The Chinese Abacus 7

Multiplication and Division

These processes are a little more complex than addition and subtraction. It requires the user to be familiar with the multiplication tables. In essence, the abacus becomes only a recording device where interim results are tabulated. It is only in the final movement of the last manipulated bead, that the answer is revealed.

The MULTIPLICAND is the number we start off with. The MULTIPLIER is the number we multiply the multiplicand with. The PRODUCT is the result of the multiplication.

  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

The multiplication tables shown above is colour coded to show the multiplier, multiplicand, and also those products which are single digit values, and double digit values.

When we tackle the idea of multiplication, the learner to needs to take into account the number and position of the rods at each stage. For the most part, two rods are must be used for each step of the manipulation, but occasionally, three rods may be involved.

The process of multiplication and division is complicated by the position of the decimal point. We have to make room on the abacus for it, and there are three important cases which arise with mumbers. A whole number is known as an INTEGRAL. Therefore, DECIMALS are non-integral because they contain part of a whole number.

Since decimals are base 10 numerals, the decimal point does not take too much manipulation. We shall first explore the method for integral multiplicand and multiplier first, then extend the idea towards the other two cases.

Multiplication

713 x 289 = ?

The integer multiplicand 713 is multiplied by the multiplier 289. We must first set out space on the abacus for our calculation.

In our example, both 713 and 289 have three digits each so that makes a maximum number of 6 digits in the resultant product. Hence, for the calculation, six rods on the abacus is needed.

In this case, count three rods along from the right side of the abacus, then another three and place the multiplier in the latter three rods. We shall see that the production will eventually replace the multiplier as each of its digits are considered in the calculation. Some where away from the six rods, you place your multiplicand for reference.

The process of multiplication requires the learner to start with the rightmost digit of the multiplier, and mulitiply the multiplicand digit by digit from the left of the multiplicand to the right. The second rightmost digit of the multiplier then repeats the process with thee multiplicand in the same manner until the leftmost digit of the multiplier is done. The product will be that which is shown at the end of the manipulation.

So, the calculation follows the following route.

713 x 289

Step (1) The right most digit of the multiplier is "9". Multiply the leftmost digit of the multiplicand, "7".
Step (2) Next, 9 multiplies the next digit of the multiplicand, "1".
Step (3) Then, 9 multiplies the last digit of the multiplicand, "3".

We then repeat this process with the second digit of the multiplier "8", and finally the leftmost digit "2"

As we said, the manipulation of the beads on the abacus requires one to think about two rods at a time. For example, in step (1), the product of this minor calculation was 63. We note the position of digit nine of the multiplier and the rod to its right. This forms the tens and units place in which the numeral 63 is inserted, replacing the digit 9 by the 6 and placing 3 to the rod on its right. In step (2), the product is nine, a single digit. Again, we consider this as a double digit number 09, and insert increment 0 beads in the tens (where the digit 3 appears), and put in the value of 9 on the abacus to the next rod on the right. In step (3), the product of 9 by 3 is 27. This move requires us to consider more than two rods. This is because we must put in the value 2 to the position of 9 on the rods, but this would make 11, hence the a bead on the next left rod is incremented by one, and the rod containing 9 would be ammended to 1, and finally the digit 7 would be put into the rightmost column as the units column.

The next step with the second digit of the multiplier, viz, "8", would give the products 56, 8 and 24. The first digit of 56, "5" replaces the the multiplier digit "8" and the second digit is added to the value on the rod immediately to the right, and so on.

Finally, the products of the last digit of the multiplier, viz, "2", yields the products, 14, 2, 6. The first digit of 14, "1", replaces the first digit of the original multiplier, and the second digit is added to the rod immediately to its right. The other numbers are placed in in the same way described above.

What we have is the following representation


*** are unused rods. 

--- are places which have not been considered yet, 

    but are about to change because of carrying overflow.

+++ are digits which are not part of the final product

713 289*** Abacus is loaded thus Abacus reads 289*** ++ 63 The product of 9 and 7 Abacus reads 2863** 9 The product of 9 and 1 Abacus reads 28639* 27 The product of 9 and 3 Abacus reads 286417 + 56 The product of 8 and 7 Abacus reads 25---- then 262417 8 The product of 8 and 1 Abacus reads 262--- then 263217 24 The product of 8 and 3 Abacus reads 263457 14 The product of 2 and 7 Abacus reads 1----- then 203457 2 The product of 2 and 1 Abacus reads 205457 6 The product of 2 and 3 Abacus reads 205--- then 206057 The answer is the product 206057.

Manipulation of the decimal point

We present here a simple method for manipulating the decimal point which basically stems from arithmetic, rather from abacus tradition. Since we are using base10 it is just a matter of noting how many decimal significant places there are in both the multiplicand and multiplier.

Our reason for doing this is itself simple. The abacus does not physically dictate where the last integer and first decimal digits are. At the end of the day, the calculations produce a bead representation of the final product or result. All that is needed if there were decimals involved would be ascertaining where the decimal point should be placed.

A decimal number contains three parts, the integer part (I), the decimal point (.), and finally the decimal part (D) or I.D

Supposing we had a number which can be represented by III.DDDDD (e.g. 123.12345 ).

We see that the integer part contains three digits. The decimal part contains five digits. When placing the number onto the abacus, we cannot indicate the decimal point in anyway. We would have just loaded in 12312345 as in the example above.

Example


2742 391***     Abacus is loaded thus   Abacus reads 391****

                                                     ++

       *2       The product of 1 and 2  Abacus reads 3902***

        *7      The product of 1 and 7  Abacus reads 39027**

         *4     The product of 1 and 4  Abacus reads 390274*

          *2    The product of 1 and 2  Abacus reads 3902742

                                                     +

      18        The product of 9 and 2  Abacus reads 3182742

       63       The product of 9 and 7  Abacus reads 3245742

        36      The product of 9 and 4  Abacus reads 3249342

         18     The product of 9 and 2  Abacus reads 3249522



     *6         The product of 3 and 2  Abacus reads 0849522

      21        The product of 3 and 7  Abacus reads 1059522

       12       The product of 3 and 4  Abacus reads 1071522

        *6      The product of 3 and 2  Abacus reads 1072122



The answer is the product 1072122. However, we must now put back in the number of decimal significant places in our original problem (27.42 x 3.91) We note that there are four decimal significant places in total, therefore, we must count four digits from the right to obtain the correct answer : 107.2122 = 27.42 x 3.91

There is one special case worth mentioning. That is when one or both numbers has zero as the integer part of the decimal. In this case, use only the significant numbers in the multiplication, and then follow the method for finding the correct position of the decimal point as usual.


0.028 x 0.11

28 x 11 = 308

There are five decimal significant places, therefore the answer is 0.00308 = 0.028 x 0.11 as the correct answer. That's all there is to decimal multiplication. The section on division can be found in Page 8.

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© Dylan W.H.Sung 1996-onwards
This page was created on Sunday 28th June 1998. It was recently updated on Monday 22nd April 2002