# Basic concepts

When humans gathered in groups large enough for barter or trade operations to acquire some importance, the need arose for basic accounting which in turn required being able to count up to high numbers, perform basic arithmetic operations and keep a permanent record of transactions. Thus both arithmetic and writing seem to have a common origin in this need.
As for the basic arithmetic operations, “these seem to have been carried out universally using an abacus of some kind” and perhaps the first historical testimony of its use is found in the proto-cuneiform character: SANGA, which appeared as part of the signature of Sumerian scribes on clay tablets some 5000 years ago and which the Assyriologists identify with such a device.

## What is an abacus?

An abacus is a tool or instrument in which numbers are physically represented in a way that allows them to be manipulated to mechanically simulate arithmetic operations.
In an abacus, the numbers are represented by "counters" or "tokens" (pebbles, seeds, shells, coins and the like, rods, etc.) to which a numerical value is assigned. The counters do not have to be all identical or have the same assigned value. To represent a number we arrange together the necessary counters on a table or any suitable surface in a similar way to how we would take a series of coins to reach a certain amount of money; it is the same process.
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. Consider the simplest case in which we only use identical counters with an assigned value of one.
In the image above we have arranged four counters of value one to represent the number 4 (left-a), after attaching another three counters (left-b) representing the number 3 we have a representation of the number 7 (left-c); that is, the sum 4 + 3. Similarly, if we start from the representation of the number 7 (right-a) and remove a set of counters that represent the number 4 (right-b), what remains on the table is three or the result of the subtraction: 7-4 ( right-c).
Please note that to perform the above operations it is not necessary to know anything about the addition or subtraction tables, in particular you do not need to know that 4 + 3 = 7 or 7 - 4 = 3, you only need to know how to manipulate the counters; on the contrary, it is the abacus that will allow you to "discover" that the result of 4 + 3 is 7 and that of 7 - 4 is 3! This is an essential point about the use of the abacus that we will return to in the chapter Addition and Subtraction.
It is commonly considered that in Arithmetic there are four fundamental operations: Addition, Subtraction, Multiplication and Division, and that any other calculation (e.g. obtaining a square root) ultimately reduces to a sequence of these four fundamental operations. But multiplication can be considered as a repeated addition in the same way that a division can be considered as a repeated subtraction, so that any arithmetic calculation is ultimately reduced to a sequence of addition and subtraction (and, going further from a modern perspective, addition and subtraction are just two aspects of the same additive composition law of numbers). Therefore, with an abacus, any arithmetic calculation can be performed in principle. But this would be extremely difficult or perhaps impossible without a few refinements to our rudimentary abacus.
With the abacus used above (only identical counters with assigned value one), it is evident that if we start working with progressively larger numbers, our table (abacus) will be cluttered with counters, making their use and interpretation impractical. We need a way to reduce the number of physical objects, counters, to manipulate and keep it within some limits that are comfortable for us. There are a couple of solutions:
* Using physically different counters with different assigned values. It is the most primitive system, already used by the Sumerians more than 5000 years ago ... and still used today since the use of coins and banknotes of different nominal values in any current monetary system corresponds perfectly with this concept of abacus.
* Define spatial regions in our table (abacus) so that a counter represents one value or another according to the region it occupies.
Let's look at an example. In the figure above we have added 7 + 7 (a and b) with our primitive abacus, and 14, the result, is shown as a cluttered table full of counters (c). We can replace some of these counters with a physically different one that has a higher value assigned, for example 10 (the replacement value). With this, the state of our abacus is easier to interpret (d), it has been simplified as 10 1-counters have been replaced by only one 10-counter.
Alternatively, we can consider the abacus divided into two spatial regions and use identical counters to which we will associate one value or another according to the region in which we place it. At (e) in the figure above, the abacus has been divided into two regions, left and right separated by the double vertical line. If we assign a value of one to the counters located on the right and 10 to the ones located on the left, the number 14 would be represented as illustrated. This way of proceeding is preferable to the previous one since we can repeat the process, defining as many regions as we need with the replacement values that suit us, allowing us to represent arbitrarily large numbers with counters of a single type, for example, in (f) we have depicted 114 using three regions and two replacement values of 10; we only needed 6 counters. We are witnessing here the birth of Positional notation.
Before continuing, it is convenient to indicate that there are two main types of abacos:
* Free-counters or Table Abacus: The counters are independent and are normally kept apart in a box or bag and are placed or removed from the table as needed. It is the most primitive type and the one that we have considered here so far.
* Fixed-Bead Abacus: The counters, called beads in this context, are always present, integrated into a frame and can be slid from an inactive position to an active one along grooves, rails, strings, wires or rods. This is the most sophisticated, portable, compact type that allows a faster calculation and, as we will see, the eastern abacus to which this book is dedicated is of this type.
Now we can mention the Russian Abacus (Schoty), the Iranian Abacus (Chortkeh) and the School Abacus as examples of fixed bead abacos that conform to what we have explained so far. Both consist of a wooden frame with horizontally arranged wires along which ten beads are strung in the case of the Russian abacus and nine in the Iranian abacus. The beads can slide from an inactive position (right) to an activated position (left) and each wire represents one of the regions mentioned above, with a replacement value of 10, so that a bead in each of the wires has a associated value ten times higher than that of the beads on the immediately lower wire.
These abaci have everything necessary to allow arithmetic operations with numbers expressed in decimal notation: several rods to represent successive powers of ten and 9 beads to represent the digits from 0 to 9 (for convenience, the Russian abacus has one more bead than is strictly necessary). You can try a Russian abacus model at this link.
But we still need one last refinement to fully understand the East Asian abacus.
Subitizing is the rapid, accurate, and confident judgments of numbers performed for small numbers of items. We can make such a judgment if the number of objects to be counted is 4 or 5 at most; from there, we will have to invest time in counting. In the Russian and Iranian abacos we have 9 or 10 beads per rod, so the reading of the number represented may be beyond the subitization limit. This is alleviated by using beads of two different colors as illustrated in the previous images, but there are also a couple of additional techniques that not only allow us to stay within the limits of subitization but also reduce the number of beads needed in the abacus.
In the upper image (a) we have the number 18 represented in two regions (rods); one of them contains 8 counters that are above the subitization limit. To simplify the reading of the abacus we can:
* Use a different type of counter with a replacement value of, for example, five (b).
* Subdivide the regions or rods into two zones: one in which a counter takes the value one and the other in which it takes the value five (c,d).
In either case, we do not need to have more than four identical counters per region to be able to represent numbers in decimal notation, so we are guaranteed the fast reading of the abacus. When using 5 as the second replacement value, we are using a bi-quinary decimal notation for numbers. Examples of both solutions are counting rods and the eastern abacus.

## Counting rods

Counting rods were a table type abacus or free counters abacus in which the counters were small rods of wood, bamboo, bone, etc. that were arranged on a flat surface, using a checkerboard or not. By the way, this abacus that dominated Chinese mathematics for at least 14 centuries and Japanese mathematics (Wasan) until the Meiji restoration, is probably the most versatile that has ever existed, although unfortunately it is also very slow to handle.
In the figure above (a) we use vertically arranged rods as counters of value one to represent the number 18. In (b) we use a horizontally arranged rod as a counter of value five and in (c) we use a more compact rod arrangement with alternation of orientation or not depending on whether we use a smooth table or a checkerboard (see details in Wikipedia). Digits from 1 to 9 are represented as:
Zero was represented by an empty cell on the checkerboard or by a space or other object (for example, a Go token) on the table. For example, the number 1547 would be expressed:
It is interesting to mention that this is the only abacus that is known to use the orientation of the counters to assign them one value or another; but we find a parallel to this concept, if not a precedent, many centuries before the appearance of counting rods in the Babylonian numerals used to write numbers in sexagecimal notation. Each sexagesimal digit was constituted by a series of vertical impressions of the edge of a reed stylus on the fresh clay tablet with unit value (, , , , ..., ), and impressions made with the stylus turned 45 degrees or more in an anticlockwise direction of value 10 (, , , , ). The Decimal number 1547 is expressed in sexagesimal in the form 25:47, where "25" and "47" are two sexagesimal digits written as: and
The very appearance of these digits suggests their immediate representation in a table abacus using counting rods.

## The Eastern Abacus

The second solution is the one adopted both in the Roman abacus and the abacuses that appeared in China.
While a few examples of Roman abacus such as the one in the figure are known, where the beads slide along grooves, nothing is known for certain about the origin of the eastern abacus. A confusing phrase from the Shushu Jiyi (術數紀遺) of Xu Yue (徐岳), which perhaps dates from the 2nd century, is often quoted as describing a computing device that we could identify with an abacus and that has been interpreted in different ways as in the above figure (a). In this interpretation of a first Chinese abacus as a table abacus, the central part is divided into a series of columns with two parts; the upper one would assign a value of 5 to each bead and the lower one a value of 1, while the inactive (unused) beads wait scattered above and below the central part.
It is unknown when the abacus of beads strung along rods appeared, but when this abacus replaced the use of counting rods throughout the 16th century it did not have four lower beads and one upper one like the Roman abacus (we will refer to this disposition as a 4 + 1 type abacus) but five in the lower part and two in the upper part (5 + 2 type abacus), separated by a horizontal beam. The additional beads, not necessary for the calculation with decimal numbers, were introduced for convenience to adapt the calculation algorithms that had been developed with the counting rods to the abacus. Historically, the four types of abacus described in the figure below have been used.
Symbolically, the upper and lower areas of the abacus have been designated Heaven (天, Tiān in Chinese, Ten in Japanese) and Earth (地, De in Chinese, Chi in Japanese).
In this book we will focus on the use of the 4 + 1 type abacus or modern abacus, following what we will call the modern abacus method. If you have understood the principles on which any abacus is based and you learn to use the modern abacus, you will have no difficulty in imagining how you can use any other type of abacus, at least for elementary operations of addition and subtraction. This could include, why not? the following abacus for sexagesimal calculations conjectured by Woods as the Babylonian abacus based on what we know about mathematics in Mesopotamia ... and the mistakes the scribes made!
Finally, if after achieving some experience with the modern abacus following this book you want to know about the traditional techniques and the use of the 5 + 2 abacus, you can continue with the book: Traditional Abacus and Bead Arithmetic.

