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Introduction
Forewords edit
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 and traditional abacus edit
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.
Modern and traditional methods edit
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 ebooks 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 edit
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 edit
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:
 finger or bead mouvements
 hand displacements
 changes of direction
 skipping rods (i.e. changing hand position from a starting rod to other nonadjacent 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 edit
As usual, in this book we will use tables to describe procedures on the abacus, for example:
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, 
+8  add 8 to F 
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 edit
Soroban Trainer edit
If you are interested in traditional techniques but do not have a traditional abacus yet, you can use the JavaScript application
Soroban Trainer
 You can run it directly from GitHub in your browser
 or you can download it to your computer from the repository on GitHub.
References edit
 ↑ 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 ParisDiderot (Paris 7). p. 40.
{{cite book}}
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 ↑ Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 9780804800037
 ↑ 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}}
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 ↑ Sunzi 孫子 (3rd to 5th centuries AD). 孫子算經 (in Chinese).
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Addition and subtraction
Introduction edit
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.
5th lower bead edit
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 nonfundamental 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 edit
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 wellknown CanadianAmerican 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 edit
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 Englishspeaking Filipinos according to the author. Here are rules/rhymes/verses that appear on it and whose interpretation is left to the reader:
Addition rules  Subtraction rules 

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 nontrivial 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 edit
Use of the 5th lower bead edit
The specialized use of the 5th lower bead and the nonunique representation of numbers 5, 10 and 15 to simplify operations.
Extending the 123456789 exercise edit
A plethora of addition and subtraction exercises that can be done without an exercise sheet.
References edit
 ↑ 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).
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{{cite book}}
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{{citation}}
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(help)  ↑ Suzuki, Hisao (鈴木 久男) (1982). "Chuugoku ni okeru shuzan kagenhou 中国における珠算加減法". Kokushikan University School of Political Science and Economics (in Japanese). 57 (3). ISSN 05869749 – via Kokushikan.
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Addition and subtraction/Use of the 5th lower bead
T  T  T  T  T  T  T  T  1 
Introduction edit
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.
Yifu Chen, as part of his doctoral thesis^{[1]}, has systematically analyzed 16 classic works on the abacus: twelve Chinese books from the late Ming and Qing dynasties and four Japanese books from the Edo period in which addition and subtraction are studied. As a result, Chen finds four different modes of using the fifth bead in addition and two in subtraction. These modes range from intensive or systematic use of the fifth bead to sporadic use or no use at all. From all those texts, only one Chinese book from the late Ming dynasty make full use of the fifth bead, belonging to Chen’s Mode 1 in both addition and subtraction: Computational Methods with the Beads in a Tray (Pánzhū Suànfǎ 盤珠算法) by Xú Xīnlǔ 徐心魯 (1573)^{[2]}, by the way, the oldest extant book entirely devoted to the abacus. This fact should not lead us to the erroneous conclusion that the use of the fifth bead was a rarity of old times, since the books analyzed are neither treatises nor compendia on stateoftheart abacus computation, but introductory manuals or textbooks for learning its use. Rather, it seems that the use or not of the fifth bead in these books corresponds more precisely to the didactic objective pursued by the authors, and that only Xu Xinlu considered it interesting to demonstrate it thoroughly from the beginning and included it in the syllabus of his course. It is mainly thanks to this work that we can rescue the traditional use of the fifth bead to simplify operations.
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 edit
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).
Rules for addition edit
 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 9783319733968 {{citation}}
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The rationale behind edit
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 leftright hand displacement span
 to avoid any “carry run” to the left (think of 99999+1=999T0 instead of 99999+1=100000)
The advantage edit
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 edit
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 edit
 s1 Always use lower fives (F) instead of upper fives (5). For instance: 72 = F
 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. 277 = 1T
should be preferred to 277 = 20
.
Remark: These two rules do not apply on rods where you are borrowing from, i.e. 1127 = 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 627 = 5F (not FF).
A  B  A  B  A  B  

