# Using an Abacus/Modern division

## Introduction and first methods

### Euclidean division

If we consider two natural numbers ${\displaystyle a}$ and ${\displaystyle b}$, the division of ${\displaystyle a}$ by ${\displaystyle b}$ (indicated as ${\displaystyle a/b}$ or ${\displaystyle a\div b}$) answers the question of how many times the number ${\displaystyle b}$ is contained in number ${\displaystyle a}$. Number ${\displaystyle a}$ in ${\displaystyle a/b}$ is called the dividend and ${\displaystyle b}$ the divisor. The answer is called the quotient.

Let's take ${\displaystyle a=1225}$ and ${\displaystyle b=35}$ 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 ${\displaystyle 35}$ is contained exactly ${\displaystyle 35}$ times in ${\displaystyle 1225}$, since we cannot continue to subtract ${\displaystyle 35}$ without starting to deal with negative numbers. Therefore, in this example, the quotient is: ${\displaystyle q=35}$.

As we can see, in this case we can write ${\displaystyle 1225=35\times 35}$ or

${\displaystyle a=q\times b}$

which we cannot expect in the general case. If we repeat the process with ${\displaystyle a=1240}$, we would see that after subtracting ${\displaystyle 35}$ times ${\displaystyle 35}$ we would have ${\displaystyle 15}$ left on the abacus, from which we cannot continue subtracting ${\displaystyle 35}$ without entering negative numbers. Therefore we have that ${\displaystyle 1240=35\times 35+15}$; that is, the result of dividing ${\displaystyle 1240}$ by ${\displaystyle 35}$ is a quotient of ${\displaystyle 35}$ leaving a remainder of ${\displaystyle 15}$. In general we will have:

${\displaystyle a=q\times b+r}$

where:

• ${\displaystyle a}$: dividend
• ${\displaystyle b}$: divisor
• ${\displaystyle q}$: quotient
• ${\displaystyle r}$: 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 ${\displaystyle b}$ (${\displaystyle 35}$) 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[1]:

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

 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[2] 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:

1. Choose an interim quotient digit.
2. 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.

A B C D E F 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.

A B C D E F 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
Abacus Comment
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
Abacus Comment
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 GHa
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 20 remainder greater or equal to divisor!
+1 revise up G
-20 subtract divisor from remainder HI
35   34 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 ${\displaystyle 3058}$ as the first digits of ${\displaystyle 1/327}$, but ${\displaystyle 1/327\approx 1/300=1/(3\cdot 100)=(1/3)\cdot 0.01\approx 0.33\cdot 0.01=0.0033}$ so our result is actually ${\displaystyle 0.003058}$. 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 9*3=27 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: 79283

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: 79283

## The unit rod and decimals

The counterpart of the rule to find the unit rod of multiplication is the following rule for division:

The column of the units of the quotients is located ${\displaystyle n+1}$ columns to the left of the column of the units of the dividend; where ${\displaystyle n}$ is the number of digits of the divisor to the left of its decimal point (which could be negative!).

The following table shows the ${\displaystyle n}$ 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

1. Wilson, Jeff. "Long Division Teaching Aid, "Double Division"". Double Division. Archived from the original on March 02, 2021.
2. "The Definitive Higher Math Guide on Integer Long Division (and Its Variants)". Math Vault. Archived from the original on May 14, 2021.

Exercise sheets