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

Chinese Division Rules on a Soroban

When solving problems of division on a 2:5 bead suan pan I often make use of the Chinese division table. The table is intuitive, easy to learn and takes much of the guesswork out of doing problems of division on the abacus. But it's really best used on a suan pan. Or so I thought... a recent discussion in the Yahoo Soroban/Abacus news group proved me wrong.

In his book 'The Japanese Abacus, its use and theory' Takashi Kojima writes, "There are two fundamental methods of division on the abacus. The older method, though still favored by some, has fallen out of general use since about 1930 because it requires the memorization of a special division table." 

The Chinese rule advantage is easily adapted for use on a 1:4 bead soroban. Let's see how how it works.

Example 1: 259 7 = 37

This example follows the rules:

Step 1: Rod I will be the unit rod. