focus on C and use rule: 7/9>7+7 i.e. change 7 in B to 7 (nothing to do) and add 7 to D
897052    989 Subtract C×L=7×8=56 from DE
897989    989 Subtract C×M=7×9=63 from EF
898000    989 Revise up: add 1 to C and subtract 989 from DEF. Remainder in DEF is zero, so that 888122/989 = 898. Done!

### Division of 888122 by 898

Division of 888122 by 898
Abacus Comment
ABCDEFGHIJKLM Dividend 888122in A-F, divisor 898 in K-M
888122    898
968122    898 Focus on A and use rule: 8/8>9+8, i.e. change 8 in A to 9 and add 8 to B
987122    898 Subtract A×L=9×9=81 from BC
979922    898 Subtract A×M=9×8=72 from CD
985922    898 Focus on B and use rule: 7/8>8+6, i.e. change 7 in B to 8 and add 6 to C
988722    898 Subtract B×L=8×9=72 from CD
988082    898 Subtract B×M=8×8=64 from DE
989882    898 Focus on C and use rule: 8/8>9+8, i.e. change 8 in C to 9 and add 8 to D
989072    898 Subtract C×L=9×9=81 from DE
989000    898 Subtract C×M=9×8=72 from EF. Remainder in DEF is zero, so that 888122/898 = 989. Done!

### Division of 412 by 896

Division of 412 by 896
Abacus Comment
ABCDEFGHIJKLM
896 412 This time the divisor goes to the left and the dividend to the right
896 512 Column E: rule 4/8>5+0, change 4 in E into 5, add 0 to F
896 492 cannot subtract E×B=5×9=45 from FG, revise down E: subtract 1 from E, add 8 to F
896 456 subtract E×B=4×9=36 from FG
896 4536 subtract E×C=4×6=24 from GH
896 4656 Column F: rule 5/8>6+2, change 5 in F into 6, add 2 to G
896 4602 subtract F×B=6×9=54 from GH
896 4582 cannot subtract F×C=6×6=36 from HI, revise down F: subtract 1 from F, add 8 to G
896 4591 and add 9 to H to return the excess 89 subtracted from GH
896 4588 Continue normally and subtract F×C=3×6=30 from HI
896 45916 Column G: rule 8/8>9+8, change 8 in G into 9, add 8 to H
896 45979 subtract G×B=9×9=81 from HI
896 459736 subtract G×C=9×6=54 from IJ
896 459896 Column H: rule 7/8>8+6, Change 7 in H into 8, add 6 to I
896 459824 subtract H×B=8×9=72 from IJ
896 4598192 subtract H×C=8×6=48 from JK
896 4598112 Column I: rule 1/8>1+2, change 1 in I into 1, add 2 to J
896 4598103 subtract I×B=1×9=9 from JK
896 45981024 subtract I×C=1×6=6 from KL
896 45982128 revise up I: add 1 to I, subtract 896 from JKL
896 45982148 Column J: rule 1/8>1+2, Change 1 in J into 1, add 2 to K
896 45982139 subtract J×B=1×9=9 from KL
896 459821384 subtract J×C=1×6=6 from LM
896 459821344 Column K: rule 3/8>3+6, change 3 in K into 3, add 6 to L
896 459821317 subtract K×B=3×9=27 from LM
896 459821315 subtract K×C=3×6=18 from M … from now it is approximateda
896 459821425 revise up K: add 1 to K, subtract 896 from LM…
896 459821429 Column L: rule 2/8>2+4, Change 2 in L into 2, add 4 to M
896 459821427 subtract L×B=2×9=18 from M…
896 459821428 Column M: rule 7/8>8+6, Change 7 in M into 8, add 4 to … Done! 412/896=0.459821428

Note: ^a See chapter: [[../../Abbreviated operations/|Abbreviated operations]]

## References

1. Chéng Dàwèi (程大位) (1993) [1592]. Suànfǎ Tǒngzōng (算法統宗) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
2. Chen, Yifu (2013). L’étude des Différents Modes de Déplacement des Boules du Boulier et de l’Invention de la Méthode de Multiplication Kongpan Qianchengfa et son Lien avec le Calcul Mental (PhD thesis) (in French). Université Paris-Diderot (Paris 7). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
3. Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
4. Yoshida, Mitsuyoshi (吉田光由) (1634). Jinkoki (塵劫記) (in Japanese). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
5. Xú Xīnlǔ (徐心魯) (1993) [1573]. Pánzhū Suànfǎ (盤珠算法) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
6. "The Definitive Higher Math Guide on Integer Long Division (and Its Variants)". Math Vault. Archived from the original on May 14, 2021. Retrieved August 4, 2021.
7. Murakami, Masaaki (2020-06-29). "The 5th lower bead". (Web link). Retrieved on 2021-08-13.
8. Xú Xīnlǔ (徐心魯) (1993) [1573]. Pánzhū Suànfǎ (盤珠算法) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
9. Yoshida, Mitsuyoshi (吉田光由) (1634). Jinkoki (塵劫記) (in Japanese). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)

## External resources

You can practice traditional division online with Soroban Trainer (see chapter: [[../../Introduction#External resources|Introduction]]) using this file kijoho-1digit.sbk that you should download to your computer and then submit it to Soroban Trainer (It is a text file that you can inspect with any text editor and that you can safely download to your computer).

# Division/Special division tables

## Principle

Suppose we have to perform a large number of divisions by 36525, which could be the case if we do calendar calculations. Then we can simplify the task by creating a specialized division table for this divisor. Following what is stated in the chapter: Guide to traditional division, we will start by calculating the following three Euclidean divisions:

Creating a special division table for 36525
100000÷36525 200000÷36525 300000÷36525
Quotient Remainder Quotient Remainder Quotient Remainder
2 26950 5 17375 8 07800

Which can be summarized in the following specialized division table:

 1/36525>2+26950 2/36525>5+17375 3/36525>8+07800

And now we can use this table to do divisions without touching the multiplication table. For example, how many Julian centuries of 36525 days can fit in 1 000 000 days?

1000000/36525
Abacus Comment
ABCDEFGHIJKLM
36525 1000000 Use rule: 1/36525>2+26950 on column G
36525 2000000 change 1 in G into 2
36525 2269500 Use rule: 2/36525>5+17375 on column H
36525 2569500 change 2 in H into 5
36525 2586875 revise up
+1
-36525
36525 2650350 revise up
+1
-36525
36525 2713825 Done! 1000000÷36525=27, remainder 13825

And we have done a division by a five-digit divisor without using the multiplication table!

## Two-digit division tables

In the past, special division tables were used for divisors between 11 and 99[1].

