ABACUS: MYSTERY OF THE BEAD
The Bead Unbaffled - An Abacus Manual

Long Division Techniques - Chinese Suan Pan

This technique was first shown to me by Torsten Reincke. It's the second chapter in how to solve problems of division using traditional Chinese techniques. Before continuing it's suggested that one should have an understanding of the techniques and rules that govern Chinese short division.

This method of long division uses a traditional 2:5 bead suan pan. In order to solve the problem it's best to have an understanding of two techniques illustrated in Traditional Multiplication Techniques;

1. The Extra Bead Technique which uses the top most heaven bead. It has a value of 5. (See Fig.2 - the extra bead is placed on rod G.)
    
2. The Suspended Bead Technique which also uses the top most heaven bead but in this case it is set  ½ way down the rod so that the bead neither touches the frame above nor the bead below. It has a value of 10. (See Fig.5 - the suspended bead is placed on rod H.)

The following example uses the rules;

a) rule 5 /9= 5 plus5
b) rule 8 /9= 8 plus 8
c) rule 9 /9= forward 1

Example:  58174 ÷ 986 = 59

Step 1: Set the dividend 58174 on rods FGHIJ and the divisor 986 on ABC.  (Fig.1)

Fig. 1

Step 1 A B C D E F G H I J K L M       9 8 6 0 0 5 8 1 7 4 0 0 0

 Step 2: Following   rule 5 /9= 5 plus 5     ===>  place the quotient 5 on rod F and the remainder 5 on G, leaving 5 "13" 1 7 4 on FGHIJ. (Note: This is where we use the  *The Extra Bead Technique*  which allows for 13 to be placed on rod G.) (Fig.2)

Fig. 2

Step 2 A B C D E F G H I J K L M       9 8 6 0 0 5 8 1 7 4 0 0 0 plus 5 Step 2 9 8 6 0 0 5"13"1 7 4 0 0 0

2a: Multiply 5 on F by 8 on B. Subtract the product 40 from rods GH, leaving, 59174 on rods FGHIJ. (Fig.3)

Fig. 3

Step 2a A B C D E F G H I J K L M       9 8 6 0 0 5"13"1 7 4 0 0 0 - 4 0 Step 2a 9 8 6 0 0 5 9 1 7 4 0 0 0

2b: Multiply 5 on F by 6 on C. Subtract the product 30 from rods GH, leaving, 58874 on rods FGHIJ. (Fig.4)

Fig. 4

Step 2b A B C D E F G H I J K L M       9 8 6 0 0 5 9 1 7 4 0 0 0 - 3 0 Step 2b 9 8 6 0 0 5 8 8 7 4 0 0 0

 Step 3: Following   rule 8 /9= 8 plus 8     ===>  place the quotient 8 on rod G and the remainder 8 on H.  (Note: Since rod H already carries a large number we don't have 8 available. This is where we'll use  *The Suspended Bead Technique*  which counts for 10.)  Slide the top most bead ½ way down on rod H for a value of 10 then subtract 2 giving us the required 8. This  leaves 5 8 "16" 7 4 on rods FGHIJ. (Fig.5)

Fig. 5

Step 3 A B C D E F G H I J K L M       9 8 6 0 0 5 8 8 7 4 0 0 0 suspended "10" - 2 Step 3 9 8 6 0 0 5 8"16"7 4 0 0 0

3a: Multiply the 8 on G by the 8 on B and subtract the product 64 from rods HI, leaving 5 8 "10" 3 4 on FGHIJ. (Fig.6)  ***see a variation below***

Fig. 6

Step 3a A B C D E F G H I J K L M       9 8 6 0 0 5 8"16"7 4 0 0 0 - 6 4 Step 3a 9 8 6 0 0 5 8"10"3 4 0 0 0

3b: Multiply the 8 on G by the 6 on C and subtract the product 48 from rods IJ, leaving 58986 on rods FGHIJ. (Fig.7)

Fig. 7

Step 3b A B C D E F G H I J K L M       9 8 6 0 0 5 8"10"3 4 0 0 0 - 4 8 Step 3b 9 8 6 0 0 5 8 9 8 6 0 0 0

3c:  Here we make a revision. Changing the 8 on G to a 9 we can now subtract a further 986 from rods HIJ. This leaves 59 on rods FG which is the answer. (Fig.8)

Fig. 8

Step 3c A B C D E F G H I J K L M       9 8 6 0 0 5 9 0 5 4 0 0 0 - 5 4 Step 3c 9 8 6 0 0 5 9 0 0 0 0 0 0

 

*** In the above step 3a one could have exercised the option to do a revision using  rule 9 /9= forward 1     ===>  thereby changing the 8 on rod G to a 9. Multiplying the revised 9 on G by the 8 on B followed by the 6 on C and subtracting the products from rods HIJ, one could have come to the answer in a slightly different way.***

 

▪ Abacus: Mystery of the Bead
▪ Advanced Abacus Techniques

 

 

© June, 2004
Totton Heffelfinger  Toronto Ontario  Canada
Email
totton[at]idirect[dot]com