## References

* 1 2 3 Woods, Christopher (2017), "The Abacus in Mesopotamia: Considerations from a Comparative Perspective", The First Ninety Years: A Sumerian Celebration in Honor of Miguel Civil, De Gruiter, ISBN 9781501511738 {{citation}}: Unknown parameter |editor1first= ignored (|editor-first1= suggested) (help); Unknown parameter |editor1last= ignored (|editor-last1= suggested) (help); Unknown parameter |editor2first= ignored (|editor-first2= suggested) (help); Unknown parameter |editor2last= ignored (|editor-last2= suggested) (help); Unknown parameter |editor3first= ignored (|editor-first3= suggested) (help); Unknown parameter |editor3last= ignored (|editor-last3= suggested) (help)
* ↑ Martzloff, Jean-Claude (2006), A history of chinese mathematics, Springer, ISBN 978-3-540-33782-9
* ↑ Kojima, Takashi (1963), Advanced Abacus: Theory and Practice, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0003-7

## External resources
### Soroban Trainer

If you are interested in trying the abacus but do not have an abacus yet, you can use the JavaScript application:
* You can run it directly from GitHub in your browser
* or you can download it to your computer from the repository on GitHub.
It may be used as an 4+1, 5+1 or 5+2 type abacus.

# Close-up of the eastern abacus
## Description of the Eastern abacus

An abacus consists of the following parts made of wood, bamboo, metal, plastic, etc.:
* A rectangular frame.
* The frame supports a variable (odd) number (from 9 to 27) of rods parallel to its shorter sides along which the beads are strung.
* A bar or beam, parallel to the longer sides of the frame, divides the abacus and the rods into two desks: a narrower upper one, the Heaven , and a wider one below, the Earth.
* Beads of the upper region of the rods (the upper beads) to which a value of 5 is assigned when they are activated.
* Beads of the lower region of the rods (the lower beads) to which a value of 1 is assigned when they are activated.
* Some modern abacos may include a reset button to return the beads to their inactive position (see below).
* 23 rods Japanese style 4+1 beads modern abacus (soroban) with reset button Modern abacuses also usually present some type of unit rod marks every three rods to facilitate the alignment of numbers as well as the reading of them. They are convenient for abacus with a high number of rods (17-27) but are not essential. For some it is a nuisance.

## Active and inactive beads

The beads are considered inactive while they are separated from the central bar or beam. The abacus in the following figure has been reset or cleared and all of its beads are inactive. All bars can be considered to have been filled with zeros.
A cleared o reset abacus :
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
When we move the beads towards the central bar we consider them active and then they acquire the assigned value 5 (the upper ones) or 1 (the lower ones). This is what allows us to represent numbers. With a modern 4 + 1 abacus we can form exactly the ten digits from 0 to 9 necessary to perform calculations with decimal numbers, and these digits have a unique representation.
Modern 4+1 abacus number representation :
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
But with a traditional 5 + 2 abacus, and using the suspended bead, we can represent numbers up to 20 on each rod.
Traditional 5+2 abacus number representation :
| 0 | 1 | 2 | 3 | 4 | 5 | 5 | 6 | 7 | 8 | 9 | 10 |
| 10 | 11 | 12 | 13 | 14 | 15 | 15 | 16 | 17 | 18 | 19 | 20 |
Note the use of the suspended bead for numbers from 15 to 20. Also note that the numbers 5, 10 and 15 can be represented in two different ways: using the fifth lower bead or not. This fact may be used to simplify addition and subtraction operations a bit. If you are interested in the details of use of a traditional abacus consult the book: Traditional Abacus and Bead Arithmetic

## Resetting the abacus

After finishing a calculation and before starting a new one, it will be necessary to reset the abacus to its cleared state.
* If your abacus has a reset button, simply press it and you will have your abacus ready for new use.
* With a modern Japanese style abacus (soroban, with biconical beads) this is achieved in a very fast and effective way. Simply tilt the abacus towards you until all the beads drop to their lowest position and return the abacus to its usual position on the table. Then use the nail of your right index finger to push the top beads upward with a flick of your finger from left to right just above the center beam.
* With a traditional Chinese style abacus (suanpan with ellipsoidal beads) the above maneuver may not work properly, but if the abacus is large enough there is another procedure that represents a small skill challenge interesting in itself:
** Take the abacus with both hands by the short sides of the frame and tilt it towards you about 45 degrees until the beads fall down.
** From that position and without moving your forearms, force a sharp rotation to the abacus to the horizontal position with a twist of the wrists. If the axis of rotation defined by your wrists passes through the highest of the lower beads, the centrifugal force will drive the upper beads to their inactive position.
** Put the abacus back on the table.
    You will probably need some time to perfect this technique.
* Finally, if all of the above fails, as a last resort you can use the fingers of your hand like a broom to sweep the beads into their inactive position.

## Moving the beads

Until late 19th century Japan, no ancient author ever bothered to write how beads should be handled; but surely the technique was transmitted orally.
To begin with, let's say that modern abacuses are so light that you need to hold them with your left hand to stabilize them and prevent them from shifting on the table when handling the beads. This could have disastrous consequences if that displacement induces the movement of other beads than the ones we want to move. By comparison, other traditional abacos are so heavy that they remain stable on their own, allowing you to use your left hand for other purposes, such as following a list of numbers in a ledger. Furthermore, you can use the abacus as a paperweight to stabilize a stack of invoices, etc.
And yet, in some countries it is taught to manipulate the beads using both hands. But we will use only the right hand for this purpose.

### Which fingers to use

For modern abacuses with a rod length of approximately 6 cm (2.4 inches) it is recommended to use two fingers: the thumb and the index finger of the right hand.
* Use your index finger to move the upper beads up and down and to move the lower beads down.
* Use your thumb to move the lower beads up.
But some very experienced masters only use the index finger...
For larger traditional abacuses, three fingers should be used: thumb, index and middle finger of the right hand.
* Use your middle finger to move the upper beads up and down.
* Use your index finger to move the lower beads down.
* Use your thumb to move the lower beads up

### Combined movements

When the operation affects both upper and lower beads, try to follow the rules in the table below. Some movements can be done simultaneously and others must be done in rapid succession in the order indicated.
Combined movements of upper and lower beads :
| To move → and ↓ | Upper bead Up | Upper bead Down |
| Lower beads Up | Do it simultaneously | Do it simultaneously |
| Lower beads Down | First lower bead(s), then upper one | First upper bead(s), then lower one |

### Exercises

* Enter digits 1 to 9 from left to right anywhere on your abacus (modern or traditional) using the above rules.
    Write caption here! 1 2 3 4 5 6 7 8 9
    then, clear them from left to right also using the above rules.
* Enter three or more consecutive sixes from left to right and wipe them in rapid succession from left to right.
    Write caption here! A B C 6 6 6
    This exercise should be repeated daily a few times until you are able to do it almost without looking at the abacus.

# Addition and subtraction
## Introduction

As has already been stated in the introduction to this book, addition and subtraction are the only two operations that can be carried out on the abacus; everything else must be reduced to a sequence of addition and subtraction, so learning these two operations is the most fundamental step in the study of the abacus.
Learning to add and subtract with the abacus is another case of psychomotor learning, similar to learning to dance, ride a bicycle, drive or learn a musical instrument.
* In a first phase you will need a continuous cognitive effort trying to determine what is the next movement you have to do.
* Later, you will notice that progressively you have to think less while the movements arise in an increasingly automatic way.
* Finally, the movements will emerge spontaneously from you, you will have them definitely hardwired in your motor cortex and you will not have to think about them again. Although you will have a lifetime to perfect them.
Yes, it is like learning a musical instrument, but learning the abacus is much easier and faster than learning to play the viola and you will be sensitive to your progress from day to day.
In what follows we will deal with addition and subtraction together; It would be very difficult to separate the learning of one of these operations from the other since, as we will see, when we are adding we spend half the time subtracting complementary numbers and vice versa, when we are subtracting we spend half the time adding complementary numbers.
It has also been anticipated in the introduction of this book that it is not necessary to know how to add and subtract to use an abacus, only to know how to manipulate the beads or counters. In fact, for centuries the abacus was taught to people who had no previous knowledge of arithmetic and that the only knowledge they would have of it throughout their lives was going to be the use of the abacus itself. They learned to add and subtract by memorizing a long series of verses, rhymes, or rules intended to be sung as they were used. For example, taken from Xú Xīnlǔ's Pánzhū Suànfǎ, the first book entirely devoted to the abacus published in 1573 (late Ming Dynasty), and liberally translated from Chinese:
* Xú Xīnlǔ's rules for 1-digit addition
* 1 activate 1, 1 activate 5 deactivate 4, 1 subtract 9 carry 1
* 2 activate 2, 2 activate 5 deactivate 3, 2 subtract 8 carry 1
* 3 activate 3, 3 activate 5 deactivate 2, 3 subtract 7 carry 1
* 4 activate 4, 4 activate 5 deactivate 1, 4 subtract 6 carry 1
* 5 activate 5, 5 deactivate 5 carry 1
* 6 activate 6, 6 activate 1 deactivate 5 carry 1, 6 subtract 4 carry 1
* 7 activate 7, 7 activate 2 deactivate 5 carry 1, 7 subtract 3 carry 1
* 8 activate 8, 8 activate 3 deactivate 5 carry 1, 8 subtract 2 carry 1
* 9 activate 9, 9 activate 4 deactivate 5 carry 1, 9 subtract 1 carry 1
* Xú Xīnlǔ's rules for 1-digit subtraction:
* 1 deactivate 1, 1 borrow 1 add 9, 1 activate 4 deactivate 5
* 2 deactivate 2, 2 borrow 1 add 8, 2 activate 3 deactivate 5
* 3 deactivate 3, 3 borrow 1 add 7, 3 activate 2 deactivate 5
* 4 deactivate 4, 4 borrow 1 add 6, 4 activate 1 deactivate 5
* 5 deactivate 5, 5 borrow 1 add 5
* 6 deactivate 6, 6 borrow 1 add 4
* 7 deactivate 7, 7 borrow 1 add 3
* 8 deactivate 8, 8 borrow 1 add 2
* 9 deactivate 9, 9 borrow 1 add 1
Which, obviously, inform us of which beads we have to move in order to add or subtract a digit. For example, the third line of the addition table contains three rules to try to add a 3:
* 3 activate 3, i.e. just activate three lower beads.
* 3 activate 5 deactivate 2, i.e. activate one upper bead and deactivate two lower ones.
* 3 subtract 7 carry 1, i.e. subtract 7 and add 1 to the left column.
which apply, for example, to the following cases:
1+3 :
| A | | A |
| | 3 activate 3 | |
| 1 | | 4 |
3+3 :
| A | | A |
| | 3 activate 5 deactivate 2 | |
| 3 | | 6 |
9+3 :
| A | B | | A | B |
| | | 3 subtract 7 carry 1 | | |
| 0 | 9 | | 1 | 2 |
You will understand these rules better later on, but don't worry anyway, you won't have to follow these 48 rules as you will go an easier path by memorizing just six rules that can be summed up into just three.