11 12 13 14 15 16 17 18 19 1 9+01 8+04 7+09 7+02 6+10 6+04 5+15 5+10 5+05 4+16 4+12 4+08 4+04 4+00 3+22 3+19 3+16 3+13 9+11 9+02 8+16 8+08 8+00 7+18 7+11 7+04 6+26 3+07 3+04 3+01 2+32 2+30 2+28 2+26 2+24 2+22 6+14 6+08 6+02 5+30 5+25 5+20 5+15 5+10 5+05 9+21 9+12 9+03 8+28 8+20 8+12 8+04 7+34 7+27 2+18 2+16 2+14 2+12 2+10 2+08 2+06 2+04 2+02 4+36 4+32 4+28 4+24 4+20 4+16 4+12 4+08 4+04 7+13 7+06 6+42 6+36 6+30 6+24 6+18 6+12 6+06 9+31 9+22 9+13 9+04 8+40 8+32 8+24 8+16 8+08 1 1+49 1+48 1+47 1+46 1+45 1+44 1+43 1+42 1+41 3+47 3+44 3+41 3+38 3+35 3+32 3+29 3+26 3+23 5+45 5+40 5+35 5+30 5+25 5+20 5+15 5+10 5+05 7+43 7+36 7+29 7+22 7+15 7+08 7+01 6+52 6+46 9+41 9+32 9+23 9+14 9+05 8+52 8+44 8+36 8+28 1+39 1+38 1+37 1+36 1+35 1+34 1+33 1+32 1+31 3+17 3+14 3+11 3+08 3+05 3+02 2+66 2+64 2+62 4+56 4+52 4+48 4+44 4+40 4+36 4+32 4+28 4+24 6+34 6+28 6+22 6+16 6+10 6+04 5+65 5+60 5+55 8+12 8+04 7+59 7+52 7+45 7+38 7+31 7+24 7+17 9+51 9+42 9+33 9+24 9+15 9+06 8+64 8+56 8+48 1+29 1+28 1+27 1+26 1+25 1+24 1+23 1+22 1+21 2+58 2+56 2+54 2+52 2+50 2+48 2+46 2+44 2+42 4+16 4+12 4+08 4+04 4+00 3+72 3+69 3+66 3+63 5+45 5+40 5+35 5+30 5+25 5+20 5+15 5+10 5+05 7+03 6+68 6+62 6+56 6+50 6+44 6+38 6+32 6+26 8+32 8+24 8+16 8+08 8+00 7+68 7+61 7+54 7+47 9+61 9+52 9+43 9+34 9+25 9+16 9+07 8+76 8+68 1+19 1+18 1+17 1+16 1+15 1+14 1+13 1+12 1+11 2+38 2+36 2+34 2+32 2+30 2+28 2+26 2+24 2+22 3+57 3+54 3+51 3+48 3+45 3+42 3+39 3+36 3+33 4+76 4+72 4+68 4+64 4+60 4+56 4+52 4+48 4+44 6+14 6+08 6+02 5+80 5+75 5+70 5+65 5+60 5+55 7+33 7+26 7+19 7+12 7+05 6+84 6+78 6+72 6+66 8+52 8+44 8+36 8+28 8+20 8+12 8+04 7+84 7+77 9+71 9+62 9+53 9+44 9+35 9+26 9+17 9+08 8+88 1+09 1+08 1+07 1+06 1+05 1+04 1+03 1+02 1+01 2+18 2+16 2+14 2+12 2+10 2+08 2+06 2+04 2+02 3+27 3+24 3+21 3+18 3+15 3+12 3+09 3+06 3+03 4+36 4+32 4+28 4+24 4+20 4+16 4+12 4+08 4+04 5+45 5+40 5+35 5+30 5+25 5+20 5+15 5+10 5+05 6+54 6+48 6+42 6+36 6+30 6+24 6+18 6+12 6+06 7+63 7+56 7+49 7+42 7+35 7+28 7+21 7+14 7+07 8+72 8+64 8+56 8+48 8+40 8+32 8+24 8+16 8+08 9+81 9+72 9+63 9+54 9+45 9+36 9+27 9+18 9+09

## Some examples

Dividing by numbers that start with 1 is awkward, the following table may be used to divide by 19[2].

 19 1 5+05
 99 1 1+01 2 2+02 3 3+03 4 4+04 5 5+05 6 6+06 7 7+07 8 8+08 9 9+09
9801÷99
Abacus Comment
ABCDEFGHI
9801   99 Dividend AD, divisor HI
9891   99 A: Rule 9/99>9+09
9899   99 B: Rule 8/99>8+08
+1 revising up
-99
99     99 Done! No remainder, quotient: 99

Dividing by 𝝅 is common in applications, here are the tables for two approximations of this irrational number.

 314 31416 1 3+058 1 3+05752 2 6+116 2 6+11504 3 9+174 3 9+17256

Finally, the division by 666 table.

 666 1 1+334 2 3+002 3 4+336 4 6+004 5 7+338 6 9+006

However, It is not advisable to divide by this number; results can be unpredictable… and uncontrollable! In any case, remember the advice:

I say to you againe, doe not call up Any that you can not put downe; by the Which I meane, Any that can in Turne call up somewhat against you, whereby your Powerfullest Devices may not be of use.

:)👿

• Murakami, Masaaki (2020). "Specially Crafted Division Tables" (PDF). 算盤 Abacus: Mystery of the Bead. Archived from the original (PDF) on August 1, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

# Division/Division by powers of two

## Introduction

A fraction whose denominator only contains 2 and 5 as divisors has a finite decimal representation. This allows an easy division by powers of two or five if we have the fractions ${\displaystyle 1/n,2/n,\cdots 9/n}$  tabulated (or memorized) where ${\displaystyle n}$  is one of such powers of two or five.

For instance, given

${\displaystyle 137=1\cdot 10^{2}+3\cdot 10^{1}+7\cdot 10^{0}}$

Then

${\displaystyle {\frac {137}{8}}={\frac {1\cdot 10^{2}+3\cdot 10^{1}+7\cdot 10^{0}}{8}}={\frac {1}{8}}\cdot 10^{2}+{\frac {3}{8}}\cdot 10^{1}+{\frac {7}{8}}\cdot 10^{0}=}$

${\displaystyle =(0.125)\cdot 10^{2}+(0.375)\cdot 10^{1}+(0.875)10^{0}=12.5+3.75+0.875=17.125}$

Which can easily be done on the abacus by working from right to left. For each digit of the numerator:

1. Clear the digit
2. Add the fraction corresponding to the working digit to the abacus starting with the column it occupied
Division 137/8 using fractions
Abacus Comment
ABCDEF
--+--- Unit rod
137 enter 137 on A-C as a guide
7 clear 7 in C
130875
3 clear 3 in B
104625
1 clear 1 in A
17125 Done!
--+--- unit rod

We only need to have the corresponding fractions tabulated or memorized, as in the table below.

## Powers of two

In the past, both in China and in Japan, monetary and measurement units were used that were related by a factor of 16[3][4][5], a factor that begins with one which makes normal division uncomfortable. For this reason, it was popular to use the method presented here for such divisions.

### Table of fractions

Power of two fractions
D D/2 D/4 D/8 D/16a D/32a D/64a
1 05 025 0125 0625 03125 015625
2 10 050 0250 1250 06250 031250
3 15 075 0375 1875 09375 046875
4 20 100 0500 2500 12500 062500
5 25 125 0625 3125 15625 078125
6 30 150 0750 3750 18750 093750
7 35 175 0875 4375 21875 109375
8 40 200 1000 5000 25000 125000
9 45 225 1125 5625 28125 140625
1 1 1
Unit rod left displacement

^a Unit rod left displacement.

### Examples of use

68.5 ABCD --+-b 137 7 +35 3 +15 1 +05 --+-b 0685
34.25 ABCDE --+--b 137 7 +175 3 +075 1 +025 --+--b 03425
17.125 ABCDEF --+---b 137 7 +0875 3 +0375 1 +0125 --+---b 017125
8.5625 ABCDEF --+---b 137 7 +4375 3 +1875 1 +0625 -+----b 085625
4.28125 ABCDEFG --+----b 137 7 +21875 3 +09375 1 +03125 -+-----b 0428125
 2.140625 ABCDEFGH --+-----b 137 7 Clear 7 in C +109375 3 Clear 3 in B +046875 1 Clear 1 in A +015625 -+------b 02140625

^b "+" indicates the unit rod position.

### Division by 2 in situ

The fractions for divisor 2 are easily memorizable and this method corresponds to the division by two "in situ" or "in place" explained by Siqueira[6] as an aid to obtaining square roots by the half-remainder method (半九九法, hankukuho in Japanese, Bàn jiǔjiǔ fǎ in Chinese, see Chapter: [[../../Roots/Square root/|Square root]]), it is certainly a very effective and fast method of dividing by two. Fractions for other denominators are harder to memorize.

Being a particular case of what was explained in the introduction above, to divide in situ a number by two we proceed digit by digit from right to left by:

1. clearing the digit
2. adding its half starting with the column it occupied

For instance, 123456789/2:

123456789÷2 in situ
Abacus Comment
ABCDEFGHIJ
123456789
9 Clear 9 in I
+45 Add its half to IJ
1234567845
8 Clear 8 in H
+40 Add its half to HI
1234567445
7 Clear 7 in G
+35 Add its half to GH
1234563945
6 Clear 6 in F
+3 Add its half to FG
1234533945
5 Clear 5 in E
+25 Add its half to EF
1234283945
4 Clear 4 in D
+2 Add its half to DE
1232283945
3 Clear 3 in C
+15 Add its half to CD
1217283945
2 Clear 2 in B
+1 Add its half to BC
1117283945
1 Clear 1 in A
+05 Add its half to AB.
617283945 Done!

The unit rod does not change in this division.

