Using an Abacus/Modern multiplication

You should achieve some basic skills in addition and subtraction before going further with multiplication. Otherwise you will be calling for frustration.

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:

Mechanical calculators of the past used the procedure on the left to multiply

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

The Tsinghua Bamboo Slips, Chinese Warring States era decimal multiplication table of 305 BC

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

   47
  ×23
-----
   21
  12   (×10)
  14   (×10)
+ 8    (×100)
-----
 1081

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:

The column of the units of the product is located $n+1$ columns to the right of the column of the units of the multiplicand; where $n$ is the number of digits of the multiplier.

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:

The column of the units of the product is located $n+1$ columns to the right of the column of the units of the multiplicand; where $n$ is the number of digits of the multiplier to the left of its decimal point (which could be negative!).

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 $0.052$.

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)