## 1-digit Addition and subtraction

The first step in learning addition and subtraction with an abacus is learning to add or subtract one of the 9 digits 1, 2, ..., 9 to / from any other 0, 1, 2, ..., 9; in total 180 cases that we will go through in our daily practice until we have them integrated into our motor memory. After this, adding or subtracting multi-digit numbers will be as simple as iterating this process in an orderly fashion for all the digits of the addend or subtrahend.

### What do you need to know

To deal with the 180 cases mentioned above without memorizing the 48 rules of Panzhu Suanfa we need to memorize some almost trivial data:
* the beads necessary to form a digit.
* the complements to 5 of the digits 1, 2, 3, 4 and 5
* the complements to 10 of the digits 1, 2, ..., 10

#### Beads needed to form a digit

Remember what was said in the Introduction of this book: "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". So we need to know the beads that make up each digit to be able to add or subtract them, but we already know this from the figure:
Modern 4+1 abacus number representation :
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
or in table form:
Beads needed to form a digit :
| Digit | Upper | Lower |
| 1 | 0 | 1 |
| 2 | 0 | 2 |
| 3 | 0 | 3 |
| 4 | 0 | 4 |
| 5 | 1 | 0 |
| 6 | 1 | 1 |
| 7 | 1 | 2 |
| 8 | 1 | 3 |
| 9 | 1 | 4 |

#### Complementary numbers

We also need to memorize two types of digit pairing:
* 5-complements
* 10-complements
These are the digit pairs that together add up to 5 or 10. We can always find them mentally with our knowledge of addition and subtraction, but with practice they will end up solidly installed in our memory without the need to mentally "calculate" them. They are the basis of the mechanics of the abacus.
5-complements :
| 0 - 5 | 1 - 4 | 2 - 3 |
10-complements :
| 0 - 10 | 1 - 9 | 2 - 8 | 3 - 7 | 4 - 6 | 5 - 5 |
At a later stage, to deal with negative numbers, you will also need to handle 9-complements:
9-complements :
| 0 - 9 | 1 - 8 | 2 - 7 | 3 - 6 | 4 - 5 |
But for now you can safely forget about them.

### The rules to use

The mechanics of addition and subtraction are based on three rules to be tried in sequence with the following protocol:
* Only if a rule fails (because we do not have the necessary beads to complete the operation) we proceed to try the next rule.
* The second of the rules only works for digits 1, 2, 3 and 4.
* The third rule decomposes the operation into two other "simpler" ones: a carry to the column directly to the left or a borrow from that column plus an operation of the opposite type (i.e. a subtraction if we are adding or a sum if we are subtracting). This case raises the following points to take into account:
** The first operation (carry or borrow) is trivial most of the time.
** The second operation (the opposite of the starting one) is guaranteed to culminate using rules 1 or 2 (never 3) of the opposite operation.
** We need to decide in what order we do these two operations.

#### Addition rules

These are the rules for the addition of a digit:
| 1 | Try to add the beads needed |
| 2 | Try to add 5 and subtract the complementary number to 5 |
| 3 | Carry 1 to the left and subtract the complementary number to 10 |

#### Subtraction rules

And these are the rules for subtraction:
| 1 | Try to subtract the beads needed |
| 2 | Try to subtract 5 and add the complementary number to 5 |
| 3 | Borrow from the left and add the complementary number to 10 |

#### Joint rules for addition and subtraction

The above rules for addition and subtraction are of identical structure so we can merge them into:
| 1 | Try to add/subtract the beads needed |
| 2 | Try to add/subtract 5 and subtract/add the complementary number to 5 |
| 3 | Carry or Borrow and subtract/add the complementary number to 10 |
and we have only three rules to memorize!

#### Order of operation

Before moving on to some preliminary examples we have to decide what order of operation to use in case we reach rule 3, which will happen half the time. This rule leads us to a split of the original problem into two hopefully simpler ones: a carry or borrow and a operation of the opposite type to the one we are performing. What do we do first?...(sakidama, atodama...)
The Japanese standard method currently taught since the end of the 19th century proposes to first carry out the borrow and then the addition of the complementary number in the case of subtraction (sakidama 先珠), while in the case of addition, the subtraction of the complementary number is done first and then carry to the left column (atodama 後珠). This seems inspired by the structure of Chinese rules / verses / rhymes used for teaching the abacus since ancient times, but there does not seem to be any compelling logical reason to do so and not everyone agrees.
As we will see, with the abacus one works from left to right during the addition and subtraction of multi-digit numbers, so it seems natural to try to respect this movement from left to right of the hand without disturbing it with continuous comings and goings to the column of the left. Always using sakidama (carries and borrows first) seems the most natural thing to do.
It goes without saying that if you have a teacher or coach you should scrupulously follow their directions, but if you are self-taught feel free to experiment until you find your way.
By the way, in some Asian countries it is taught to use the left hand for carries and borrows.

#### Some preliminary examples

Ex: Enter 1 to a column of your abacus and add 3 to it:
* Do we have at our disposal (inactive) the necessary beads (3 lower ones) to add them to the 1 in our abacus? Yes!
** then we activate them and we have completed the operation with the first rule.
1+3 :
| A | | A |
| | Rule 1! | |
| 1 | | 4 |
Ex: Enter 3 to a column of your abacus and add 3 to it:
* Do we have at our disposal (inactive) the necessary beads (3 lower ones) to add them to the 3 in our abacus? No!
** then we go to the second rule.
* As the addend 3 is less than 5 we can try the second rule: Do we have at our disposal (inactive) an upper bead? Yes!
** then we apply the second rule: we activate the upper bead and retire two lower beads (the 5-complement of the addend 3).
3+3 :
| A | | A |
| | Rule 2! | |
| 3 | | 6 |
Ex: Enter 9 to a column of your abacus and add 3 to it:
* Do we have at our disposal (inactive) the necessary beads (3 lower ones) to add them to the 3 in our abacus? No!
** then we go to the second rule.
* As the addend 3 is less than 5 we can try the second rule: Do we have at our disposal (inactive) an upper bead? no!
** the we proceed to the third rule:
* Carry one to A and subtract 7 (the 10-complement of 3) from B
** Do we have at our disposal (active, we are subtracting now!) the necessary beads (one upper bead and 2 lower ones) to retire them from the 9 on our abacus? Yes!
*** then retire them and we have completed this part of the operation with the first rule.
9+3 :
| A | B | | A | B |
| | | Rule 3! | | |
| 0 | 9 | | 1 | 2 |
As you can see, the rules used here are the same that appeared in Xu Xinlu's Panzhu Suanfa, but condensed into only three rules thanks to the concept of complementary numbers!
Let us see now the reverse mouvements for subtraction:
Ex: Enter 4 to a column of your abacus and subtract 3 from it:
* Do we have at our disposal (active) the necessary beads (3 lower ones) to retire them from the 4 in our abacus? Yes!
** then we deactivate them and we have completed the operation with the first rule.
4-3 :
| A | | A |
| | Rule 1! | |
| 4 | | 1 |
Ex: Enter 6 to a column of your abacus and subtract 3 from it:
* Do we have at our disposal (active) the necessary beads (3 lower ones) to subtract them from the 6 in our abacus? No!
** then we go to the second rule.
* As the subtrahend 3 is less than 5 we can try the second rule: Do we have at our disposal (active) an upper bead? Yes!
** then we apply the second rule: we deactivate the upper bead and add two lower beads (the 5-complement of the subtrahend 3).
6-3 :
| A | | A |
| | Rule 2! | |
| 6 | | 3 |
Ex: Enter 12 to a pair of columns of your abacus (AB) and subtract 3 from B:
* Do we have at our disposal (active) the necessary beads (3 lower ones) to retire them from B? No!
** then we go to the second rule.
* As the subtrahend 3 is less than 5 we can try the second rule: Do we have at our disposal (active) an upper bead? no!
** the we proceed to the third rule:
* Borrow 1 from A and ADD 7 (the 10-complement of 3) to B
** Do we have at our disposal (inactive, we are adding now!) the necessary beads (one upper bead and 2 lower ones) to add them to the 2 on B? Yes!
*** then activate them and we have completed this part of the operation with the first rule.
12-3 :
| A | B | | A | B |
| | | Rule 3! | | |
| 1 | 2 | | 0 | 9 |

### Types of one-digit addition and subtraction

The following table shows for each of the 180 elementary operations of addition and subtraction which rule solves the problem. It can be useful during your first practice, to choose which digits to add or subtract.
Types of one-digit addition and subtraction :
| Addition | Addition | Addition | Addition | Addition | Addition | Addition | Addition | Addition | Addition | Addition | Subtraction | Subtraction | Subtraction | Subtraction | Subtraction | Subtraction | Subtraction | Subtraction | Subtraction | Subtraction | Subtraction |
| | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| +1 | 1 | 1 | 1 | 1 | 2 | 1 | 1 | 1 | 1 | 3 | -1 | 3 | 1 | 1 | 1 | 1 | 2 | 1 | 1 | 1 | 1 |
| +2 | 1 | 1 | 1 | 2 | 2 | 1 | 1 | 1 | 3 | 3 | -2 | 3 | 3 | 1 | 1 | 1 | 2 | 2 | 1 | 1 | 1 |
| +3 | 1 | 1 | 2 | 2 | 2 | 1 | 1 | 3 | 3 | 3 | -3 | 3 | 3 | 3 | 1 | 1 | 2 | 2 | 2 | 1 | 1 |
| +4 | 1 | 2 | 2 | 2 | 2 | 1 | 3 | 3 | 3 | 3 | -4 | 3 | 3 | 3 | 3 | 1 | 2 | 2 | 2 | 2 | 1 |
| +5 | 1 | 1 | 1 | 1 | 1 | 3 | 3 | 3 | 3 | 3 | -5 | 3 | 3 | 3 | 3 | 3 | 1 | 1 | 1 | 1 | 1 |
| +6 | 1 | 1 | 1 | 1 | 3 | 3 | 3 | 3 | 3 | 3 | -6 | 3 | 3 | 3 | 3 | 3 | 3 | 1 | 1 | 1 | 1 |
| +7 | 1 | 1 | 1 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | -7 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 1 | 1 | 1 |
| +8 | 1 | 1 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | -8 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 1 | 1 |
| +9 | 1 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | -9 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 1 |
As you can see, the tables for addition and subtraction are mirror images of each other. Also note how half the cases correspond to rule three, that is, they require carries and borrows and from them, those marked in bold, end with an opposite type 2 operation. Also check how rule 2 only affects the addition / subtraction of digits less than 5.