## Powers of five

### Table of fractions

Power of five fractions
D D/5 D/25 D/125 D/625
1 0.2 0.04 0.008 0.0016
2 0.4 0.08 0.016 0.0032
3 0.6 0.12 0.024 0.0048
4 0.8 0.16 0.032 0.0064
5 1 0.2 0.04 0.008
6 1.2 0.24 0.048 0.0096
7 1.4 0.28 0.056 0.0112
8 1.6 0.32 0.064 0.0128
9 1.8 0.36 0.072 0.0144

# Multiplication

## How many multiplication methods are there?

Let's take an example: ${\displaystyle 345\times 6789=2342205}$ . We do this multiplication by adding the 12 partial products that result from the expansion:

${\displaystyle 345\times 6789=(300+40+5)\times (6000+700+80+9)=300\times 6000+300\times 700+\cdots +5\times 9}$

That is, all the products listed in this table:

Partial products of 345✕6789
6000 700 80 9
300 1800000 210000 24000 2700
40 240000 28000 3200 360
5 30000 3500 400 45

But these products can be added in any of the ${\textstyle 12!}$  (12 factorial) ${\textstyle =1\times 2\times 3\times \cdots \times 12=479\,001\,600}$  ways of ordering them, so we could say that there are, at least, almost 500 million ways to calculate the product of the two given numbers.

But it is clear that, of this immense number of ways of sequentially adding partial products, only a few can be efficiently and safely generated and followed by the human brain. But these few are still a lot ... especially if we think that we can also choose whether or not to enter multiplicand and multiplier in the abacus and where to start adding the partial products with respect to said operands. In what follows we will focus on this last aspect.

## Inverse operations

Addition and subtraction are inverse operations in the sense that each undoes the effect of the other by reverting the result to the first operand; for example: ${\displaystyle 422+313=735}$  and now subtracting ${\displaystyle 735-313=422}$ . On the abacus:

Abacus Comment
ABC
422
+1
+3
735 Result
-3 Subtract 313 to ABC
-1
-3
422 Result reverted

and, as we can see, we not only obtain the starting value but also obtain it in the original position. In turn, multiplication and division are also inverse operations; i.e: if ${\displaystyle a/b=q,r}$  where ${\displaystyle q}$  is the quotient of dividing ${\displaystyle a}$  by ${\displaystyle b}$  and ${\displaystyle r}$  is the remainder, we can reverse the operation in the form: ${\displaystyle a=q\times b+r}$  for example: ${\displaystyle 4727/72=65,47}$  where 65 is the quotient and 47 the remainder and we can reverse the operation in the form ${\displaystyle 4727=65\times 72+47}$ . On the abacus, using modern division and multiplication methods:

4727÷72, modern method
Abacus Comment
ABCDEFGHI 4727÷72
72   4727 Dividend:F-I, divisor:AB
72  64727 Try 6 as interim quotient
-42 Subtract 6✕7=42 from FG
-12 Subtract 6✕2=12 from GH
72  6 407
72  65407 Try 5 as interim quotient
-35 Subtract 5✕7=35 from GH
-10 Subtract 5✕2=10 from HI
72  65 47 Stop: quotient=65, remainder=47
72  65 47 Reverting by multiplication
72  65407
72  6 407 Clear F
72  64727 Clear E
72   4727 Done!

We have reversed the operation and returned the abacus to its original state. Note the relative position of operands and results using the modern method:

Relative position of operands and results (modern method)
Abacus Comment
ABCDEFGHI 4727÷72
72   4727 Divisor & dividend
72  65 47 Divisor: AB, quotient: EF & remainder: HI

Now let's try the same with the traditional method of division (TD) and the traditional division arrangement (TDA).

Abacus Comment
ABCDEFGHI 4727÷72
72   4727 Dividend:F-I, divisor:AB
72   5227 Rule: 4/7>5+5 (overflow!)
-10 Subtract 5✕2=10 from GH
72   5127
+1 Revise up F
-72 Subtract 72 from GH
72   6407
72   6557 Rule: 4/7>5+5
-10 Subtract 5✕2=10 from HI
72   6547 Stop: quotient=65, remainder=47

now the relative position of operands and results using the traditional method is different:

Relative position of operands and results (traditional method)
Abacus Comment
ABCDEFGHI 4727÷72
72   4727 Divisor & dividend
72   6547 Divisor: AB, quotient: FG & remainder: HI

If we want to reverse the operation by multiplication, we could first proceed by memorizing the digit of the multiplicand to use and clearing it, then we would proceed to add the partial products:

Abacus Comment
ABCDEFGHI
72   6547 Reverting by multiplication
72   6 47 Clear G and remember 5
72   6407
72    407 Clear F and remember 6
72   4727 Done!

and we have also reversed the operation and returned the abacus to its original state. In this way we proceed exactly the same as with modern multiplication, previously freeing up and reusing the space occupied by the digit in use of the multiplicand. However, memorizing and keeping something in memory while working with the abacus opens a door for errors and it is desirable to minimize this possibility by trying to keep the digit in memory for as little time as possible. This is achieved by altering the order in which we add the partial products:

Abacus Comment
ABCDEFGHI
72   6547 Reverting by multiplication
+35 Clear G and add 5✕7=35 to GH
72   6407
72    407 Clear F and remember 6
+42 Clear F and add 6✕7=42 to FG
72   4727 Done!

As we can see, we have delayed clearing the digit in use until the last possible moment. This is the basis of the traditional method of multiplication.

The traditional method of multiplication was first introduced using counting rods[7] and the best way to introduce it to the modern abacist is to consider that a multi-digit multiplier consists of a head (the first digit from the left) and a body (the rest of the digits); for example: 4567✕23, considering 4567 as the multiplier, its head is 4 and the body 567. So, for each digit of the multiplicand (from right to left):

• proceed as in modern multiplication with the product of the digit of the multiplicand by the body of the multiplier
• then clear the digit of the current multiplicand and add its product by the head of the multiplier to the column just cleared and the one adjacent to its right
Abacus Comment
ABCDEFGHIJKL Multiplicand:FG, Multiplier: A-D
4567  23 Head: A (4), Body: BCD (567)
+12 Clear H and add 3✕4=12 to HI
4567  213701
+08 Clear G and add 3✕4=12 to GH
4567  10F041 Done!a

note: ^a Result is 10F041 if you use the 5th lower bead,105041 otherwise.

But things are not always as simple as in the previous example; if both the multiplicand and the multiplier contain high digits (7, 8, 9) we may have overflow problems and need to deal with them (see chapter: Dealing with overflow), as in the case 999✕999=998001:

Abacus Comment
ABCDEFGHIJK Multiplicand:A-C, Multiplier: I-K
999     999 Head: I (9), Body: JK (99)
+81 Clear C and add 9✕9=81 to CD
998991  999
+81 Clear B and add 9✕9=81 to BC
988901  999 (overflow!)
+81 Clear A and add 9✕9=81 to AB
888001  999 (double overflow!)
998001  999 Normalizing result, done!

The most convenient, as in the case of division, is to have additional upper beads, that is, a 5+2 type abacus or a 5+3 if we are lucky enough. For the 4+1 and 5+1 abaci, it may be best to use the following fallback to the method outlined in the previous section, clearing the current digit of the multiplicand at the beginning (or when necessary) to have space to hold the partial results; for instance:

999✕999 Traditional method (fallback for 4+1 and 5+1 abaci)
Abacus Comment
ABCDEFGHIJK Multiplicand:A-C, Multiplier: I-K
999     999
+81 Clear C, remember 9 and add 9✕9=81 to CD
998991  999
+81 Clear B, remember 9 and add 9✕9=81 to BC
998901  999
+81 Clear A, remember 9 and add 9✕9=81 to AB
998001  999

If you practice the previous example together with the two traditional exercises: 898✕989, using 898 both as multiplier and multiplicand, you will be prepared for any traditional multiplication problem.