AB  7 =  not  
6  2  5  F  F  F 
The rationale behind edit
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 nonadjacent columns or skipping rods.
The advantage edit
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 edit
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 edit
It was common in ancient books on the abacus to demonstrate addition and subtraction using the wellknown 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 kagenhou 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 1217, 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.
Additional rules edit
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:
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
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.
About the advantage edit
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 edit
Addition edit
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 edit
ABCDEFGHI ABCDEFGHI ABCDEFGHI ABCDEFGHI ABCDEFGHI      TTTTTTTT1 9876F4312 864197523 740740734 61728394F 9TTTTTTT1 A1 8876F4312 A1 764197523 A1 640740734 A1 F1728394F A1 98TTTTTT1 B2 8676F4312 B2 744197523 B2 620740734 B2 49728394F B2 987TTTTT1 C3 8646F4312 C3 741197523 C3 617740734 C3 49428394F C3 9876TTTT1 D4 8642F4312 D4 740797523 D4 617340734 D4 49388394F D4 9876FTTT1 E5 8641T4312 E5 740747523 E5 617290734 E5 49383394F E5 9876F4TT1 F6 864198312 F6 740741523 F6 617284734 F6 49382794F F6 9876F43T1 G7 864197612 G7 740740823 G7 617283T34 G7 49382724F G7 9876F4321 H8 864197532 H8 740740743 H8 6172839F4 H8 49382716F H8 9876F4312 I9 864197523 I9 740740734 I9 61728394F I9 493827156 I9 ABCDEFGHI ABCDEFGHI ABCDEFGHI ABCDEFGHI     493827156 36T370367 246913578 123456789 393827156 A1 26T370367 A1 146913578 A1 023456789 A1 373827156 B2 24T370367 B2 126913578 B2 003456789 B2 36T827156 C3 247370367 C3 123913578 C3 000456789 C3 36T427156 D4 246970367 D4 123F13578 D4 000056789 D4 36T377156 E5 246920367 E5 123463578 E5 000006789 E5 36T371156 F6 246914367 F6 123457578 F6 000000789 F6 36T370456 G7 246913667 G7 123456878 G7 000000089 G7 36T370376 H8 246913587 H8 123456798 H8 000000009 H8 36T370367 I9 246913578 I9 123456789 I9 000000000 I9
Table 2: The 123456789 exercise over a background edit
(See also the next chapter)
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 edit
 ↑ 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.
 ↑ 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 parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ Suzuki, Hisao (1982). "Zhusuan addition and subtraction methods in China". Kokushikan University School of Political Science and Economics (in Japanese). 57 – via Kokushikan.
Further readings edit
 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}}
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External resources edit
You can practice using the fifth bead online with Soroban Trainer (see chapter: [[../../Introduction#External resourcesIntroduction]]) using this file 1234567895bead.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 edit
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 nontrivial 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 edit
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.
+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 
+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 edit
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.
+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 
+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 edit
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.
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 
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 
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 
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 edit
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 edit
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]].
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 edit
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 edit
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 multidigit 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 edit
Modern and traditional division; close relatives edit
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) edit
Here, we will see how to use the traditional method.
Learning the division table edit
It contains some indications that may make it easier for you to memorize the division table.
Dealing with overflow edit
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 edit
Division tables can be coined for multidigit divisors, allowing dividing by them without resorting to the multiplication table.
Traditional division examples edit
A basic set of examples to illustrate all of the above.
Division by powers of two edit
Another traditional division method different from 帰除法 based in fractions; a form of division in situ.
References edit
 ↑ Suzuki, Hisao (鈴木 久男) (1980). "Chūgoku ni okeru josanhō 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 05869749 – via Kokushikan.
{{cite journal}}
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ignored (transtitle=
suggested) (help)  ↑ Suzuki, Hisao (鈴木 久男) (1981). "Chūgoku ni okeru josanhō 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 05869749 – via Kokushikan.
{{cite journal}}
: Unknown parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ Momokawa, Jihei (百川治兵衛) (1645). Kamei Zan (亀井算) (in Japanese).
{{cite book}}
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 ↑ 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).
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Division/Modern and traditional division; close relatives
Introduction edit
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 edit
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.
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 
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 
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
times  chunks 

1  35 
2  70 
3  105 
4  140 
5  175 
6  210 
7  245 
8  280 
9  315 
then
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 onedigit 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) edit
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 
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 
Traditional Division (帰除法) edit
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
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) 
+135  Revising up 
35 34 35  remainder equal to divisor (35) 
+135  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 edit
When we learn the multiplication table we memorize a sequence of phrases like:
 “nine times nine , eightyone”
 “nine times eight, seventytwo”
 ...
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  

English  nine times nine  ➡  eightyone 
Chinese  九九  ➡  八十一 
Japanese  くく  ➡  はちじゅういち 
Symbolic  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 nonindexed 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) edit
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!).
edit
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.
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) 
+135  Revising up 
35 34 35  remainder equal to divisor (35) 
+135  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 edit
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 edit
 ↑ 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}}
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suggested) (help)  ↑ Shinoda, Shosaku (篠田正作) (1895). Jitsuyo Sanjutsu (実用算術) (in Japanese).
{{cite book}}
: Unknown parametertrans_title=
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suggested) (help)  ↑ Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 9780804802789
 ↑ ^{a} ^{b} "The Definitive Higher Math Guide on Integer Long Division (and Its Variants)". Math Vault. Archived from the original on May 14, 2021.
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Further readings edit
 "The Definitive Higher Math Guide on Integer Long Division (and Its Variants)". Math Vault. Archived from the original on May 14, 2021.
{{cite web}}
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suggested) (help)  *Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73 deals with traditional division
 Totton Heffelfinger (2013). "Suan Pan and the Unit Rod  Division". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 3, 2021.
{{cite web}}
: Unknown parameteraccesdate=
ignored (accessdate=
suggested) (help)  Totton Heffelfinger (2013). "Short Division Techniques  Chinese Suan Pan". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 3, 2021.
{{cite web}}
: Unknown parameteraccesdate=
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suggested) (help)  Totton Heffelfinger (2013). "Long Division Techniques  Chinese Suan Pan". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 3, 2021.
{{cite web}}
: Unknown parameteraccesdate=
ignored (accessdate=
suggested) (help)  Totton Heffelfinger (2013). "Chinese Division Rules on a Soroban". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 3, 2021.
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Division/Guide to traditional division (帰除法)
Introduction edit
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 digitbydigit 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 edit
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
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:
 the dividend is greater than or equal to the divisor (ex. )
 the dividend is less than the divisor and the first digit of the divisor is equal to the first digit of the dividend (ex. )
 the dividend is less than the divisor and the first digit of the divisor is greater than the first digit of the dividend (ex. )
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 nondiagonal 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.
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? edit
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 onedigit 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) edit
It is am assumed throughout this textbook that the reader has already studied the modern abacus method, as typified in the work of Takashi Kojima^{[2]}. In the following examples we will illustrate traditional division using the same division layout that you are already familiar with so that you can more easily follow them and use your usual 4+1 type abacus if you want. We will call this layout Modern Division Arrangement (MDA), but this is not the way division was traditionally organized on the abacus. Later, I will introduce the Traditional Division Arrangement (TDA) which, as we will see, it has some advantages and some disadvantages, including the need (or at least the convenience) of using a specialized abacus with additional upper beads.
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
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
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 edit
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 edit
Abacus  Comment 

ABCDEFGHIJ  (Divisor not indicated) 
123456789  Rule 1/9>1+1 
+1  enter quotient 1 into A 
1  clear B 
+1  add remainder 1 to adjacent digit 
1 33456789  Rule 3/9>3+3 
13 6456789  Rule 6/9>6+6 
1361056789  
+19  revising up 
137 156789  Rule 1/9>1+1 
1371 66789  Rule 6/9>6+6 
1371612789  
+19  revising up 
13717 3789  Rule 3/9>3+3 
1371731089  
+19  revising up 
137174 189  Rule 1/9>1+1 
1371741 99  
+19  revising up 
1371742 9  
+19  revising up 
13717421  Done! 
Example 123456789÷5=24691357.8 edit
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  
+15  revising up 
2469 6789  
+15  revising up 
24691 1789  Rule 1/5>2+0 
246912 789  
+15  revising up 
246913 289  Rule 2/5>4+0 
2469134 89  
+15  revising up 
2469135 39  Rule 3/5>6+0 
24691356 9  
+15  revising up 
24691357 4  Rule 3/5>6+0 
246913578  Done! 
Example 123456789÷2=61728394.5 edit
Abacus  Comment 

ABCDEFGHIJ  (Divisor not indicated) 
123456789  Rule 1/2>5+0 
5 23456789  
+12  revising up 
6 3456789  
+12  revising up 
61 1456789  Rule 1/2>5+0 
615 456789  
+24  revising up twice 
617 56789  
+24  revising up twice 
6172 16789  Rule 1/2>5+0 
61725 6789  
+36  revising up three times 
61728 789  
+36  revising up twice 
617283 189  Rule 1/2>5+0 
6172835 89  
+48  revising up four times 
6172839 9  
+48  revising up four times 
61728394 1  Rule 1/2>5+0 
617283945  Done! 
Multi Digit divisors edit
Consider, for example, , 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, , where is the divider (9) and 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:
 First we consider only the divider 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 and add the remainder (from the rule) to the adjacent column
 Then we subtract the chunk from the remainder if we can; otherwise we have to revise down and restore to the remainder using the following rules:
While dividing by  Revise q to  Add to remainder 

1  q1  +1 
2  q1  +2 
3  q1  +3 
4  q1  +4 
5  q1  +5 
6  q1  +6 
7  q1  +7 
8  q1  +8 
9  q1  +9 
These rules are for twodigit divisors, for divisors with more digits things may be more complicated, as in MD (see example below). Let us see the above case
Example 359936÷9728=37 edit
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=2184^{a} from IL 
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 JM 
9728 36 9728  revising up 
+19728  
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… edit
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… edit
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. 
Traditional Division Arrangement (TDA) edit
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]},
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]},
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 13rod 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 edit
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:
Abacus  Comment 

ABCDEFGHI  
16 128  Divide in situ by 2 
8 64  Rule 6/8>7+4 
8 7 8  
+18  revising up 
8 8  Done! 
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 edit
 ↑ "割り算九九". Japanese Wikipedia.
{{cite web}}
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ignored (language=
suggested) (help); Unknown parameteraccesdate=
ignored (accessdate=
suggested) (help); Unknown parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ ^{a} ^{b} Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 9780804802789
 ↑ 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 parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ Yoshida, Mitsuyoshi (吉田光由) (1634). Jinkoki (塵劫記) (in Japanese).
{{cite book}}
: Unknown parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ Shinoda, Shosaku (篠田正作) (1895). Jitsuyo Sanjutsu (実用算術) (in Japanese).
{{cite book}}
: Unknown parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ 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 9789811565250
 ↑ 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 parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ Kwa Tak Ming (1922), The Fundamental Operations in Bead Arithmetic, How to Use the Chinese Abacus (PDF), San Francisco: Service Supply Co.
Further readings edit
 Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73 deals with traditional division
 Totton Heffelfinger (2013). "Suan Pan and the Unit Rod  Division". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 3, 2021.
{{cite web}}
: Unknown parameteraccesdate=
ignored (accessdate=
suggested) (help)  Totton Heffelfinger (2013). "Short Division Techniques  Chinese Suan Pan". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 3, 2021.
{{cite web}}
: Unknown parameteraccesdate=
ignored (accessdate=
suggested) (help)  Totton Heffelfinger (2013). "Long Division Techniques  Chinese Suan Pan". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 3, 2021.
{{cite web}}
: Unknown parameteraccesdate=
ignored (accessdate=
suggested) (help)  Totton Heffelfinger (2013). "Chinese Division Rules on a Soroban". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 3, 2021.
{{cite web}}
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Division/Learning the division table
Memorization of the division table. edit
The division table contains 45 rules, including the 9 diagonal elements for multidigit 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: ( : quotient, : remainder, digits from 1 to 9) or, equivalently
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 edit
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.
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 edit
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 subseries 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 edit
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 edit
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 multiplicationdivision table edit
What follows is a simple historical note with little or no practical relevance.
The multiplication table in the English language contains all the 81 twodigit 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 multiplicationdivision 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.
While dividing by  Revise q to  Add to remainder 

1  q1  +1 
2  q1  +2 
3  q1  +3 
4  q1  +4 
5  q1  +5 
6  q1  +6 
7  q1  +7 
8  q1  +8 
9  q1  +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 edit
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 
Please try them sometime during your practice!
Division/Dealing with overflow
Introduction edit
Excluding the socalled "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]},
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]}
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 13rod 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]}.
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 edit
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 edit
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:
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.
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 edit
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 edit
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)
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 edit
And now on a 4+1 abacus. We need to use the suspended group of four lower beads as a code for 9:
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 edit
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 edit
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:
 All way, effectively adding it as a carry to the quotient
 Only half way, leaving a suspended lower bead
 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.
Division/Traditional division examples
Onedigit divisors (short division) edit
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 edit
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 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend in AI, 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 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend in AI, 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 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend in AI, 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 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend in AI, 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 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend in AI, 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 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend in AI, 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 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend in AI, 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 
Multidigit divisors (long division) edit
Division of 998001 by 999 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend in AF, divisor 8 in KM 
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 

On a 5+2 abacus

On a 5+1 abacus

On a 5+3 abacus
Division of 888122 by 989 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend 888122 in AF, divisor 989 in KM 
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 edit
Abacus  Comment 

ABCDEFGHIJKLM  Dividend 888122in AF, divisor 898 in KM 
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 edit
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 approximated^{a} 
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 edit
 ↑ 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 parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ 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 ParisDiderot (Paris 7).
{{cite book}}
: Unknown parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 9780804802789
 ↑ Yoshida, Mitsuyoshi (吉田光由) (1634). Jinkoki (塵劫記) (in Japanese).
{{cite book}}
: Unknown parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ 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 parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ "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.
 ↑ Murakami, Masaaki (20200629). "The 5th lower bead". (Web link). Retrieved on 20210813.
 ↑ 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 parametertrans_title=
ignored (transtitle=
suggested) (help)  ↑ Yoshida, Mitsuyoshi (吉田光由) (1634). Jinkoki (塵劫記) (in Japanese).
{{cite book}}
: Unknown parametertrans_title=
ignored (transtitle=
suggested) (help)
External resources edit
You can practice traditional division online with Soroban Trainer (see chapter: [[../../Introduction#External resourcesIntroduction]]) using this file kijoho1digit.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 edit
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:
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?
Abacus  Comment 

ABCDEFGHIJKLM  
36525 1000000  Use rule: 1/36525>2+26950 on column G 
36525 2000000  change 1 in G into 2 
+26950  add 26950 to HL 
36525 2269500  Use rule: 2/36525>5+17375 on column H 
36525 2569500  change 2 in H into 5 
+17375  add 17375 to IM 
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 fivedigit divisor without using the multiplication table!
Twodigit division tables edit
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 
21  22  23  24  25  26  27  28  29  
1  4+16  4+12  4+08  4+04  4+00  3+22  3+19  3+16  3+13 
2  9+11  9+02  8+16  8+08  8+00  7+18  7+11  7+04  6+26 
31  32  33  34  35  36  37  38  39  
1  3+07  3+04  3+01  2+32  2+30  2+28  2+26  2+24  2+22 
2  6+14  6+08  6+02  5+30  5+25  5+20  5+15  5+10  5+05 
3  9+21  9+12  9+03  8+28  8+20  8+12  8+04  7+34  7+27 
41  42  43  44  45  46  47  48  49  
1  2+18  2+16  2+14  2+12  2+10  2+08  2+06  2+04  2+02 
2  4+36  4+32  4+28  4+24  4+20  4+16  4+12  4+08  4+04 
3  7+13  7+06  6+42  6+36  6+30  6+24  6+18  6+12  6+06 
4  9+31  9+22  9+13  9+04  8+40  8+32  8+24  8+16  8+08 
51  52  53  54  55  56  57  58  59  
1  1+49  1+48  1+47  1+46  1+45  1+44  1+43  1+42  1+41 
2  3+47  3+44  3+41  3+38  3+35  3+32  3+29  3+26  3+23 
3  5+45  5+40  5+35  5+30  5+25  5+20  5+15  5+10  5+05 
4  7+43  7+36  7+29  7+22  7+15  7+08  7+01  6+52  6+46 
5  9+41  9+32  9+23  9+14  9+05  8+52  8+44  8+36  8+28 
61  62  63  64  65  66  67  68  69  
1  1+39  1+38  1+37  1+36  1+35  1+34  1+33  1+32  1+31 
2  3+17  3+14  3+11  3+08  3+05  3+02  2+66  2+64  2+62 
3  4+56  4+52  4+48  4+44  4+40  4+36  4+32  4+28  4+24 
4  6+34  6+28  6+22  6+16  6+10  6+04  5+65  5+60  5+55 
5  8+12  8+04  7+59  7+52  7+45  7+38  7+31  7+24  7+17 
6  9+51  9+42  9+33  9+24  9+15  9+06  8+64  8+56  8+48 
71  72  73  74  75  76  77  78  79  
1  1+29  1+28  1+27  1+26  1+25  1+24  1+23  1+22  1+21 
2  2+58  2+56  2+54  2+52  2+50  2+48  2+46  2+44  2+42 
3  4+16  4+12  4+08  4+04  4+00  3+72  3+69  3+66  3+63 
4  5+45  5+40  5+35  5+30  5+25  5+20  5+15  5+10  5+05 
5  7+03  6+68  6+62  6+56  6+50  6+44  6+38  6+32  6+26 
6  8+32  8+24  8+16  8+08  8+00  7+68  7+61  7+54  7+47 
7  9+61  9+52  9+43  9+34  9+25  9+16  9+07  8+76  8+68 
81  82  83  84  85  86  87  88  89  
1  1+19  1+18  1+17  1+16  1+15  1+14  1+13  1+12  1+11 
2  2+38  2+36  2+34  2+32  2+30  2+28  2+26  2+24  2+22 
3  3+57  3+54  3+51  3+48  3+45  3+42  3+39  3+36  3+33 
4  4+76  4+72  4+68  4+64  4+60  4+56  4+52  4+48  4+44 
5  6+14  6+08  6+02  5+80  5+75  5+70  5+65  5+60  5+55 
6  7+33  7+26  7+19  7+12  7+05  6+84  6+78  6+72  6+66 
7  8+52  8+44  8+36  8+28  8+20  8+12  8+04  7+84  7+77 
8  9+71  9+62  9+53  9+44  9+35  9+26  9+17  9+08  8+88 
91  92  93  94  95  96  97  98  99  
1  1+09  1+08  1+07  1+06  1+05  1+04  1+03  1+02  1+01 
2  2+18  2+16  2+14  2+12  2+10  2+08  2+06  2+04  2+02 
3  3+27  3+24  3+21  3+18  3+15  3+12  3+09  3+06  3+03 
4  4+36  4+32  4+28  4+24  4+20  4+16  4+12  4+08  4+04 
5  5+45  5+40  5+35  5+30  5+25  5+20  5+15  5+10  5+05 
6  6+54  6+48  6+42  6+36  6+30  6+24  6+18  6+12  6+06 
7  7+63  7+56  7+49  7+42  7+35  7+28  7+21  7+14  7+07 
8  8+72  8+64  8+56  8+48  8+40  8+32  8+24  8+16  8+08 
9  9+81  9+72  9+63  9+54  9+45  9+36  9+27  9+18  9+09 
Some examples edit
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 
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.
:)👿
Further reading edit
 Murakami, Masaaki (2020). "Specially Crafted Division Tables" (PDF). 算盤 Abacus: Mystery of the Bead. Archived from the original (PDF) on August 1, 2021.
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Division/Division by powers of two
Introduction edit
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 tabulated (or memorized) where is one of such powers of two or five.
For instance, given
Then
Which can easily be done on the abacus by working from right to left. For each digit of the numerator:
 Clear the digit
 Add the fraction corresponding to the working digit to the abacus starting with the column it occupied
Abacus  Comment 

ABCDEF  
+  Unit rod 
137  enter 137 on AC as a guide 

clear 7 in C 
+0875  add 7/8 to CF 
130875  

clear 3 in B 
+0375  add 3/8 to BE 
104625  

clear 1 in A 
+0125  add 1/8 to AD 
17125  Done! 
+  unit rod 
We only need to have the corresponding fractions tabulated or memorized, as in the table below.
Powers of two edit
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 edit
D  D/2  D/4  D/8  D/16^{a}  D/32^{a}  D/64^{a} 

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 edit
ABCD 
+^{b} 
137 

+35 

+15 

+05 
+^{b} 
0685 
68.5 

ABCDE 
+^{b} 
137 

+175 

+075 

+025 
+^{b} 
03425 
34.25 

ABCDEF 
+^{b} 
137 

+0875 

+0375 

+0125 
+^{b} 
017125 
17.125 

ABCDEF 
+^{b} 
137 

+4375 

+1875 

+0625 
+^{b} 
085625 
8.5625 

ABCDEFG 
+^{b} 
137 

+21875 

+09375 

+03125 
+^{b} 
0428125 
4.28125 

ABCDEFGH  
+^{b}  
137  

Clear 7 in C 
+109375  

Clear 3 in B 
+046875  

Clear 1 in A 
+015625  
+^{b}  
02140625  
2.140625 

^b "+" indicates the unit rod position.
Division by 2 in situ edit
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 halfremainder 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:
 clearing the digit
 adding its half starting with the column it occupied
For instance, 123456789/2:
Abacus  Comment 

ABCDEFGHIJ  
123456789  

Clear 9 in I 
+45  Add its half to IJ 
1234567845  

Clear 8 in H 
+40  Add its half to HI 
1234567445  

Clear 7 in G 
+35  Add its half to GH 
1234563945  

Clear 6 in F 
+3  Add its half to FG 
1234533945  

Clear 5 in E 
+25  Add its half to EF 
1234283945  

Clear 4 in D 
+2  Add its half to DE 
1232283945  

Clear 3 in C 
+15  Add its half to CD 
1217283945  

Clear 2 in B 
+1  Add its half to BC 
1117283945  

Clear 1 in A 
+05  Add its half to AB. 
617283945  Done! 
The unit rod does not change in this division.
Powers of five edit
Table of fractions edit
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? edit
Let's take an example: . We do this multiplication by adding the 12 partial products that result from the expansion:
That is, all the products listed in this table:
✕  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 (12 factorial) 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 edit
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: and now subtracting . On the abacus:
Abacus  Comment 

ABC  
422  
+3  Add 313 to ABC 
+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 where is the quotient of dividing by and is the remainder, we can reverse the operation in the form: for example: where 65 is the quotient and 47 the remainder and we can reverse the operation in the form . On the abacus, using modern division and multiplication methods:
Abacus  Comment 

ABCDEFGHI  4727÷72 
72 4727  Dividend:FI, 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 
+35  Add 5✕7=35 to GH 
+10  Add 5✕2=10 to HI 
72 65407  
72 6 407  Clear F 
+42  Add 6✕7=42 to FG 
+12  Add 6✕2=12 to GH 
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:
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:FI, 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:
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 
+35  Add 5✕7=35 to GH 
+10  Add 5✕2=10 to HI 
72 6407  
72 407  Clear F and remember 6 
+42  Add 6✕7=42 to FG 
+12  Add 6✕2=12 to GH 
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 
+10  Add 5✕2=10 to HI 
+35  Clear G and add 5✕7=35 to GH 
72 6407  
72 407  Clear F and remember 6 
+12  Add 6✕2=12 to GH 
+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.
Traditional multiplication method edit
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 multidigit 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: AD 
4567 23  Head: A (4), Body: BCD (567) 
+15  Add 3✕5=15 to IJ 
+18  Add 3✕6=18 to JK 
+21  Add 3✕5=21 to KL 
+12  Clear H and add 3✕4=12 to HI 
4567 213701  
+10  Add 2✕5=10 to HI 
+12  Add 2✕6=12 to IJ 
+14  Add 2✕7=14 to JK 
+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:AC, Multiplier: IK 
999 999  Head: I (9), Body: JK (99) 
+81  Add 9✕9=81 to DE 
+81  Add 9✕9=81 to EF 
+81  Clear C and add 9✕9=81 to CD 
998991 999  
+81  Add 9✕9=81 to CD 
+81  Add 9✕9=81 to DE 
+81  Clear B and add 9✕9=81 to BC 
988901 999  (overflow!) 
+81  Add 9✕9=81 to BC 
+81  Add 9✕9=81 to CD 
+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:
Abacus  Comment 

ABCDEFGHIJK  Multiplicand:AC, Multiplier: IK 
999 999  
+81  Clear C, remember 9 and add 9✕9=81 to CD 
+81  Add 9✕9=81 to DE 
+81  Add 9✕9=81 to EF 
998991 999  
+81  Clear B, remember 9 and add 9✕9=81 to BC 
+81  Add 9✕9=81 to CD 
+81  Add 9✕9=81 to DE 
998901 999  
+81  Clear A, remember 9 and add 9✕9=81 to AB 
+81  Add 9✕9=81 to BC 
+81  Add 9✕9=81 to CD 
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.
The 123456789 exercise in multiplication edit
References edit
 ↑ Martzloff, JeanClaude (2006), A history of chinese mathematics, Springer, p. 221, ISBN 9783540337829
 ↑ Cabrera, Jesús (2021), "Tide Abacus", jccAbacus, retrieved 4 August 2021
 ↑ Williams, Samuel Wells; Morrison, John Robert (1856), A Chinese commercial guide, Canton: Printed at the office of the Chinese Repository, p. 298
 ↑ 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 parameteraccesdate=
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 ↑ 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 parameteraccesdate=
ignored (accessdate=
suggested) (help)  ↑ 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 9783319682938
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Further reading edit
 Kojima, Takashi (1963), "III Other multiplication methods", Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 9780804800037
 Totton Heffelfinger (2004). "Traditional Multiplication techniques for Chinese Suan Pan  The Extra Bead and the Suspended Bead". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021.
{{cite web}}
: Unknown parameteraccesdate=
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 Totton Heffelfinger (2013). "Suan Pan and the Unit Rod  Multiplication". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021.
{{cite web}}
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Roots
Introduction edit
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 edit
This part of the book consists of the following chapters:
Checking your exercises edit
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 edit
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 edit
File knott.bc edit
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/rootslaknott.html
as a reference.
Jesús Cabrera, June 2021
CC0 1.0 Universal (CC0 1.0) Public Domain Dedication
Use at your oun risk!
*/
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=b1;print "Revising down, b = ",b, "\n"
tf = div b*a b^2/2
}
print "New root: ", a+b,", New halfremainder: ", 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 edit
/* 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 halfremainder: ", 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 halfremainder: 2.00000000000000000000 ================== New dividend: 203.5 Revising down, b = 4 New root: 44, New halfremainder: 35.5 ================== New dividend: 3551.5 Revising down, b = 7 New root: 447, New halfremainder: 447.0 ================== New dividend: 44700.0 Revising down, b = 9 New root: 4479, New halfremainder: 4429.5 ================== New dividend: 442950.0 New root: 44799, New halfremainder: 39799.5 ================== New dividend: 3979950.0 New root: 447998, New halfremainder: 395998.0 ================== New dividend: 39599800.0 New root: 4479988, New halfremainder: 3759928.0 ==================
File cr488931400152.bc edit
/* 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 edit
 ↑ 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 edit
Let be the number of which we want to obtain the square root ; Let's consider its decimal expansion, for example: . Let's separate its digits into groups of two around the decimal point in the following way
or, in other words, let's define the sequence of integers
and let's build the sequence recursively from
and let be the integer part of the square root of
i.e. is the largest integer whose square is not greater than . Finally, let us call remainders to the differences
For our example we have:
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, grows as (two more digits in each step), in fact the sequence : (0, 400, 456, 456.78, 456.7890, etc.) tends to or . By comparison, , as the integer part of the square root of , grows only as (one digit more each step). As is the largest integer whose square is not greater than we have as above but
by definition of , or
multiplying by
but as grows only as , the second term tends to zero as . With this
and so that we have
For other numbers, the above factors are: and , where is the number of twodigit groups to the left of the decimal point, negative if it is followed by 00 groups (ex. for , for , etc.).
This is the basis for traditional manual square root methods.
Procedure edit
We start with , , , .
First digit edit
1  1 
2  4 
3  9 
4  16 
5  25 
6  36 
7  49 
8  64 
9  81 
For and , it is trivial to find such that its square does not exceed 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 .
Rest of digits edit
For , we have as defined above and we try to build in the way:
where is a onedigit integer ranging from 0 to 9. To obtain we have to choose the greatest digit from 0 to 9 such that:
or
if we write . Expanding the binomial we have
or
The left side of the above expression may be seen as simply the previous remainder with the next twodigits 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 we have 56 on the left and the above expression is
which holds only for or so that 1 is our next root digit but, how can we proceed in the general case without having to systematically explore every possibility ( )?
Here Knott^{[1]} distinguishes two different approaches:
 Preparing the divisor
 Preparing the dividend.
Preparing the divisor edit
This corresponds with the above expression
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
but since we don't know b yet, we approximate it using only the main part of the divisor
This gives us a guess as to what the value of b might be, but we need:
 Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
 Obtain the next remainder to prepare the calculation of the next digit of the root.
Both steps require subtracting or and from ; checking that we are not going to negative values and that what remains is less than (otherwise we would have to revise up or down). If we do this correctly, what we are left with is the new remainder . It should be noted that, as we proceed in the calculations ( increasing) is progressively a smaller and smaller contribution to the divisor ; 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:
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 EH 
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 edit
Starting again with
dividing by 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 halfremainders 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 we obtain a guess of by simply dividing the extended halfremainder by the previous root (in fact ). After that, we need again:
 Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
 Obtain the next halfremainder to prepare for the next digit of the root.
This is done by subtracting and from the half remainder, and this makes it convenient to memorize the following table of semisquares:
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 edit
Here three examples are presented, for additional examples please see Further reading and specially External resources below.
Square root of 961 edit
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 semiremainder.
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).
Using traditional division 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 BE 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 situinsitu]] division, as explained in chapter: [[../../Division/Division by powers of twoDivision by powers of two]], to obtain the semiremainder.
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:
Abacus  Comment 

ABCDE  
30305  
+1  Divide halfremainder BE by 3. Revising up B 
3  
31005  
05  Subtract b^2/2 =0.5 from D 
31000  Halfremainder is 0, Done! Root is 31 
31  Root is 31 
Square root of 998001 edit
Abacus  Comment 

ABCDEFG  
998001  Enter the radicand 
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 edit
Our example above...
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 rootAbbreviated operations]] to see how to quickly approximate the next 4 digits.
Conclusion edit
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 Halfremainder method and is by far the most convenient to use, at least for two reasons:
 The root, and not its double, replaces the operand (radicand) as in the rest of operations on the abacus.
 (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 halfremainder or preparing the dividend method is undeniable.
References edit
 ↑ Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73
 ↑ Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 9780804800037
 ↑ 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.
Further reading edit
 Siqueira, Edvaldo; Heffelfinger, Totton. "Kato Fukutaro's Square Roots". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021.
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 Treadwell, Steve (2015). "Improvements to the Kato Method for Finding Square Roots" (PDF). 算盤 Abacus: Mystery of the Bead. Archived from the original (PDF) on August 1, 2021.
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suggested) (help)  Square root (半九九法); the traditional way using traditional division in jccAbacus
External resources edit
 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 edit
Let be the number of which we want to obtain the cube root ; Let's consider its decimal expansion, for example: . Let's separate its digits into groups of three around the decimal point in the following way
or, in other words, let's define the sequence of integers
and let's build the sequence recursively from
and let be the integer part of the cube root of
i.e. is the largest integer whose cube is not greater than . Finally, let us call remainders to the differences
For our example we have:
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, grows as (three more digits in each step), in fact the sequence , i.e. 0, 400, 456, 456.789, 456.789012, etc. tends to ( ). By comparison, , as the integer part of the cube root of , grows only as (one digit more each step). As is the largest integer whose square is not greater than we have as above but
by definition of , or
multiplying by
but as grows only as , the second term tends to zero as .
and so that we have
For other numbers, the above factors are: and , where is the number of threedigit groups to the left of the decimal point, negative if it is followed by 000 groups (ex. for , for , etc.).
This is the basis for traditional manual cube root methods.
Procedure edit
We start with .
First digit edit
1  1 
2  8 
3  27 
4  64 
5  125 
6  216 
7  343 
8  512 
9  729 
For . It is trivial to find such that its square does not exceed through the use of the following table of cubes that can be easily retained in memory. In the case of the example it is
Rest of digits edit
For , we have as defined above and we try to build in the way:
where is a onedigit integer ranging from 0 to 9. To obtain we have to choose the greatest digit from 0 to 9 such that:
or
if we write . Expanding the binomial we have
or
The left side of the above expression may be seen as simply the previous remainder with the next threedigits group appended to it. If we evaluate the term on the right for each value of and compare with the left term we have:
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 ( )?
Here Knott^{[1]} distinguishes two different approaches:
 Preparing the divisor
 Preparing the dividend
Preparing the divisor edit
This correspond with the above expression
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, is the first digit of the division
but since we don't know b yet, we approximate it using only the main part of the divisor
This gives us a guess as to what the value of might be, but we need:
 Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
 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 edit
Starting again with
we prepare the dividend by dividing (the next threedigits group appended to the previous remainder) by
As usual, we don't know and we can't evaluate the parentheses on the right, but we can get a clue about by approximating the parentheses by its main part and use it as a trial divisor.
so that
After that, we need again:
 Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
 Obtain the next remainder to prepare for the obtention of the next digit of the root by evaluating .
Please note that:
 Divisor 3 is involved in the prepared dividend and this leads to nonfinite decimal fractions.
 The division by not only worsens the above, but also makes the prepared dividend specific to the current step, since the value of 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 preparationrestoration of the dividend that, following Knott, can be represented by the following scheme:
 a) Divide by .
 b) Divide by 3.
 c) Obtain as the first digit of the division of the above by .
 d) Subtract (Equivalent to subtracting and in ).
 e) Multiply by 3.
 f) Multiply by .
 g) Subtract .
In our example ( ), using traditional division and traditional division arrangement (like Knott does), working the two first digits:
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 BF by 7^{1} 
7162554  b) Divide BF by 3^{2} 
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 threedigit 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 edit
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 edit
The following examples are all worked using traditional division and traditional division arrangement. Components of the dividend preparationrestoration cycle are labelled with a), b), etc as detailed above.
Cube root of 157464 edit
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 CF by 5 (G will be the division remainder) 
5 64924  b) Divide CF 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 edit
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 CF by 9 (G will be the division remainder) 
9112871  b) Divide CF 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 edit
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 BF by 8 (A) 
8192500  b) Divide BF by 3 
8641662  c) Divide B by 8 (A) 
8781662  d) Subtract BxB=49 from CD 
8732662  e) Multiply CF by 3 in CG 
87 9800  f) Multiply CF by 8 (A) in CG 
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 ( )
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: ) 
Cube root of 237176659 (three digits) edit
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 BF by 6 (A) 
6 35292659  b) Divide BF by 3 
6117633659  c) Divide B by 6 (A) 
6157633659  d) Subtract BxB=1 from CD 
6156633659  e) Multiply CF by 3 in CG 
6116992659  f) Multiply CF by 8 (A) in CG 
6110196659  g) Subtract B^3=343 from EFG 
6110195659  Root so far 61, Remainder 10195 
  
6110195659  Append third group 
6110195659  a) Divide CH by 61 (AB) 
6116714158  b) Divide CH 3 
6155713678  c) Divide C by 61 (AB) 
6190813678  d) Subtract CxC=81 from EF 
619 3678  e) Multiply DH by 3 in DI 
619 1158  f) Multiply DH by 61 (AB) in DJ 
619 729  g) Subtract C^3=729 from HIJ 
619 000  Done, no remainder! 
  Root is 619 
Cube root of to eight digits edit
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:
Abacus  Comment 

ABCDEFG  Cube root of 110591 
110591  Enter number aligning first group with B 
64  Subtract 6^3=216 from BCD 
46591  46: first remainder 
46591  Enter 4 in A as first root digit and append second group 
4 46591  OK 1st digit! 
Second digit:
Abacus  Comment 

ABCDEFG  
4 46591  a) Divide BF by 4 (A) 
4116473  b) Divide BF by 3 
4388234  c) Divide B by 4 (A) 
4868234  d) Subtract BxB=64 from CD 
48 4234  e) Multiply CF by 3 in CG 
48 1273  f) Multiply CF by 4 (A) in CG 
48 511  g) Cannot subtract 8^3=512 from EFG! Going back (See note at the end) 
48 511  f) Divide CF by 4 (A) 
48 1273  e) Divide CF by 3 
48 4234  d) Add 8x8=64 in CD 
4868234  c) Revise down B 
1  
+4  
47T8234  d) Subtract BxB=49 from CD (T=10) 
4759234  e) Multiply CF by 3 in CG 
4717773  f) Multiply CF by 4 (A) in CG 
47 7111  g) Subtract B^3=343 from EFG 
47 6768  OK 2nd digit! Remainder 6768 
Third digit:
Abacus  Comment 

ABCDEFGHIJ  
47 6768000  Append 000 to previous remainder 
47 6768000  a) Divide CH by 47 (AB) 
4714400000  b) Divide CH 3 
4748000000  c) Divide C by 47 (AB) 
4795700000  d) Subtract C^2=81 from EF 
4794890000  e) Multiply DH by 3 in DI 
4792298300  f) Multiply DH by 47 (AB) in DJ 
479 689490  g) Subtract C^3=729 from HIJ 
479 688761  OK 3rd digit! Remainder 688761 
Fourth digit:
Abacus  Comment 

ABCDEFGHIJKLM  
479 688761000  Append 000 to previous remainder 
479 688761000  a) Divide DJ by 479 
4791437914194  b) Divide DJ by 3 
4794793046394  c) Divide D by 479 1d 
4799482046394  d) Subtract 9^2=81 from GH 
4799473946394  e) Multiply EJ by 3 in EK 
4799142184194  f) Multiply EJ by 479 in EM 
4799 68106330  g) Subtract D^3=729 from KLM 
4799 68105601  Ok 4th digit! Remainder 68105601 
Now we finish the calculation using [[../../Abbreviated operations/abbreviated operations]]. We need to divide the remainder (68105601) by three times the square of the current root (4799). The first four digits of the result are appended after the ones already obtained; for instance:
Abacus  Comment 

ABCDEFGHIJKLM  
4799 68105601  Divide EM by 4799 
479914191623  Divide EM by 4799 
47992957204  Divide EM by 3 
47999857  Compare this to 
As we can see, we have obtained a result with 7 correct figures.
Note: We found above that with root 48 we could not subtract , or we had a negative remainder (1). This might seem unfortunate since it forced us to undo part of the work and correct the new root figure downwards, but in practice what we find is a fortunate result: the small remainder (1) tells us that 48 was a excellent approximation (by excess) to the root, opening a new way to solve the problem. In fact, what we have is:
or
where we can use
so that