### Examples

Ex: Enter 5 to a column of your abacus and add 7 to it:
In this example, which requires a carry, the subtraction of the complementary number in turn requires the use of rule 2, affecting the upper bead.
5+7 :
| A | B | | A | B |
| | | Rule 3, then rule 2! | | |
| 0 | 5 | | 1 | 2 |
Ex: Enter 95 into your abacus and add 7 to it:
Now the carry leads to another type 3 operation, requiring in turn a new carry. Three columns of the abacus are affected in this operation.
95+7 :
| A | B | C | | A | B | C |
| | | | Rule 3! | | | |
| 0 | 9 | 5 | | 1 | 0 | 2 |
Ex: Enter 999995 into your abacus and add 7 to it:
This is an extreme situation, extrapolation from the previous case, which you should study carefully. The carry spreads or runs through the columns to the left until it finds a hole to lodge!
999995+7 :
| A | B | C | D | E | F | G | | A | B | C | D | E | F | G |
| | | | | | | | Rule 3! | | | | | | | |
| 0 | 9 | 9 | 9 | 9 | 9 | 5 | | 1 | 0 | 0 | 0 | 0 | 0 | 2 |
Note that, if we had at our disposal a lower fifth bead, in the case of the traditional abacus, we could have avoided this "carry run" at least temporarily.
999995+7 on a 5+2 abacus :
| A | B | C | D | E | F | | A | B | C | D | E | F |
| | | | | | | Rule 3! | | | | | | |
| 9 | 9 | 9 | 9 | 9 | 5 | | 9 | 9 | 9 | 9 | T | 2 |
For details on the use of the lower fifth ball you can consult the book: Traditional Abacus and Bead Arithmetic once you have acquired practice in addition and subtraction.
Ex: Enter 50 into your abacus and subtract 3 from it it: In this case of a type 3 operation, the borrow in turn requires a type 2 operation (affecting the upper bead).
50 - 3 :
| A | B | | A | B |
| 5 | 0 | | 4 | 7 |
Ex: Enter 10006 into your abacus and subtract 7 from it it:
Finally, this is a case of "borrow run" where we have to travel far to the left to find something to subtract from! Study this case carefully as well.
10006 - 7 :
| A | B | C | D | E | | A | B | C | D |
| 1 | 0 | 0 | 0 | 6 | | 0 | 9 | 9 | 9 | 9 |

### Two advices

So far our theoretical or intellectual explanations about the abacus. Now you know what the eastern abacus "is" and you are on your way. This intellectual knowledge will be your guide during your first steps, but with practice the movements of the beads will become second nature to you and you will never think about all these rules again (at least, until you write your first book on the abacus). To achieve this you will need to practice and practice and we offer you a couple of important tips to help you complete the path you are taking now.
* Never read intermediate results. This is a bad habit that does not lead to anything, only to wasting time and wasting mental energy, and what you want is to acquire speed and comfort in the use of the abacus. Your abacus is there, and you have paid for it, just to keep your numbers safe for you without you having to worry about them. You only have to "react" to the arrangement of the beads without having to be aware of what number they represent.
* Forget the addition and subtraction tables, except what we have extracted from them in the form of numbers complementary to five and ten. In particular, never think: "I have to add 7 + 8, this gives 15, then a fifteen has to appear on the abacus." If you do this, you will be "thinking" while adding and subtracting and that will tire you out and slow you down. If you have to think about something, think about the rules of movement of beads and not numbers, until you are able to add and subtract mechanically while thinking about anything else.
If you do not follow these tips, you will develop a bad habit that can be very difficult to correct later, as with the bad habits that are acquired when studying a musical instrument.

### And now the practice

Your first exercises should be as simple as possible and nothing seems easier than randomly choosing two digits, for example: 6 and 8, and trying to add or subtract them (perhaps adding a one in front of the first digit if when subtracting you will need to borrow). You can use the table of types of operations explained earlier in this chapter to know in advance the type of operations to be carried out.
Subsequently, you should proceed to a systematic practice of all 180 cases of addition and subtraction of a single digit, for which the following exercise is proposed, which will also serve as an introduction to the addition and subtraction of multi-digit numbers.
Start with the abacus in the following state and add the same digit to each of the nine columns B-J proceeding from left to right
Starting addition exercise :
| A | B | C | D | E | F | G | H | I | J |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
For example, to add 1 to each digit of 123456789 follow the steps indicated in the following table
Adding digit 1 to B-J :
| Abacus | Comment |
| ABCDEFGHIJ | |
| 123456789 | Start with this |
| +1 | Add 1 to B (Type 1) |
| +1 | Add 1 to C (Type 1) |
| +1 | Add 1 to D (Type 1) |
| +1 | Add 1 to E (Type 1) |
| +1 | Add 1 to F (Type 1) |
| +1 | Add 1 to G (Type 1) |
| +1 | Add 1 to H (Type 1) |
| +1 | Add 1 to I (Type 1) |
| +1 | Add 1 to J (Type 3 with "carry run") |
| 234567900 | Result |
| ABCDEFGHIJ | |
and you should arrive at the result: 234567900; that is, 123456789+111111111. The following table shows the results of adding 111111111, 222222222, ... 999999999 to 1234568789.
123456789 +ddddddddd :
| d | Result |
| 1 | 234567900 |
| 2 | 345679011 |
| 3 | 456790122 |
| 4 | 567901233 |
| 5 | 679012344 |
| 6 | 790123455 |
| 7 | 901234566 |
| 8 | 1012345677 |
| 9 | 1123456788 |
For subtraction, add an additional 1 to column A for future borrows:
Starting subtraction exercise :
| A | B | C | D | E | F | G | H | I | K |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
and proceed similarly
Subtracting digit 1 from B-J :
| Abacus | Comment |
| ABCDEFGHIJ | |
| 1123456789 | Start with this |
| -1 | Subtract 1 from B (Type 1) |
| -1 | Subtract 1 from C (Type 1) |
| -1 | Subtract 1 from D (Type 1) |
| -1 | Subtract 1 from E (Type 1) |
| -1 | Subtract 1 from F (Type 2) |
| -1 | Subtract 1 from G (Type 1) |
| -1 | Subtract 1 from H (Type 1) |
| -1 | Subtract 1 from I (Type 1) |
| -1 | Subtract 1 from J (Type 3 with "carry run") |
| 1012345678 | Result |
| ABCDEFGHIJ | |
and the result is 1012345678 = 1123456789-111111111. For the rest of digits, the following table shows the results of subtracting 111111111, 222222222, ... 999999999 from 1123456789.
1123456789 -ddddddddd :
| d | Result |
| 1 | 1012345678 |
| 2 | 901234567 |
| 3 | 790123456 |
| 4 | 679012345 |
| 5 | 567901234 |
| 6 | 456790123 |
| 7 | 345679012 |
| 8 | 234567901 |
| 9 | 123456790 |
During a time you should practice these exercises daily until you notice that little by little you are replacing your intellectual work (thinking about the rules to use) by an instinctive mechanical response. Then you can say that you have started to learn the abacus.

## Multi-digit addition and subtraction
### Always work from left to right

In English, numbers are stated starting with the highest power of ten; 327 is "three hundred twenty-seven" and not "seven-twenty-three hundred". This is the case for many other languages, including Chinese and Japanese, but not for others such as those of the Semitic family. This is the main reason why in the abacus the sum of multi-digit numbers is worked from left to right; everything will be much easier for you, whether you have to read the numbers on a list or if someone else is dictating them to you.
For instance, let us obtain 44+78. Start with a cleared abacus and introduce the first addend 44 anywhere on the abacus (aligned to a unit rod mark if you wish, this is convenient but not essential)
Clear your abacus :
| A | B | C |
| 0 | 0 | 0 |
then enter 4 (40) in B
Enter 40 :
| A | B | C |
| 0 | 4 | 0 |
then enter 4 in C
Enter 4 :
| A | B | C |
| 0 | 4 | 4 |
now add 7 (70) to B
Add 70 :
| A | B | C |
| 1 | 1 | 4 |
and, finally, add 8 to C
Add 8 :
| A | B | C |
| 1 | 2 | 2 |
The result: 122 appears on ABC.
In a more synthetic form:
Caption text :
| Abacus | Comment |
| ABC | |
| . | Unit rod |
| | Start with this |
| 4 | Enter 4 in B (40) |
| 44 | Enter 4 in C |
| +7 | Add 7 to B (70) |
| 114 | |
| +8 | Add 8 to C |
| 122 | Result |
| . | Unit rod |
Another example. Suppose we have to obtain the total of these amounts in euros:
Add :
* 7.77 €
* 11.99 €
* 69.62 €
* 54.43 €
* -96.99 €
* Total
* 46.82 €
Start by clearing your abacus and enter the first number (from left to right). Align it with some of the unit dots markers if you wish.
After entering first addend :
| A | B | C | D | E | F |
| 0 | 0 | 7 | 7 | 7 | 0 |
Caption text :
| Abacus | Comment |
| ABCDEF | |
| . . | Unit rod |
| 777 | Enter 7.77 € |
| +1 | Add 11.99 € |
| +1 | |
| +9 | |
| +9 | |
| 1976 | Intermediate result |
| +6 | Add 69.62 € |
| +9 | |
| +6 | |
| +2 | |
| 8938 | Intermediate result |
| +5 | Add 54.43 € |
| +4 | |
| +4 | |
| +3 | |
| 14381 | Intermediate result |
| -9 | Subtract 96.99 # |
| -6 | |
| -9 | |
| -9 | |
| 4682 | Total: 46.82 € |
| . . | Unit rod |
Final result on the abacus :
| A | B | C | D | E | F |
| 0 | 4 | 6 | 8 | 2 | 0 |