## References

1. Martzloff, Jean-Claude (2006), A history of chinese mathematics, Springer, p. 221, ISBN 978-3-540-33782-9
2. Cabrera, Jesús (2021), "Tide Abacus", jccAbacus, retrieved 4 August 2021
3. Williams, Samuel Wells; Morrison, John Robert (1856), A Chinese commercial guide, Canton: Printed at the office of the Chinese Repository, p. 298
4. Murakami, Masaaki (2020). "Specially Crafted Division Tables" (PDF). 算盤 Abacus: Mystery of the Bead. Archived from the original (PDF) on August 1, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
5. Kwa Tak Ming (1922), The Fundamental Operations in Bead Arithmetic, How to Use the Chinese Abacus (PDF), San Francisco: Service Supply Co.
6. Siqueira, Edvaldo; Heffelfinger, Totton. "Kato Fukutaro's Square Roots". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
7. Volkov, Alexei (2018), "Visual Representations of Arithmetical Operations Performed with Counting Instruments in Chinese Mathematical Treatises", Researching the History of Mathematics Education - An International Overview, Springer Publishing, ISBN 978-3-319-68293-8 {{citation}}: Unknown parameter |editor1first= ignored (|editor-first1= suggested) (help); Unknown parameter |editor1last= ignored (|editor-last1= suggested) (help); Unknown parameter |editor2first= ignored (|editor-first2= suggested) (help); Unknown parameter |editor2last= ignored (|editor-last2= suggested) (help)

• Kojima, Takashi (1963), "III Other multiplication methods", Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0003-7

# Roots

## Introduction

Obtaining square and cubic roots are the most complex operations studied within Elemental Arithmetic. The eastern abacus is very well adapted to obtaining square roots by a direct and efficient procedure, but unfortunately the same cannot be said with reference to cube roots which, although possible, require a tortuous path full of comings and goings that it is prone to errors.

Cargill Gilston Knott (1856 - 1922), one of the fathers of modern seismology, was a Scottish physicist and mathematician who served nine years as a professor of mathematics, acoustics and electromagnetism at the Imperial University of Tokyo; after which he was awarded the Order of the Rising Sun by Emperor Meiji in 1891. During his stay in Japan he came into contact with the Japanese abacus which he studied in depth and without a doubt he used professionally in his own work as a teacher and researcher. The result of this study was a famous 55 pages article[1] written in 1885 about it, which for a long time has been the best informed account in English, and obligatory reference, on the history and foundations of soroban. The next two chapters in this book develop and expand on Knott's vision of the traditional methods of obtaining square and cube roots, the vision of a western scientist and mathematician, offering both a theoretical and practical approach illustrated with several examples.

## Chapters

This part of the book consists of the following chapters:

Obtaining square and cubic roots with the abacus can be a somewhat long process and during the learning phase it is interesting to have a tool that allows us to control whether we are doing it correctly.

### Square root

For square roots you can try the excellent Murakami's Square root tutor with Kijoho, a JavaScript application that you can run directly in your browser or download to your computer from its GitHub repository. You just have to enter the root in the small input box on the left and repeatedly press the "next" button on the screen to see the development of the process step by step.

### Cube root

#### File knott.bc

And mainly for cube roots, the following BC code may help, copy and paste it to a text file and call it knott.bc:

/*
Functions to help to learn/verify square and cube roots a la Knott
with the abacus, soroban, suanpan.

See: https://jccabacus.blogspot.com/2021/06/roots-la-knott.html
as a reference.

Jesús Cabrera, June 2021
CC0 1.0 Universal (CC0 1.0) Public Domain Dedication

*/

define int(x)
{

# Integer part of x

auto os,r
os=scale; scale=0
r=x/1
scale= os
return (r)
}

define cbrt(x)
{

# Cube root of x

return (e(l(x)/3))
}

define knott2(r0, y0, alpha)
{

/*
Square root following Cargill G. Knott steps

See example of use in file sr200703.bc
use: $sr200703.bc |bc -l knott.bc */ auto so, div so = scale; /* Store old scale value */ scale = 1 a = 10*y0 div = 100*r0 + alpha/2 print "New dividend: ",div/1,"\n" b = int(div/(a)) tf = div -b*a -b^2/2 if (tf<0){ b=b-1;print "Revising down, b = ",b, "\n" tf = div -b*a -b^2/2 } print "New root: ", a+b,", New half-remainder: ", tf/1 print "\n==================\n\n" scale = so; /* restore old scale value */ return } define knott3(r0, y0, alpha) { /* Cube root following Cargill G. Knott steps See example of use in file cr488931400152.bc use:$ cat cr488931400152.bc |bc -l knott.bc

*/
auto so, div, ta, tb, tc, td, te

so = scale; /* Store old scale value */
scale = 0

a = 10*y0
div = 1000*r0 + alpha
print "New dividend: ",div,"\n\n"

ta = div/y0; rem1 = div % y0
print "a) /a:   ", ta, "   rem1: ", rem1, "\n"
tb = (10*ta)/3; rem2 = (10*ta) % 3
print "b) /3:   ", tb, "   rem2: ", rem2, "\n"
b = tb/(100*a)
print "     b = ",b,"\n"
tc = tb - b*(a+b)*100
print "d)   :   ",tc,"\n"
b = tb/(100*(a+b))
print "     b = ",b,"\n"
tc = tb - b*(a+b)*100
print "d)   :   ",tc,"\n"
if(b==10){
/* Trick to avoid some problems */
b = 9
print "b: ",b,"\n"
tc = tb - b*(a+b)*100
print "d) tc:   ",tc,"\n"
}
td = tc*3 +rem2
print "e) *3:   ",td,"\n"
te = (td/10)*y0 +rem1
print "f) *a:   ",te,"\n"
tf = te - b^3
print "g) -b^3: ",tf,"\n"
print "\nNew root: ",(a+b)," New remainder: ",tf,"\n\n"
print "==================\n\n"
scale = so; /* restore old scale value */

return
}


#### File: sr200703.bc

/*
Example: square root of 200703

Use:
$cat sr200703.bc |bc -l knott.bc or$ bc -l knott.bc < sr200703.bc
*/

print "\nSquare root of ", 200703, " = ", sqrt(200703), "\n\n"

/*
Decompose in pairs of digits (will be alpha): 20, 07, 03

Initialize (first step)
*/
alpha = 20
b = int(sqrt(alpha))
r0 = alpha - b^2
a = 0
tf = r0/2
print "First root: ", b, ", First half-remainder: ", tf, "\n"
print "==================\n\n"

/*
Main:
Repeat for each pair of digits (alpha)...
*/

alpha =07
mute=knott2(tf, a+b, alpha)
alpha =03
mute=knott2(tf, a+b, alpha)
/*
For additional digits continue with alpha = 00
*/
alpha =00
mute=knott2(tf, a+b, alpha)
alpha =00
mute=knott2(tf, a+b, alpha)
alpha =00
mute=knott2(tf, a+b, alpha)
alpha =00
mute=knott2(tf, a+b, alpha)


Output:

Square root of 200703 = 447.99888392718122931160

First root: 4, First half-remainder: 2.00000000000000000000
==================

New dividend: 203.5
Revising down, b = 4
New root: 44, New half-remainder: 35.5
==================

New dividend: 3551.5
Revising down, b = 7
New root: 447, New half-remainder: 447.0
==================

New dividend: 44700.0
Revising down, b = 9
New root: 4479, New half-remainder: 4429.5
==================

New dividend: 442950.0
New root: 44799, New half-remainder: 39799.5
==================

New dividend: 3979950.0
New root: 447998, New half-remainder: 395998.0
==================

New dividend: 39599800.0
New root: 4479988, New half-remainder: 3759928.0
==================


#### File cr488931400152.bc

/*
Example: cube root of 488931400152

Use:
$cat cr488931400152.bc |bc -l knott.bc or$ bc -l knott.bc < cr488931400152.bc
*/

print "\nCube root of ", 488931400152, " = ", cbrt(488931400152), "\n\n"

/*
Decompose in triplets (will be alpha): #   488, 931, 400, 152

Initialize (first step)
*/

alpha = 488
b = int(cbrt(alpha))
r0 = alpha - b^3
a = 0
tf = r0
print "First root: ", b, ", First remainder: ", r0, "\n"
print "==================\n\n"

/*
Main:
Repeat for each triplet (alpha)...
*/

alpha = 931
mute = knott3(tf, a+b, alpha)
alpha = 400
mute = knott3(tf, a+b, alpha)
alpha = 152
mute = knott3(tf, a+b, alpha)

/*
For additional digits continue with alpha = 000
*/


Output

Cube root of 488931400152 = 7877.99999999999999999871

First root: 7, First remainder: 145
==================

New dividend: 145931

a) /a:   20847   rem1: 2
b) /3:   69490   rem2: 0
b = 9
d)   :   -1610
b = 8
d)   :   7090
e) *3:   21270
f) *a:   14891
g) -b^3: 14379

New root: 78 New remainder: 14379

==================

New dividend: 14379400

a) /a:   184351   rem1: 22
b) /3:   614503   rem2: 1
b = 7
d)   :   63603
b = 7
d)   :   63603
e) *3:   190810
f) *a:   1488340
g) -b^3: 1487997

New root: 787 New remainder: 1487997

==================

New dividend: 1487997152

a) /a:   1890720   rem1: 512
b) /3:   6302400   rem2: 0
b = 8
d)   :   0
b = 8
d)   :   0
e) *3:   0
f) *a:   512
g) -b^3: 0

New root: 7878 New remainder: 0

==================


## References

1. Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73

# Roots/Square root

## Theory

Let ${\displaystyle x}$  be the number of which we want to obtain the square root ${\textstyle y={\sqrt {x}}}$ ; Let's consider its decimal expansion, for example: ${\textstyle x=456.7890123}$ . Let's separate its digits into groups of two around the decimal point in the following way

${\displaystyle x=456.7890123=4\cdot 56\cdot 78\cdot 90\cdot 12\cdot 30\cdot 00\cdot 00\cdots }$

or, in other words, let's define the sequence of integers ${\displaystyle \alpha _{i}}$

${\displaystyle \alpha _{1}=4,\alpha _{2}=56,\alpha _{3}=78,\alpha _{4}=90,\alpha _{5}=12,\alpha _{6}=30,\alpha _{7}=00,\cdots }$

and let's build the sequence ${\textstyle x_{i}}$  recursively from ${\textstyle x_{0}=0}$

${\displaystyle x_{i}=100x_{i-1}+\alpha _{i}}$

and let ${\displaystyle y_{i}}$  be the integer part of the square root of ${\displaystyle x_{i}}$

${\displaystyle y_{i}=\lfloor {\sqrt {x}}_{i}\rfloor }$

i.e.  ${\displaystyle y_{i}}$  is the largest integer whose square is not greater than ${\displaystyle x_{i}}$ . Finally, let us call remainders to the differences

${\displaystyle r_{i}=x_{i}-y_{i}^{2}\geq 0}$

For our example we have:

${\displaystyle i}$  ${\displaystyle \alpha _{i}}$  ${\displaystyle x_{i}}$  ${\displaystyle y_{i}}$  ${\displaystyle r_{i}}$
0 0 0 0
1 4 4 2 0
2 56 456 21 15
3 78 45678 213 309
4 90 4567890 2137 1121
5 12 456789012 21372 26628

Let's see that, by construction, ${\displaystyle x_{i}}$  grows as ${\displaystyle 10^{2i}}$  (two more digits in each step), in fact the sequence ${\displaystyle x_{i}\cdot 10^{4-2i}}$ : (0, 400, 456, 456.78, 456.7890, etc.) tends to ${\displaystyle x}$  ${\textstyle \left(x_{i}\cdot 10^{4-2i}\rightarrow x\right)}$  or ${\textstyle x=\left(\lim _{i\to \infty }x_{i}\cdot 10^{4-2i}\right)}$ . By comparison, ${\displaystyle y_{i}}$  , as the integer part of the square root of ${\displaystyle x_{i}}$ , grows only as ${\displaystyle 10^{i}}$  (one digit more each step). As ${\displaystyle y_{i}}$  is the largest integer whose square is not greater than ${\displaystyle x_{i}}$  we have ${\displaystyle r_{i}=x_{i}-y_{i}^{2}\geq 0}$  as above but

${\displaystyle x_{i}-(y_{i}+1)^{2}=x_{i}-y_{i}^{2}-2y_{i}-1<0}$

by definition of ${\displaystyle y_{i}}$ , or

${\displaystyle r_{i}=x_{i}-y_{i}^{2}<2y_{1}+1}$

multiplying by ${\displaystyle 10^{4-2i}}$

${\displaystyle x_{i}\cdot 10^{4-2i}-(y_{i}\cdot 10^{2-i})^{2}<(2y_{i}+1)\cdot 10^{4-2i}}$

but as  ${\displaystyle y_{i}}$  grows only as ${\displaystyle 10^{i}}$ , the second term tends to zero as ${\displaystyle 10^{-i}}$ .  With this

${\displaystyle x_{i}\cdot 10^{4-2i}-(y_{i}\cdot 10^{2-i})^{2}\rightarrow 0}$

and ${\displaystyle x_{i}\cdot 10^{4-2i}\rightarrow x}$  so that we have

${\displaystyle y_{i}\cdot 10^{2-i}\rightarrow y={\sqrt {x}}}$

For other numbers, the above factors are: ${\displaystyle 10^{2k-2i}}$  and ${\displaystyle 10^{k-i}}$ , where ${\displaystyle k}$  is the number of two-digit groups to the left of the decimal point, negative if it is followed by 00 groups (ex. ${\displaystyle k=-1}$  for ${\displaystyle x=0.00456}$ , ${\displaystyle k=-2}$  for ${\displaystyle x=0.00000456}$ , etc.).

This is the basis for traditional manual square root methods.

## Procedure

We start with ${\displaystyle i=0}$ , ${\displaystyle x_{0}=0}$ , ${\displaystyle y_{0}=0}$ , ${\displaystyle r_{0}=0}$ .

### First digit

Squares table
${\displaystyle b}$  ${\displaystyle b^{2}}$
1 1
2 4
3 9
4 16
5 25
6 36
7 49
8 64
9 81

For ${\displaystyle i=1}$  and ${\displaystyle x_{1}=\alpha _{1}}$ , it is trivial to find ${\displaystyle y_{1}}$  such that its square does not exceed ${\displaystyle x_{1}}$  through the use of the following table of squares that is easily retained in memory as it is only a subset of the multiplication table. In the case of the example we find ${\displaystyle y_{1}=2}$ .

### Rest of digits

For ${\displaystyle i>1}$ , we have ${\displaystyle x_{i}=100x_{i-1}+\alpha _{i}}$  as defined above and we try to build ${\displaystyle y_{i}}$  in the way:

${\displaystyle y_{i}=10y_{i-1}+b}$

where ${\displaystyle b}$  is a one-digit integer ranging from 0 to 9. To obtain ${\displaystyle b}$  we have to choose the greatest digit from 0 to 9 such that:

${\displaystyle x_{i}-y_{i}^{2}=x_{i}-(10y_{i-1}+b)^{2}\leq 0}$

or

${\displaystyle x_{i}-(a+b)^{2}\leq 0}$

if we write ${\displaystyle a=10y_{i-1}}$ . Expanding the binomial we have

${\displaystyle x_{i}-a^{2}-2ab-b^{2}=100x_{i-1}+\alpha _{i}-(10y_{i-1})^{2}-2ab-b^{2}\leq 0}$

or

${\displaystyle 100r_{i-1}+\alpha _{i}\geq 2ab+b^{2}=(2a+b)b=(20y_{i-1}+b)b}$

The left side of the above expression may be seen as simply the previous remainder with the next two-digits group appended to it, and the parenthesis of the last term as twice the previous root with digit b appended to it. In our example, for ${\displaystyle i=2}$  we have 56 on the left and the above expression is

${\displaystyle 56\geq (40+b)b}$

which holds only for ${\displaystyle b=0}$  or ${\displaystyle b=1}$  so that 1 is our next root digit but, how can we proceed in the general case without having to systematically explore every possibility (${\displaystyle b=0,1,\cdots ,9}$ )?

Here Knott[1] distinguishes two different approaches:

• Preparing the divisor
• Preparing the dividend.

#### Preparing the divisor

This corresponds with the above expression

${\displaystyle 100r_{i-1}+\alpha _{i}\geq (20y_{i-1}+b)b}$

And it is the strategy usually used with paper and pencil and can also be implemented, of course, on the abacus. In the above expression, if we see the left part as the dividend and the parenthesis as the divisor, b is the first digit of the division

${\displaystyle b=(100r_{i-1}+\alpha _{i})/(20y_{i-1}+b)}$

but since we don't know b yet, we approximate it using only the main part of the divisor

${\displaystyle b=(100r_{i-1}+\alpha _{i})/(20y_{i-1})}$

This gives us a guess as to what the value of b might be, but we need:

1. Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
2. Obtain the next remainder to prepare the calculation of the next digit of the root.

