## How many multiplication methods are there?

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

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

That is, all the products listed in this table:

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

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

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

## Inverse operations

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The traditional method of multiplication was first introduced using counting rods[1] and the best way to introduce it to the modern abacist is to consider that a multi-digit multiplier consists of a head (the first digit from the left) and a body (the rest of the digits); for example: 4567✕23, considering 4567 as the multiplier, its head is 4 and the body 567. So, for each digit of the multiplicand (from right to left):

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

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

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

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

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

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

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

## References

1. Volkov, Alexei (2018), "Visual Representations of Arithmetical Operations Performed with Counting Instruments in Chinese Mathematical Treatises", Researching the History of Mathematics Education - An International Overview, Springer Publishing, ISBN 978-3-319-68293-8 {{citation}}: Unknown parameter |editor1first= ignored (|editor-first1= suggested) (help); Unknown parameter |editor1last= ignored (|editor-last1= suggested) (help); Unknown parameter |editor2first= ignored (|editor-first2= suggested) (help); Unknown parameter |editor2last= ignored (|editor-last2= suggested) (help)