### Ways of practicing addition and subtraction
#### With exercise sheets

You should start your practice by adding and subtracting short series of small integers; for example, 3 to 5 numbers of 2 or 3 digits. For instance:
| 594 807 -660 -466  275 | 880 343 -181 -580  462 | 480 879 -472 19  906 | 336 309 450 -335  760 | 480 -269 -122 780  869 | 963 744 -154 -811  742 | 29 261 909 186  1385 | 373 -163 423 -445  188 |
Progressively increase the size of these series until you reach 10 numbers and, from here, progressively increase the size of the numbers to be added / subtracted to 5 or 6 digits. For instance:
| 514299 127127 774517 -895449 907858 67913 -918061 930513 -582082 -722266  204369 | 375287 611780 -312229 618415 -78719 -467463 -406146 481087 958663 216295  1996970 | 351129 806691 600755 -368489 815758 573731 51556 668536 -609796 713031  3602902 | 882678 876701 -365479 -157706 17497 999762 -262868 -910991 -56430 -333692  689472 | 758320 769094 991286 -49973 74914 -590317 644711 -900673 -449638 -380293  867431 | 562337 315480 -540643 513724 -651332 359925 285750 883744 -591941 75119  1212163 | 388730 -287030 -11891 323483 212117 373242 118641 -693301 442672 -370874  495789 | 798306 -483827 572862 840450 452414 -298427 503089 175358 918199 315118  3793542 |
You will therefore need collections of problems of this type that you can generate with some free utilities on the internet
With this type of practice you will develop two different skills
* efficiently add and subtract with the abacus.
* read numbers at a glance and keep them in memory long enough to work on the abacus.
The latter is essential to, for example, make use of the abacus in accounting.

#### Without exercise sheets
##### The 123456789 exercise

It was common in ancient books on the abacus to demonstrate addition and subtraction using a well-known exercise that consists of adding the number 123456789 nine times to a cleared abacus until the number 1111111101 is reached, and then erase it again by subtracting the same number nine times. This is a convenient exercise because it uses many of the 1-digit addition and subtraction cases (but not all) and allows you to practice addition and subtraction without a paper worksheet, but it is not an elementary exercise given its length. You will need some practice time to complete it without error; consider it rather as a test of your proficiency in addition and subtraction.
Throughout this exercise the following partial results are obtained:
* 000000000
* 123456789
* 246913578
* 370370367
* 493827156
* 617283945
* 740740734
* 864197523
* 987654312
* 1111111101
For more details, please refer to the chapter: Extending the 123456789 exercise of the book: Traditional Abacus and Bead Arithmetic.

## Addition and subtraction with other abacuses

Everything you learn about the eastern abacus will work well with other types of abacus or at least simplify your learning. Remember that the basic operations of the abacus are addition and subtraction and everything else must be reduced to a sequence of these two operations and to the problem of how to organize such a sequence of operations on the abacus.

### Counting rods

Counting rods are another example of a bi-quinary abacus, so the same three rules of addition and subtraction studied here apply to it. You just have to bear in mind that the concepts of activating / deactivating beads translate to placing / removing rods from the table and that "having at our disposal" rods to add does not refer to the pile of rods ready to be used that we have in the box, but to "fit" within the limits of representation of numbers (one 5-rod and 5 1-rods maximum).

### Russian abacus

The Russian abacus (Schoty) is not a bi-quinary abacus; so the second of the addition and subtraction rules given here does not work, everything is solved with the help of the first and third rules only.

## External resources

* Uitti, Stephen. "Soroban Sheets (Addition and subtraction)". Soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* Uitti, Stephen. "Soroban Sheets (Multiplication)". Soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* "The generator". Practicing the soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

## References

* ↑ Xú Xīnlǔ (徐心魯) (1993) [1573]. Pánzhū Suànfǎ (盤珠算法) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
* ↑ Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
* ↑ Abraham, Ralph (2011). "Smart Moves". The Soroban Site of the Visual Math Institute. Archived from the original on January 18, 2020. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* ↑ Uitti, Stephen. "Soroban Sheets (Addition and subtraction)". Soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* ↑ Uitti, Stephen. "Soroban Sheets (Multiplication)". Soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* ↑ "The generator". Practicing the soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

# Modern multiplication
## Introduction

The basic concept of multiplication for natural numbers is that of a repeated addition.
$a\times b=\underbrace {b+\cdots +b} _{a{\text{ times}}}$
For example, to multiply 47 by 23 it is only necessary to add 23 47 times or add 47 23 times; we can do it with our abacus:
Multiplication as repeated addition :
| Abacus | Comment |
| ABCDEFHIJ | |
| . . . | Unit rod |
| +1 +47 | Add 1 to C and 47 to IJ |
| 1 47 | |
| +1 +47 | Add 1 to C and 47 to IJ |
| 2 94 | |
| +1 +47 | Add 1 to C and 47 to IJ |
| 3 141 | |
| ... | Continue in the same way 19 times...! |
| 22 1034 | |
| +1 +47 | Add 1 to C and 47 to IJ |
| 23 1081 | End. 23×47=1081 |
| . . . | Unit rod |
Where we repeat 23 times the sum of 47 to the IJ columns while we add 1 to C to have a "counter" at our disposal. But this is desperately slow! A more effective way to do the same thing can be the following:
A more efficient multiplication :
| Abacus | Comment |
| ABCDEFHIJ | |
| . . . | Unit rod |
| +1 +47 | Add 1 to C and 47 to IJ |
| 1 47 | |
| +1 +47 | Add 1 to C and 47 to IJ |
| 2 94 | |
| +1 +47 | Add 1 to C and 47 to IJ |
| 3 141 | |
| +1 +47 | Add 1 to B and 47 to HI |
| 13 511 | |
| +1 +47 | Add 1 to B and 47 to HI |
| 23 1081 | End. 23×47=1081 |
| . . . | Unit rod |
Where this time, after adding 47 three times to IJ (and 1 to C) we have moved one column to the left and we have started adding 47 to columns HI (and 1 to B). Adding 47 in HI is equivalent to adding 470 = 10×47 to HIJ (10 to BC) drastically reducing the number of operations to be carried out, because after doing it twice only we reach 23 in the counter BC and 1081 in GHIJ, the final result. This way of multiplying was the usual one in mechanical calculators that appeared at the end of the 19th century and that continued in use until the 1970s. But this is still excessively slow.
Think that the abacus as we know it now allows adding very quickly, but that before its invention Chinese mathematicians used counting rods which are extraordinarily slow to handle. It is not surprising therefore that Chinese mathematicians, seeking to abbreviate calculations, eventually invented the decimal multiplication table, as we know it, a few centuries before our era.

## The multiplication table

This is the decimal multiplication table as we learn it in school:
Decimal multiplication table :
| × | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 |
| 3 | 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 |
| 4 | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 |
| 5 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 |
| 6 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 |
| 7 | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 |
| 8 | 8 | 16 | 24 | 32 | 40 | 48 | 56 | 64 | 72 |
| 9 | 9 | 18 | 27 | 36 | 45 | 54 | 63 | 72 | 81 |
But living in the computer age, the most likely thing is that we will soon start using an electronic calculator and in adulthood we will do little multiplication by hand. Often many of us, even mathematicians, do not have the multiplication table "fresh" in memory and this can be bad news for you: if you want to multiply (and divide) efficiently with an abacus, you necessarily have to refresh the multiplication table in your memory!
Using the multiplication table we can solve the multiplication problem $47\times 23$ in the form:
$47\times 23=(40+7)\times (20+3)=$
$=40\times 20+40\times 3+7\times 20+7\times 3=$
$=(4\times 2)\times 100+(4\times 3)\times 10+(7\times 2)\times 10+(7\times 3)$
i.e. we only have to retrieve the partial products: $(4\times 2)=8,(4\times 3)=12,(7\times 2)=14,(7\times 3)=21$ from the multiplication table and add them in the correct places, as we do with paper and pencil
This is absolutely parallel to the multiplication method that we are going to follow with the abacus.

## The modern multiplication method

When we multiply two numbers $a$ and $b$, we call both numbers factors and product to the result $a\times b$, but it is also common to call multiplicand to one of the factors and multiplier to the other. Nevertheless, when it comes to multiplying with the abacus:

### Multiplicand

    It is the number that we are going to manipulate on the abacus and that will guide us to obtain the partial products in an orderly manner and to align them correctly for their addition in the correct positions.

### Multiplier

    It is the factor that we are not going to manipulate on the abacus. in fact it is not mandatory to even enter it (but it is convenient). It will usually be the factor of the two with the fewest digits.

### Multiplication arrangement

There are two ways of entering both factors in the abacus that can be considered practically equivalent; Each of them has its own advantages and disadvantages. The same can be said of the division that we will study in the next chapter. Feel free to experiment with both arrangements.

#### Traditional Chinese arrangement

The multiplicand is located to the left of the abacus and the multiplier far enough away from the multiplicand. At least as many columns as digits have the multiplier plus two or better three must be left free.

##### Example

Traditional Chinese arrangement multiplicand: 345, multiplier: 67 :
| A | B | C | D | E | F | G | H | I | K | J | L | M |
| 3 | 4 | 5 | | | | | | | | | 6 | 7 |
or in table form:
Traditional Chinese arrangement multiplicand: 345, multiplier: 67 :
| Abacus | Comment |
| ABCDEFGHIJKLM | |
| 345 67 | |

#### Traditional Japanese arrangement

This is the reverse way. The multiplier is on the left and the multiplicand on the right, leaving at least two empty columns in between. We need to have at least as many free columns to the right of the multiplier as the number of digits in the multiplier plus one.
Traditional Japanese arrangement multiplicand: 345, multiplier: 67 :
| A | B | C | D | E | F | G | H | I | K | J | L | M |
| 6 | 7 | | | 3 | 4 | 5 | | | | | | |
or in table form:
Traditional Japanese arrangement multiplicand: 345, multiplier: 67 :
| Abacus | Comment |
| ABCDEFGHIJKLM | |
| 67 345 | |
This is the form that has been most popular in Japan and also ended up being imported to China. It is also the form that we will use in this book.

### 1-digit × 1-digit multiplication

Of course this is so trivial that we don't need an abacus, but it serves to introduce the rest of the examples. Suppose we have to multiply $7\times 8$, let's take 7 as a multiplier, 8 as multiplicand and adopt the Japanese arrangement just explained; that is, we start from:
7×8=56 :
| A | B | C | D | E | F | G |
| 7 | | | 8 | | | |
7×8=56 :
| Abacus | Comment |
| ABCDEFG | |
| 7 8 | Setting up the problem |
| +56 | Multiply D×A and add it to EF |
| 7 856 | |
| 7 856 | Clear D |
| 7 56 | Result: 7×8=56 |
Final result :
| A | B | C | D | E | F | G |
| 7 | | | | 5 | 6 | |
Yes, you are right; it is you who did the multiplication, not the abacus. In the following example, the abacus begins to show its usefulness.

### 1-digit × 2-digit multiplication

Let us multiply $7\times 83$, the multiplicand will be 83.
7×83 :
| A | B | C | D | E | F | G | H |
| 7 | | | 8 | 3 | | | |
7×83 :
| Abacus | Comment |
| ABCDEFGH | |
| 7 83 | Setting up the problem |
| +21 | Multiply E by A and add it to FG |
| 7 8321 | |
| 7 8321 | Clear E |
| 7 8 21 | |
| +56 | Multiply D by A and add it to EF |
| 7 8581 | |
| 7 8581 | Clear D |
| 7 581 | Result: 7×83=581 |
7×83 Result :
| A | B | C | D | E | F | G | H |
| 7 | | | | 5 | 8 | 1 | |
At least, the abacus has served to add the two partial products in FG and EF.

### 2-digit × 2-digit multiplication

Now, let us multiply $79\times 83$.
79×83 :
| A | B | C | D | E | F | G | H | I |
| 7 | 9 | | | 8 | 3 | | | |
Caption text :
| Abacus | Comment |
| ABCDEFGHI | |
| 79 83 | Setting up the problem |
| +21 | Multiply F by A and add it to GH |
| +27 | Multiply F by B and add it to HI |
| 79 83237 | |
| 79 83237 | Clear F |
| 79 8 237 | |
| +56 | Multiply E by A and add it to FG |
| +72 | Multiply E by B and add it to GH |
| 79 86557 | |
| 79 86557 | Clear E |
| 79 6557 | Result: 79×83=6557 |
79×83 Result :
| A | B | C | D | E | F | G | H | I |
| 7 | 9 | | | | 6 | 5 | 5 | 7 |

### Multi-digit multiplication

Generalizing what was seen in the previous examples:

#### For each digit of the multiplicand, starting from the right

   * Multiply the current digit of the multiplicand by the digits of the multiplier (from left to right), adding the first partial product to the two columns to the right of the current digit of the multiplicand, and the rest of the products by successively shifting one column to right every time.
   * Clear the current multiplicand digit.
Let us see it with the following example: $799\times 835=667165$:
799×835 :
| A | B | C | D | E | F | G | H | I | K | J | L |
| 7 | 9 | 9 | | | 8 | 3 | 5 | | | | |
799×835 :
| Abacus | Comment |
| ABCDEFGHIJKL | |
| 799 835 | Setting up the problem |
| +35 | Multiply H by A and add it to IJ |
| +45 | Multiply H by B and add it to JK |
| +45 | Multiply H by C and add it to KL |
| 799 8353995 | |
| 799 8353995 | Clear H |
| 799 83 3995 | |
| +21 | Multiply G by A and add it to HI |
| +27 | Multiply G by B and add it to IJ |
| +27 | Multiply G by C and add it to JK |
| 799 8327965 | |
| 799 8327965 | Clear G |
| 799 8 27965 | |
| +56 | Multiply F by A and add it to GH |
| +72 | Multiply F by B and add it to HI |
| +72 | Multiply F by C and add it to IJ |
| 799 8667165 | |
| 799 8667165 | Clear F |
| 799 667165 | Result: 799×835=667165 |
799×835 Result :
| A | B | C | D | E | F | G | H | I | K | J | L |
| 7 | 9 | 9 | | | | 6 | 6 | 7 | 1 | 6 | 5 |

### Embedded zeroes

3075×2707 :
| A | B | C | D | E | F | G | H | I | K | J | L | M | N | O |
| 3 | 0 | 7 | 5 | | | 2 | 7 | 0 | 7 | | | | | |
3075×2707 :
| Abacus | Comment |
| ABCDEFGHIJKLMNO | |
| 3075 2707 | Set up problem |
| +21 | Multiply JxA, add it to KL |
| +49 | Multiply JxC, add it to MN! |
| +35 | Multiply JxD, add it to NO |
| 3075 270721525 | |
| 3075 270721525 | Clear J |
| 3075 27 21525 | |
| +21 | Multiply HxA, add it to IJ |
| +49 | Multiply HxC, add it to KL! |
| +35 | Multiply HxD, add it to LM |
| 3075 272174025 | |
| 3075 272174025 | Clear H |
| 3075 2 2174025 | |
| +06 | Multiply GxA, add it to HI |
| +14 | Multiply GxC, add it to JK! |
| +10 | Multiply GxD, add it to KL |
| 3075 2 8324025 | |
| 3075 2 8324025 | Clear G |
| 3075 8324025 | Result: 3075×2707=8324025 |
3075×2707 Result :
| A | B | C | D | E | F | G | H | I | K | J | L | M | N | O |
| 3 | 0 | 7 | 5 | | | | | 8 | 3 | 2 | 4 | 0 | 2 | 5 |

## The unit rod and decimals

Please, review all the examples seen so far and check that, in all cases:
$n+1$
$n$
This is a general rule for the multiplication of natural numbers following the modern method of multiplication that we are studying. It is convenient to keep this rule in mind since the product could have zeros at the end, as in the case $32\times 1625=52000$; which could confuse you. For instance
32×1625 :
| A | B | C | D | E | F | G | H | I | K | J | L |
| 3 | 2 | | | 1 | 6 | 2 | 5u | | | | |
In the above diagram, the unit rod of multiplicand is column H (signaled with a white dot on the bar). After multiplication, the abacus shows:
32×1625 Result: 52000 :
| A | B | C | D | E | F | G | H | I | K | J | L |
| 3 | 2 | | | | | 5 | 2 | 0 | 0 | 0u | |
You need to know that the unit rod of the result is $n+1=2+1=3$ rods to the right of H (i.e. in J) to correctly read the result 52000.
We can extend this rule to decimal numbers:
$n+1$
$n$
The following table shows the $n$ values for some multipliers:
| Multiplier | n |
| 32.7 | 2 |
| 3.27 | 1 |
| 0.327 | 0 |
| 0.00327 | -2 |
Let us multiply $0.0032\times 16.25$; The unit rod of multiplicand is F.
0.0032×16.25 :
| A | B | C | D | E | F | G | H | I | K | J | L |
| 3 | 2 | | | 1 | 6 | 2 | 5 | | | | |
and for the multiplier $0.0032$, we have $n=-2$
0.0032×16.25 Result: 0.052 :
| A | B | C | D | E | F | G | H | I | K | J | L |
| 3 | 2 | | | 0 | 0 | 5 | 2 | | | | |
so that the unit rod of the product is $n+1=-2+1=-1$ rods to the right of F, i.e. one rod to its left (E) and the result must be read as $0.052$.

## References

* ↑ Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9

## External resources

Exercise sheets
* Uitti, Stephen. "Soroban Sheets (Multiplication)". Soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* "The generator". Practicing the soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

## Further reading

* Kojima, Takashi (1954), "Multiplication", The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
* Heffelfinger, Totton (2004). "Multiplication". 算盤 Abacus: Mystery of the Bead. Archived from the original on June 29, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

# Modern division
## Introduction and first methods
### Euclidean division

If we consider two natural numbers $a$ and $b$, the division of $a$ by $b$ (indicated as $a/b$ or $a\div b$) answers the question of how many times the number $b$ is contained in number $a$. Number $a$ in $a/b$ is called the dividend and $b$ the divisor. The answer is called the quotient.
Let's take $a=1225$ and $b=35$ as an example. There is no simpler way to proceed to answer the question than by repeated subtraction counting the number of times we subtract the divisor from the dividend. We can do it directly on the abacus using a column as counter:
1225÷35 = 35, primitive approach :
| Abacus | Comment |
| ABCDEFGHIJKL | |
| 35 1225 | |
| +1 -35 | subtract 35 from KL, add 1 to counter F |
| 35 1 1190 | |
| +1 -35 | subtract 35 from KL, add 1 to counter F |
| 35 2 1155 | |
| +1 -35 | subtract 35 from KL, add 1 to counter F |
| 35 3 1120 | |
| ... | Continue 33 times more... |
| 35 33 70 | |
| +1 -35 | subtract 35 from KL, add 1 to counter F |
| 35 34 35 | |
| +1 -35 | subtract 35 from KL, add 1 to counter F |
| 35 35 00 | Done, quotient is 35 in EF! |
Thus we discover that the number $35$ is contained exactly $35$ times in $1225$, since we cannot continue to subtract $35$ without starting to deal with negative numbers. Therefore, in this example, the quotient is: $q=35$.
As we can see, in this case we can write $1225=35\times 35$ or
$a=q\times b$
which we cannot expect in the general case. If we repeat the process with $a=1240$, we would see that after subtracting $35$ times $35$ we would have $15$ left on the abacus, from which we cannot continue subtracting $35$ without entering negative numbers. Therefore we have that $1240=35\times 35+15$; that is, the result of dividing $1240$ by $35$ is a quotient of $35$ leaving a remainder of $15$. In general we will have:
$a=q\times b+r$
where:
* $a$: dividend
* $b$: divisor
* $q$: quotient
* $r$: remainder
In the case that the remainder is zero, the dividend is a multiple of the divisor.
This is the concept of Euclidian division for natural numbers to which the division of numbers with decimal fractions can be reduced.

### Some improvements: Chunking methods

The procedure followed in the previous section is the simplest possible conceptually, but it is extraordinarily long and inefficient. Instead of starting directly by subtracting the divisor $b$ ($35$) from the dividend, let's start by asking what power of 10 times the divisor we can subtract from the dividend; in our case: can we subtract 3500, 350, or only 35? Clearly we can subtract 350 and we will start subtracting 350 chunks, and when we cannot continue, we will start subtracting 35 chunks as follows:
1225÷35 = 35, a great improvement :
| 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 31140 | subtract 35 from HI, add 1 to counter E, |
| 35 32105 | 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! |
Which has been a lot faster (We have intentionally reduced the distance between the counter and the dividend as much as possible. This obscures the process somewhat but brings us closer to what we will routinely do with the modern division method. Please study the above calculation carefully using your own abacus). Let's continue from here looking for even more efficiency.
If we can easily double the divisor and retain it in memory, we can shorten the operation by subtracting one or two times the divisor chunks.
| times | chunks |
| 1 | 35 |
| 2 | 70 |
1225÷35 = 35, something more sophisticated :
| 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 32105 | 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:
| times | chunks |
| 1 | 35 |
| 2 | 70 |
| 4 | 140 |
| 8 | 280 |
1225÷35 = 35, a very effective method :
| 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 00 | subtract 35 from HI, add 1 to counter E. |
| 35 35 | No remainder. Done, quotient is 35! |
which is somewhat shorter and, clearly, nothing could be faster than having a complete multiplication table of the divisor

### Multiplication table

Multiplication table by 35 :
| times | chunks |
| 1 | 35 |
| 2 | 70 |
| 3 | 105 |
| 4 | 140 |
| 5 | 175 |
| 6 | 210 |
| 7 | 245 |
| 8 | 280 |
| 9 | 315 |
then
1225÷35 = 35, the shortest method! :
| Abacus | Comment |
| ABCDEFGHI | |
| 35 1225 | Start, counter in D, |
| 35 3 175 | subtract 105 from GH, add 3 to counter D, |
| 35 35 00 | subtract 175 from HI, add 5 to counter E. |
| 35 35 | No remainder. Done, quotient is 35! |
There is no doubt, this is an optimal division method, nothing can be faster and more comfortable ... once we have a chunk table like the one above. But calculating the chunk table is time consuming and requires paper and pencil to write it and this extra work would only be justified if we have a large number of divisions to do with the same common divisor.
In 1617 John Napier, the father of logarithms, presented his invention to alleviate this problem consisting of a series of rods, known as Napier's Bones, with the one-digit multiplication table written on them and that could be combined to get the multiplication table of any number. For example, in our case
Multiplication table by 35 using Napier's bones :
| 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 traditional 1-digit multiplication table that is learned by heart and that we are going to use as an approximation to the specific multiplication table of the divisor (the one used above), this table will guide us to choose the digit of the quotient that we must try.
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 article, you will be amazed at the variety of division methods that can be performed.

## Modern Division
### Modern vs. traditional division

The modern method of division is so called because throughout the first half of the 20th century its use has displaced that of the traditional method, but in fact it is much older than this, having been displaced by it in the 13th century. A characteristic of the modern method is the use of the 1-digit multiplication table both as a guide to the choice of the interim quotient figure that we have to try and for the calculation of the chunk that we have to subtract from the dividend.
Decimal multiplication table :
| × | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 |
| 3 | 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 |
| 4 | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 |
| 5 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 |
| 6 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 |
| 7 | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 |
| 8 | 8 | 16 | 24 | 32 | 40 | 48 | 56 | 64 | 72 |
| 9 | 9 | 18 | 27 | 36 | 45 | 54 | 63 | 72 | 81 |
By comparison, the traditional method uses both a special division table as a guide for the interim quotient figure and the multiplication table for calculating the chunk to subtract.
The main reason why the modern method began to displace the traditional method in Japan after the Meiji Restoration is that it can be learned more easily and quickly by those who already know how to divide with paper and pencil, since it does not require memorization of the complex division table. On the other hand, the traditional method makes the division a completely automated process, without the need to think, one only has to follow the rules to obtain the result, which allows the operation to be carried out without any mental fatigue. If you are interested in this topic you can consult the Wikibook: Traditional Abacus and Bead Arithmetic.

### Key point of division with the abacus

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, which makes the abacus an instrument especially suited to trial and error procedures. 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

    that is, we can subtract the chunk (interim quotient digit) ✕ (divisor) without entering negative numbers but we cannot subtract the quotient one more time because the remainder is less than the divisor.

#### It is insufficient and we must revise it up

    we can subtract the chunk (interim quotient digit) ✕ (divisor) without entering negative numbers but we can still subtract the quotient one more time because the remainder is greater/equal than the divisor. We add one to the interim quotient and subtract the divisor again from the remainder.

#### It is excessive and we must revise it down

    this is the most complex and error-prone situation. We usually discover too late (in the middle of the chunk subtraction) that the interim figure is excessive and we need to go back, subtract one from the quotient, and restore the dividend/remainder by adding what has been subtracted in excess before we can continue.
Therefore, the process of obtaining a digit of the quotient has two phases:
* Choose an interim quotient digit.
* Test if it is correct and modify it if not.
Once we have found the correct figure, we will generally have a non-zero remainder that will act as a dividend if we want to extend the division to the next digit of the quotient.
We will see all of this throughout the examples that follow, but first, we need a few words about how to organize the division on the abacus.

### Modern division arrangement

The dividend is the active term with which we are going to work in the abacus, the divisor is inactive and remains unchanged during the operation, in fact it is not essential to enter it in the abacus but it is recommended, especially for beginners. As in the case of multiplication, there are two styles to place dividend and divisor on the abacus, each with its advantages and disadvantages.

#### Traditional Chinese arrangement

The divisor goes to the far right of the abacus while the dividend is to the left, leaving at least two columns free to its left.
1225÷35 Chinese style arrangement :
| A | B | C | D | E | F | G | H | I | K | J | L | M |
| | | 1 | 2 | 2 | 5 | | | | | | 3 | 5 |

#### Traditional Japanese arrangement

The divisor goes to the extreme left of the abacus while the dividend is to its right, leaving at least four free columns between the two terms.
1225÷35 Japanese style arrangement :
| A | B | C | D | E | F | G | H | I | K | J | L | M |
| 3 | 5 | | | | | 1 | 2 | 2 | 5 | | | |
In this book we will use the Japanese style for the examples, but feel free to try both.

#### Placing the quotient figure

The interim quotient figure is placed in one of the two columns directly to the left of the dividend. To decide which one we need to compare the divisor with an equal number of the first digits of the dividend, adding zeros to its right if necessary; call this the working dividend:

##### Working dividend greater than or equal to the divisor

    this means that the divisor goes into the working dividend and the quotient, i.e. the number of times the divisor goes into the working dividend, is arranged in the second column to the left of the first digit of the dividend
    Example: 827÷46, 82, the working dividend, is greater than 46, then the interim quotient goes to the second column to the left of dividend. Multiplication table suggest us using 2 as interim quotient (simplify 827÷46 to 8÷4)
    827÷46 AbacusComment ABCDEFGHIJKLM 46 827 46 2 827 Place interim quotient 2 in E

##### Working dividend smaller than the divisor

    this means that the divisor does not go into the working dividend. In this case, we need to include the next digit of the dividend, or a zero if there are no more left, in our working dividend, and the quotient, the number of times the divisor goes into this expanded working dividend, is arranged in the column directly to the left of the first digit of the dividend
    Example: 18÷467, 180 is less than 467, then the working dividend is 1800 and the interim quotient goes to the first column to the left of dividend. Multiplication table suggest us using 4 as interim quotient after simplifying 1800÷467 to 18÷4.
    Caption text AbacusComment ABCDEFGHIJKLM 467 18 467 418 Place interim quotient 4 in G

### Examples

You should start by doing exercises with single-digit divisors and later try divisors with two, three, etc. figures. With one-digit divisors you should never have to revise up or down. For example you can divide 123456789 by the digits 2, 3, ..., 9. Let's see the division by 9 here.

#### Example: 123456789÷9 = 13717421

* Please read the "->" symbol as: "the multiplication table suggests using ...".
* As you will see, in all cases except the last one, the working dividend is less than the divisor and we need to expand it to two digits.
123456789÷9 = 13717421 :
| Abacus | Comment |
| ABCDEFGHIJKLMNO | |
| 9 123456789 | 12/9 -> 1 as interim quotient |
| 9 1123456789 | place i. quotient in E |
| -9 | subtract 9✕1=9 from FG |
| 9 1 33456789 | 33/9 -> 3 as interim quotient |
| 9 1333456789 | place i. quotient in F |
| -27 | subtract 9✕3=27 from GH |
| 9 13 6456789 | 64/9 -> 7 as interim quotient |
| 9 1376456789 | place i. quotient in G |
| -63 | subtract 9✕7=63 from HI |
| 9 137 156789 | 15/9 -> 1 as interim quotient |
| 9 1371156789 | place i. quotient in H |
| -9 | subtract 9✕1=9 from IJ |
| 9 1371 66789 | 66/9 -> 7 as interim quotient |
| 9 1371766789 | place i. quotient in I |
| -63 | subtract 9✕7=63 from JK |
| 9 13717 3789 | 37/9 -> 4 as interim quotient |
| 9 1371743789 | place i. quotient in J |
| -36 | subtract 9✕4=36 from KL |
| 9 137174 189 | 18/9 -> 2 as interim quotient |
| 9 1371742189 | place i. quotient in K |
| -18 | subtract 9✕2=18 from LM |
| 9 1371742 9 | 9/9 -> 1 as interim quotient |
| 9 13717421 9 | place i. quotient in L |
| -9 | subtract 9✕1=9 from MN |
| 9 13717421 | No remainder! Done 123456789÷9 = 13717421 |
123456789 is a curious number, it is precisely the product of 9 by 13717421, a large prime!

#### Example: 1225÷35 = 35 Two-digit divisor. Revising down and up

1225÷35 = 35 :
| Abacus | Comment |
| ABCDEFGHIJ | |
| 35 1225 | 12÷3↦4 as interim quotient |
| +4 | enter interim quotient in F |
| 35 41225 | Now try to subtract the chunk 4✕35 from GHI, |
| -12 | first 4✕3 from GH |
| 35 40025 | then 4✕5 from HI |
| -20 | Cannot subtract! |
| -1 | Revise down interim quotient digit |
| 35 30025 | |
| +3 | return the excess subtracted from GH |
| 35 30325 | |
| -15 | continue normally, subtract 3✕5 from HI |
| 35 3 175 | 17÷3↦5 as interim quotient |
| +5 | enter interim quotient in G |
| 35 35175 | Try to subtract chunk 5✕35 from HIJ |
| -15 | first 5✕3 from HI |
| 35 35025 | |
| -25 | then 5✕5 from IJ |
| 35 35 | No remainder, done! 1225÷35 = 35 |
Note:^a We have subtracted 4 × 3 = 12 from FGH, but if the correct quotient digit is 3, we should have subtracted 3 × 3 = 9, so we subtracted 3 in excess (just the first digit of the divisor). We must return this excess before continuing.
Now, suppose that after our "bad experience" revising down the first figure of the quotient, and in an excess of prudence, we choose 4 as as the second interim quotient instead of 5 as suggested by the multiplication table. Then we continue this way:
1225÷35 = 35, second quotient digit, example of revising up :
| Abacus | Comment |
| ABCDEFGHI | |
| 35 3 175 | 17÷3 -> 5, but we use 4! |
| +4 | enter interim quotient in G |
| 35 34175 | Try to subtract chunk 4✕35 from HIJ |
| -12 | first 4✕3 = 12 from HI |
| -20 | then 4✕5 = 20 from IJ |
| 35 34 35 | remainder greater or equal to divisor! |
| +1 | revise up G |
| -35 | subtract divisor from remainder HI |
| 35 35 | No remainder, Done! |

#### Example 1÷327

So far we have considered divisions between natural numbers with quotients and remainders as well as natural numbers, but we can operate with decimal numbers exactly as we do in written calculation with long division. For example, let's find the inverse of 327; that is, 1/327 in an abacus with 13 columns.
1÷327 = 0.00305810... :
| Abacus | Comment |
| ABCDEFGHIJKLM | |
| 327 1 | 10/3 -> 3 as interim quotient |
| 327 31 | enter interim quotient in G |
| -09 | subtract 3✕3=9 from HI |
| 327 3 1 | |
| -06 | subtract 3✕2=6 from IJ |
| 327 3 4 | |
| -21 | subtract 3✕7=21 from JK |
| 327 3 19 | 19/3 -> 6 as interim quotient |
| 327 30619 | enter interim quotient in I |
| -18 | subtract 6✕3=18 from JK |
| 327 306 1 | |
| -12 | cannot subtract 6✕2=12 from KL! |
| -1 | revise down I |
| +3 | return the excess subtracted from JK |
| 327 305 4 | |
| -10 | continue normally, subtract 5✕2=10 from KL |
| 327 305 30 | |
| -35 | subtract 5✕7=35 from LM |
| 327 305 265 | 36/3 -> 8 as interim quotient |
| 327 3058265 | enter interim quotient in J |
| -24 | subtract 8✕3=24 from KL |
| 327 3058 25 | |
| -16 | subtract 8✕2=16 from LM |
| 327 3058 9 | cannot continue! Result: 3058 |
We have obtained $3058$ as the first digits of $1/327$, but $1/327\approx 1/300=1/(3\cdot 100)=(1/3)\cdot 0.01\approx 0.33\cdot 0.01=0.0033$ so our result is actually $0.003058$. See below the rule to find the unit rod of the division.

#### Example: 634263÷79283 = 7,999987..., a tricky case

Finally let's get the first digit of the quotient of this especially malicious division
634263÷79283 = 7,999987... :
| Abacus | Comment |
| ABCDEFGHIJKLM | |
| 79283 634263 | 63/7 -> try 9 |
| 79283 9634263 | |
| -63 | subtract 9*7=63 from HI |
| 79283 9004263 | |
| -81 | cannot subtract 9*9=81 from IJ! |
| -1 | revise D down |
| +7 | restore excess subtracted from remainder |
| 79283 8 74263 | |
| -72 | continue subtracting 8x9=72 from IJ |
| 79283 8 02263 | |
| -16 | subtract 8*2=16 from JK |
| 79283 8 00663 | |
| -64 | subtract 8*8=64 from KL |
| 79283 8 00023 | cannot subtract 8*3=24 from LM! |
| -1 | revise D down |
| +7928 | restore excess subtracted from remainder |
| 79283 7 79303 | |
| -21 | continue subtracting 7x3=21 from LM |
| 79283 7 79282 | quotient: 7, remainder: 79282 |
There is no doubt that in this case rounding the divisor 79283 to 80000 would have given us better results since 63÷/8 suggests using 7 (the correct figure) as the interim quotient digit.
634263÷79283 = 7,999987..., rounding divisor :
| Abacus | Comment |
| ABCDEFGHIJKLM | |
| 79283 634263 | 63/8 -> try 7 |
| 7634263 | |
| -49 | subtract 7*7=49 from HI |
| 79283 7144263 | |
| -63 | subtract 7*9=63 from IJ |
| 79283 7 81263 | |
| -14 | subtract 7*2=14 from JK |
| 79283 7 79863 | |
| -56 | subtract 7*8=56 from KL |
| 79283 7 79303 | |
| -21 | subtract 7*3=21 from LM |
| 79283 7 79282 | quotient: 7, remainder: 79282 |

## The unit rod and decimals

The counterpart of the rule to find the unit rod of multiplication is the following rule for division:
$n+1$
$n$
The following table shows the $n$ values for some divisors:
| Multiplier | n |
| 32.7 | 2 |
| 3.27 | 1 |
| 0.327 | 0 |
| 0.00327 | -2 |
Example: 1/327 (we have seen it above)
1/327 unit rod :
| Abacus | Comment |
| ABCDEFGHIJKLM | |
| 327 1 | divisor has 3 digits. n=3 |
| . | dividend unit rod |
| 327 3058 9 | End of division. Result: 3058 |
| . | dividend unit rod |
| <--- | shift it n+1 = 4 positions to the left |
| . | unit rod of quotient |
| 3058 | so that this... |
| .003058 | ... should be read 0.003058 |

## Multiplication and division as inverse operations

In written calculations we can always review our calculation to make sure that we have not made mistakes and that the result obtained is correct. In calculations with the abacus this is not possible since the abacus does not keep memory of the past and of intermediate results. We can resort to some sanity tests such as casting nines or elevens out, but the traditional way of checking the results with the abacus has been either to repeat the calculations or to undo the calculations.
Undoing additions and subtractions is as simple as starting from the result and subtracting what we have added, adding what we have subtracted; If we do both the calculation and the verification correctly, we should end up with a reset abacus. To verify a multiplication we will use division and, reciprocally, to verify a division we will use multiplication, adding the remainder if there is one. After doing this we will return the abacus to its starting state with the two original operands in their initial positions. Let's see an example:
Testing 2461÷64 by multiplication :
| Abacus | Comment |
| ABCDEFGHIJ | |
| 64 2461 | 24/6 -> 4 as interim quotient |
| 42461 | enter interim quotient in F |
| -24 | subtract 4✕6=24 from GH |
| 64 4 61 | |
| -16 | cannot subtract 4✕4=16 from HI |
| -1 | revise down interim quotient digit |
| 64 3 61 | |
| +6 | return the excess subtracted from GH |
| 64 3 661 | |
| -12 | continue normally, subtract 3✕4=12 from HI |
| 64 3 541 | 54/6 -> 9, but we will use 8 |
| 64 38541 | |
| -48 | subtract 8✕6=48 from HI |
| 64 38 61 | |
| -32 | subtract 8✕4=32 from IJ |
| 64 38 29 | quotient: 38, remainder 29 |
| | revision by multiplication start here! |
| +48 | add 8✕6=48 to HI |
| 64 38509 | |
| +32 | add 8✕4=32 to IJ |
| 64 38541 | |
| 64 3 541 | clear G |
| +18 | add 3✕6=18 to GH |
| 64 32341 | |
| +12 | add 3✕4=12 to HI |
| 64 32461 | |
| 64 2461 | clear F. Initial status! |
It has been suggested in this book to use the number 123456789 for your first exercises in both multiplication and division by a single digit. Try combining them with the reverse operation; for example: divide 123456789 by 9 to get 13717421 and multiply this result by 9 to get 123456789 back to the same starting position on the abacus. Or start by multiplying 123456789 by 9 to get 1111111101 and then divide by 9 to get back to where you started. Try all the digits from 2 to 9.

## References

* ↑ Wilson, Jeff. "Long Division Teaching Aid, "Double Division"". Double Division. Archived from the original on March 02, 2021. {{cite web}}: Check date values in: |archivedate= (help); Text "year 2005" ignored (help); Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* ↑ "The Definitive Higher Math Guide on Integer Long Division (and Its Variants)". Math Vault. Archived from the original on May 14, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help); Unknown parameter |name= ignored (help)

## External resources

Exercise sheets
* "The generator". Practicing the soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

## Further reading

* Kojima, Takashi (1954), "Division", The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
* Heffelfinger, Totton (2004). "Division". 算盤 Abacus: Mystery of the Bead. Archived from the original on June 29, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* Siqueira, Edvaldo (2004). "Decimals & Division". 算盤 Abacus: Mystery of the Bead. Archived from the original on May 6, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)