Both steps require subtracting ${\displaystyle (20y_{i-1}+b)b}$  or ${\displaystyle 20y_{i-1}b}$  and ${\displaystyle b^{2}}$  from ${\displaystyle 100r_{i-1}+\alpha _{i}}$ ; checking that we are not going to negative values and that what remains is less than ${\displaystyle 2y_{i}+1}$  (otherwise we would have to revise ${\displaystyle b}$  up or down). If we do this correctly, what we are left with is the new remainder ${\displaystyle r_{i}}$ . It should be noted that, as we proceed in the calculations (${\displaystyle i}$  increasing) ${\displaystyle b}$  is progressively a smaller and smaller contribution to the divisor ${\displaystyle (20y_{i-1}+b)}$ ; so the process indicated above will look more and more like a mere division.

This is the method proposed by Takashi Kojima in his second book: Advanced Abacus - Theory and Practice[2], and that you can see described in Square roots as solved by Kojima[3] in Totton heffelfinger’s website, works to which I refer the reader to see explanations and examples. What follows here, for purposes of illustration, is a sketch of how the calculation might be started in our example: ${\displaystyle x=456.7890123}$

First 3 digits of ${\displaystyle {\sqrt {456.7890123}}}$  preparing the divisor
Abacus Comment
ABCDEFGHIJKLM
4567890123 Entering radicand starting in CD (first group)
2 First root digit in B
-4 Subtract square of B from first group
2  567890123 Null remainder
4  567890123 Doubling B. Appending next group (56)to remainder
41 567890123 5/4≈1, try 1 as next root digit
-4 Continue division by 41, subtract 1✕41 from EF
-1
41 157890123 15 as remainder
42 157890123 Double second root digit
42 157890123 Append next group (78)
423157890123 157/42≈3, try 3 as next root digit
-12 Continue division by 423, subtract 3✕423 from E-H
-06
-09
423 30990123 309 as remainder
426 30990123 Double third root digit
426 30990123 Append next group (90)
etc.

As can be seen, twice the root grows to the left of the abacus to the detriment of the radicand / remainder and the groups of two figures still unused. This is contrary to what happens with the rest of the elementary operations on the abacus, where the result sought replaces the operand (or one of them). For this reason, the traditionally preferred method for obtaining square roots seems to have been the following, where we will see the root appear directly on the abacus, not its double .

#### Preparing the dividend

Starting again with

${\displaystyle 100r_{i-1}+\alpha _{i}\geq (20y_{i-1}+b)b=20y_{i-1}b+b^{2}}$

dividing by 2

${\displaystyle (100r_{i-1}+\alpha _{i})/2\geq 10y_{i-1}b+b^{2}/2}$

This modified expression will allow us to directly obtain the square root in the abacus following practically the same procedure as above with only keeping half-remainders on our instrument. Here, with square roots, the change is almost trivial, but it will be more important when dealing with cube roots. As can be seen in the above expression, neglecting the term ${\textstyle b^{2}/2}$  we obtain a guess of ${\displaystyle b}$  by simply dividing the extended half-remainder ${\displaystyle (100r_{i-1}+\alpha _{i})/2}$  by the previous root ${\displaystyle y_{i-1}}$  (in fact ${\displaystyle 10y_{i-1}}$ ).  After that, we need again:

1. Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
2. Obtain the next half-remainder to prepare for the next digit of the root.

This is done by subtracting ${\displaystyle 10y_{i-1}b}$  and ${\textstyle b^{2}/2}$  from the half remainder, and this makes it convenient to memorize the following table of semi-squares:

Semi-squares table
${\displaystyle b}$  ${\displaystyle b^{2}/2}$
1 0.5
2 2
3 4.5
4 8
5 12.5
6 18
7 24.5
8 32
9 40.5

Fortunately, since 2 is a divisor of our base (10), the decimal fractions in the table have a finite expression; which will not happen when we try to extend this procedure to cube roots and we have to deal with thirds of cubes. According to Knott, this makes cube roots a problem that is not well suited to being treated with abacus.

## Examples

Here three examples are presented, for additional examples please see Further reading and specially External resources below.

### Square root of 961

In this example we have two groups of two figures: 09 and 61. The first group informs us that the first digit of the root is 3.

There are two ways to start square roots:

• Aligning the groups to the left of the abacus from column B and using the traditional division to obtain the semi-remainder.
A B C D E ... 0 9 6 1
This is the form used in the old books and also the one used in Murakami's Square root tutor with Kijoho (see below External resources).
Abacus Comment
ABCDE
0961 Align the radicand with B
30961 Enter first root digit in A
-9 Subtract the square of first root digit (9)
30061
30305 Divide the remainder B-E by 2 (帰除法)
• Aligning the groups to the left of the abacus from column A and using [[../../Division/Division by powers of two#Division by 2 in situ|in-situ]] division, as explained in chapter: [[../../Division/Division by powers of two|Division by powers of two]], to obtain the semi-remainder.
A B C D E ... 0 9 6 1
This method is somewhat faster.
Using division in situ
Abacus Comment
ABCDE
0961 Align the radicand with A
-9 Subtract the square of first root digit (9)
0061
0305 Divide in situ the remainder by 2
30305 Enter first root digit in A

From here, the state of the abacus coincides and we can continue:

Continuation
Abacus Comment
ABCDE
30305
+1 Divide half-remainder B-E by 3. Revising up B
-3
31005
-05 Subtract b^2/2 =0.5 from D
31000 Half-remainder is 0, Done! Root is 31
31 Root is 31

### Square root of 998001

${\displaystyle {\sqrt {998001}}=999}$
Abacus Comment
ABCDEFG
-81 Subtract 9^2=81 from first group
188001
940005 Halve the remainder in situ
9940005 Enter first root digit into A
9930005 B: Rule 9/9>9+9
-405 Subtract 9^2/2=40.5 from D
9989505
9987505 C: Rule 8/9>8+8
-72 Subtract CxB=72 from DE
998T305 Revise up C
+1
-99
9990405
-405 Subtract 9^2/2=40.5 from F
9990000 Remainder is 0. Done!
999 Root is 999

### Root of 456.7890123

Our example above...

First 4 digits of ${\displaystyle {\sqrt {456.7890123}}=21.3726229\cdots }$
Abacus Comment
ABCDEFGHIJKL
04567890123 Enter x aligning digit pairs from AB, CD, etc.
-4 Subtract 2^2 from first group
567890123
2839450615 Halve remainder and rest of digits pairs
2 2839450615 Enter first root digit in A
+1 Divide BCD by A (revise up B)
-2
-05 Subtract B^2/2=0.5 from D
21 789450615
+3 Divide CDEF by AB (revise up C three times)
-6
-3
-45 Subtract C^2/2=4.5 from F
213154950615
213554950615 Divide DEFGH by ABC. D: Rule 1/2>5+0
-5 Subtract DxB=5 from EF
-15 Subtract DxC=15 from FG
213548450615
+2 Revise up D twice
-426
213705850615
-245 Subtract 7^2/2=24.5 From H
21370560F615 Root so far: 21.37
etc.     etc.

The root 2137… (21.37…) is appearing on the left. See chapter: [[../../Abbreviated operations#Square root|Abbreviated operations]] to see how to quickly approximate the next 4 digits.

## Conclusion

The method explained as: Preparing the dividend is known as 半九九法 (Hankukuhou in Japanese, Bàn jiǔjiǔ fǎ in Chinese) that we freely translate here as the Half-remainder method and is by far the most convenient to use, at least for two reasons:

1. The root, and not its double, replaces the operand (radicand) as in the rest of operations on the abacus.
2. (The most important) Dividing by numbers that start with 1 is awkward. Think of the first group of two digits, its value is between 1 and 99 and determines the first figure of the square root. For values ​​of the first pair between 25 and 99 (75% of the cases) the first digit of the root is between 5 and 9 and its double begins with one! Therefore, if we use the method preparing the divisor, we will be dividing by numbers that begin with 1 in 75% of the cases. On the contrary, if we use the method preparing the dividend, only in the case that the first group is 1, 2 or 3 (3% of the cases) will we have to divide by numbers that start with one.

The superiority of the half-remainder or preparing the dividend method is undeniable.

## References

1. Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73
2. Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0003-7
3. Heffelfinger, Totton (2003). "Square Roots as Solved by Kojima". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021. Retrieved August 16, 2021.

• Siqueira, Edvaldo; Heffelfinger, Totton. "Kato Fukutaro's Square Roots". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

## External resources

• Murakami's Square root tutor with Kijoho, a JavaScript application that you can run directly in your browser or download to your computer from its GitHub repository. You just have to enter the root in the small input box on the left and repeatedly press the "next" button on the screen to see the development of the process step by step. So you can generate as many examples or exercises as you want.

# Roots/Cube root

## Theory

Let ${\displaystyle x}$  be the number of which we want to obtain the cube root ${\displaystyle y={\sqrt[{3}]{x}}}$  ; Let's consider its decimal expansion, for example: ${\displaystyle x=456.7890123}$ . Let's separate its digits into groups of three around the decimal point in the following way

${\displaystyle x=456.7890123=456.789\cdot 012\cdot 300\cdot 000}$

or, in other words, let's define the sequence of integers ${\displaystyle \alpha _{i}}$

${\displaystyle \alpha _{1}=456,\alpha _{2}=789,\alpha _{3}=012,\alpha _{4}=300,\alpha _{5}=000\cdots }$

and let's build the sequence ${\displaystyle x_{i}}$  recursively from ${\displaystyle x_{0}=0}$

${\displaystyle x_{i}=1000x_{i-1}+\alpha _{i}}$

and let ${\displaystyle y_{i}}$  be the integer part of the cube root of ${\displaystyle x_{i}}$

${\displaystyle y_{i}=\lfloor {\sqrt[{3}]{x}}_{i}\rfloor }$

i.e.  ${\displaystyle y_{i}}$  is the largest integer whose cube is not greater than ${\displaystyle x_{i}}$ . Finally, let us call remainders to the differences

${\displaystyle r_{i}=x_{i}-y_{i}^{3}\geq 0}$

For our example we have:

${\displaystyle i}$  ${\displaystyle \alpha _{i}}$  ${\displaystyle x_{i}}$  ${\displaystyle y_{i}}$  ${\displaystyle r_{i}}$
0 0 0 0
1 456 456 7 113
2 789 456789 77 256
3 012 456789012 770 256012
4 300 456789012300 7701 78119199
5 000 456789012300000 77014 6949021256

Let's see that, by construction, ${\displaystyle x_{i}}$  grows as ${\displaystyle 10^{3i}}$  (three more digits in each step), in fact the sequence ${\displaystyle x_{i}10^{3-3i}}$ , i.e. 0, 400, 456, 456.789, 456.789012, etc. tends to ${\displaystyle x}$  (${\displaystyle x_{i}10^{3-3i}\to x}$ ). By comparison, ${\displaystyle y_{i}}$  , as the integer part of the cube root of ${\displaystyle x_{i}}$ , grows only as ${\displaystyle 10^{i}}$  (one digit more each step). As ${\displaystyle y_{i}}$  is the largest integer whose square is not greater than ${\displaystyle x_{i}}$  we have ${\displaystyle r_{i}=xi-y_{i}^{3}\geq 0}$  as above but

${\displaystyle x_{i}-(y_{i}+1)^{3}=x_{i}-y_{i}^{3}-3y_{i}^{2}-3y_{i}-1<0}$

by definition of ${\displaystyle y_{i}}$ , or

${\displaystyle r_{i}=x_{i}-y_{i}^{3}<3y_{i}^{2}+3y_{i}+1}$

multiplying by ${\displaystyle 10^{3-3i}}$

${\displaystyle x_{i}\cdot 10^{3-3i}-(y_{i}\cdot 10^{1-i})^{3}<(3y_{i}^{2}+3yi+1)\cdot 10^{3-3i}}$

but as  ${\displaystyle y_{i}}$  grows only as ${\displaystyle 10^{i}}$ , the second term tends to zero as ${\displaystyle 10^{-i}}$ .

${\displaystyle x_{i}\cdot 10^{3-3i}-(y_{i}\cdot 10^{1-i})^{3}\to 0}$

and ${\displaystyle x_{i}\cdot 10^{3-3i}\to x}$  so that we have

${\displaystyle y_{i}\cdot 10^{1-i}\to y={\sqrt[{3}]{x}}}$

For other numbers, the above factors are: ${\displaystyle 10^{3k-3i}}$  and ${\displaystyle 10^{k-i}}$ , where ${\displaystyle k}$  is the number of three-digit groups to the left of the decimal point, negative if it is followed by 000 groups (ex. ${\displaystyle k=0}$  for ${\displaystyle x=0.00456}$ , ${\displaystyle k=-2}$  for ${\displaystyle x=0.000000456}$ , etc.).

This is the basis for traditional manual cube root methods.

## Procedure

We start with ${\displaystyle i=0,x_{0}=0,y_{0}=0,r_{0}=0}$ .

### First digit

${\displaystyle b}$  ${\displaystyle b^{3}}$
1 1
2 8
3 27
4 64
5 125
6 216
7 343
8 512
9 729

For ${\displaystyle i=1,x_{1}=\alpha _{1}}$ . It is trivial to find ${\displaystyle y_{1}}$  such that its square does not exceed ${\displaystyle x_{1}}$  through the use of the following table of cubes that can be easily retained in memory. In the case of the example it is ${\displaystyle y_{1}=7}$

### Rest of digits

For ${\displaystyle i>1}$ , we have ${\displaystyle x_{i}=1000x_{i-1}+\alpha _{i}}$  as defined above and we try to build ${\displaystyle y_{i}}$  in the way:

${\displaystyle y_{i}=10y_{i-1}+b}$

where ${\displaystyle b}$  is a one-digit integer ranging from 0 to 9. To obtain ${\displaystyle b}$  we have to choose the greatest digit from 0 to 9 such that:

${\displaystyle x_{i}-y_{i}^{3}=x_{i}-(10y_{i-1}+b)^{3}\geq 0}$

or

${\displaystyle x_{i}-(a+b)^{3}\geq 0}$

if we write ${\displaystyle a=10y_{i-1}}$ . Expanding the binomial we have

${\displaystyle x_{i}-a^{3}-3a^{2}b-3ab^{2}-b^{3}=1000x_{i-1}+\alpha _{i}-(10y_{i-1})^{3}-3a^{2}b-3ab^{2}-b^{3}\geq 0}$

or

${\displaystyle 1000r_{i-1}+\alpha _{i}\geq 3a^{2}b+3ab^{2}+b^{3}=(3a^{2}+3ab+b^{2})b}$

The left side of the above expression may be seen as simply the previous remainder with the next three-digits group appended to it. If we evaluate the term on the right for each value of ${\displaystyle b}$  and compare with the left term we have:

${\displaystyle a=10y_{i-1}=70}$
${\displaystyle b}$  ${\displaystyle (3a^{2}+3ab+b^{2})b}$  ${\displaystyle 1000r_{i-1}+\alpha _{i}}$
0 0 ≤ 113789
1 14911 ≤ 113789
2 30248 ≤ 113789
3 46017 ≤ 113789
4 62224 ≤ 113789
5 78875 ≤ 113789
6 95976 ≤ 113789
7 113533 ≤ 113789  ⬅
8 131552 > 113789
9 150039 > 113789

and it is clear that the next figure of our root is a 7 but, how can we proceed in the general case without having to systematically explore every possibility (${\displaystyle b=0,1,\cdots ,9}$ )?

Here Knott[1] distinguishes two different approaches:

• Preparing the divisor
• Preparing the dividend

#### Preparing the divisor

This correspond with the above expression

${\displaystyle 1000r_{i-1}+\alpha _{i}\geq 3a^{2}b+3ab^{2}+b^{3}=(3a^{2}+3ab+b^{2})b}$

And it is the strategy usually used with paper and pencil and can also be implemented, of course, on the abacus. In the above expression, if we see the left part as dividend and the parentese as divisor, ${\displaystyle b}$  is the first digit of the division

${\displaystyle b=(1000r_{i-1}+\alpha _{i})/(3a^{2}+3ab+b^{2})}$

but since we don't know b yet, we approximate it using only the main part of the divisor

${\displaystyle b=(1000r_{i-1}+\alpha _{i})/(3a^{2})=(1000r_{i-1}+\alpha _{i})/(300y_{i-1}^{2})}$

This gives us a guess as to what the value of ${\displaystyle b}$  might be, but we need:

1. Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
2. Obtain the next remainder to prepare the calculation of the next digit of the root.

You can see an example in Tone nikki blog[2], see also Modern approaches below.

#### Preparing the dividend

Starting again with

${\displaystyle 1000r_{i-1}+\alpha _{i}\geq (3a2+3ab+b2)b=3a\left(a+b+{\frac {b}{3a^{2}}}\right)b}$

we prepare the dividend by dividing ${\displaystyle 1000r_{i-1}+\alpha _{i}}$  (the next three-digits group appended to the previous remainder) by ${\displaystyle 3a}$

${\displaystyle (1000r_{i-1}+\alpha _{i})/3a\geq \left(a+b+{\frac {b}{3a^{2}}}\right)b}$

As usual, we don't know ${\displaystyle b}$  and we can't evaluate the parentheses on the right, but we can get a clue about ${\displaystyle b}$  by approximating the parentheses by its main part ${\displaystyle a}$  and use it as a trial divisor.

${\displaystyle \left(a+b+{\frac {b}{3a^{2}}}\right)\approx a}$

so that

${\displaystyle b\approx (1000r_{i-1}+\alpha _{i})/3a^{2}}$

After that, we need again:

1. Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
2. Obtain the next remainder to prepare for the obtention of the next digit of the root by evaluating ${\displaystyle 1000r_{i-1}+\alpha _{i}-3a^{2}b-3ab^{2}-b^{3}}$ .

• Divisor 3 is involved in the prepared dividend and this leads to non-finite decimal fractions.
• The division by ${\displaystyle a}$  not only worsens the above, but also makes the prepared dividend specific to the current step, since the value of ${\displaystyle a}$  evolves with the calculation of the different figures of the result.

This did not occur in the calculation of square roots and, as a consequence, the process of obtaining cube roots is much more complicated and requires a complex cycle of preparation-restoration of the dividend that, following Knott, can be represented by the following scheme:

• a) Divide by ${\displaystyle a}$ .
• b) Divide by 3.
• c) Obtain ${\displaystyle b}$  as the first digit of the division of the above by ${\displaystyle a}$ .
• d) Subtract ${\displaystyle b(a+b)}$  (Equivalent to subtracting ${\displaystyle 3a^{2}b}$  and ${\displaystyle 3ab^{2}}$  in ${\displaystyle 1000r_{i-1}+\alpha _{i}-3a^{2}b-3ab^{2}-b^{3}}$ ).
• e) Multiply by 3.
• f) Multiply by ${\displaystyle a}$ .
• g) Subtract ${\displaystyle b^{3}}$ .

In our example (${\displaystyle x=456.7890123}$ ), using traditional division and traditional division arrangement (like Knott does), working the two first digits:

Cube root of 456.7890123
Abacus Comment
ABCDEFG
456789 Enter number aligning first group with B
-343 -7^3=343
113789 First remainder
7113789 Enter 7 in A as first root digit and append second group
7113789 a) Divide B-F by 71
7162554 b) Divide B-F by 32
7541835 c) Divide B by A (one digit)
7751835 d) Subtract 7*7=49 from CD
77 2835 e) Multiply CDEF by 3. Add 3✕283 to CDEFG
77  854 f) Multiply CDEF by 7. Add 7✕85 to CDEFG
77  599
-343 g) Subtract 7^3=343 to CDEFG
77  256 New remainder
... Root obtained so far: 7.7

Notes:

• ^1 a) It is unnecessary to extend the division by 7 beyond the current three-digit group. The 4 in G is a division remainder meaning 4/7.
• ^2 b) The same can be said of division by 3. It is carried out up to column F and the remainder (1) is temporarily added to column G. The value (5) in said column is a strange hybrid meaning 1/3 and 4/7 . It does not matter, it will be reabsorbed in steps e) and f).

#### Modern approaches

Members of the Soroban & Abacus Group modified the technique described by Knott to adapt it to modern soroban use[3]. The result is allegedly faster at the expense of being less compact and requiring an abacus with more rods to store intermediate data. The simplicity of having the result directly substituting the radicand is also lost.

You can also find a compilation of modern methods for both square and cube roots in Tone Nikki (とね日記)[2] by a Japanese blogger (Author's name does not appear to be available).

## Examples of cube roots

The following examples are all worked using traditional division and traditional division arrangement. Components of the dividend preparation-restoration cycle are labelled with a), b), etc as detailed above.

### Cube root of 157464

Abacus Comment
ABCDEFG Cube root of 157464
157464 Enter number aligning first group with B
-125 Subtract 5^3=125 from BCD
32464 First remainder: 32
5 32464 Enter 5 in A as first root digit and append second group
5 32464 a) Divide C-F by 5 (G will be the division remainder)
5 64924 b) Divide C-F by 3
5216404 c) Divide B by 5
5416404 d) Subtract 4x4=16 from CD
54  404 e) Multiply 40x3 in EFG (adding to remainder in G)
54  124 f) Multiply 12x5 in EFG
54   64 g) Subtract 4^3=64 from FG
54 Remainder 0; Done! Root is 54

Clearly, if the remainder is zero and there are no more (not null) groups to add, the number is a perfect cube and we are done. Root is 54.

### Cube root of 830584

Abacus Comment
ABCDEFG Cube root of 830584
830584 Enter number aligning first group with B
-729 Subtract 9^3=729 from BCD
101584 101: first remainder
9101584 Enter 9 in A as first root digit and append second group
9101584 a) Divide C-F by 9 (G will be the division remainder)
9112871 b) Divide C-F by 3
9376232 c) Divide B by 9 (A)
9416232 d) Subtract 4x4=16 from CD
94  232 e) Multiply 23x3 in EFG (adding to remainder in G)
94   71 f) Multiply 07x9 in EFG
94   64 g) Subtract 4^3= 64 from FG
94 Remainder 0; Done! Root is 94

Root is 94.

### Cube root of 666

Abacus Comment
ABCDEFG Cube root of 666
666 Enter 666 in BCD
+ (Unit rod)
-512 Subtract 8^3=512 from BCD
154 First remainder
8154 Enter 8 in A as the first root digit
8154000 Append 000 as new group
8154000 a) Divide B-F by 8 (A)
8192500 b) Divide B-F by 3
8641662 c) Divide B by 8 (A)
8781662 d) Subtract BxB=49 from CD
8732662 e) Multiply C-F by 3 in C-G
87 9800 f) Multiply C-F by 8 (A) in C-G
87 7840 g) Subtract B^3=343 from EFG
87 7497 Root so far 8.7, Remainder 7.497

Now we continue using [[../../Abbreviated operations/|Abbreviated operations]]. We need to divide the remainder (7497) by three times the square of the current root (${\displaystyle 3\cdot 87^{2}=22707}$ )

Abacus Comment
ABCDEFGHIJKLM
87 7497
87 7497------ Squaring 87
+49 7^2
+112 2*7*8
+64 8^2
87 7497  7569 multiplying by 3 (adding double)
+14
+10
+12
+18
87 7497 22707 dividing 7497/22707, two digits
...
8733 Root 8.733 (Compare to: ${\displaystyle {\sqrt[{3}]{666}}=8.7328917}$ )

### Cube root of 237176659 (three digits)

Abacus Comment
ABCDEFGHIJ Cube root of 237176659
237176659 Enter number aligning first group with B
-216 Subtract 6^3=216 from BCD
21176659 21: first remainder
21176659 Enter 6 in A as first root digit and append second group
6 21176659 a) Divide B-F by 6 (A)
6 35292659 b) Divide B-F by 3
6117633659 c) Divide B by 6 (A)
6157633659 d) Subtract BxB=1 from CD
6156633659 e) Multiply C-F by 3 in C-G
6116992659 f) Multiply C-F by 8 (A) in C-G
6110196659 g) Subtract B^3=343 from EFG
6110195659 Root so far 61, Remainder 10195
----------
6110195659 Append third group
6110195659 a) Divide C-H by 61 (AB)
6116714158 b) Divide C-H 3
6155713678 c) Divide C by 61 (AB)
6190813678 d) Subtract CxC=81 from EF
619   3678 e) Multiply D-H by 3 in D-I
619   1158 f) Multiply D-H by 61 (AB) in D-J
619    729 g) Subtract C^3=729 from HIJ
619    000 Done, no remainder!
---------- Root is 619

### Cube root of ${\displaystyle 48^{3}-1=110591}$  to eight digits

The first triplet 110 is between 64 and 125, so that the cube root of 110 591 is between 40 and 50. First root digit is 4

First digit: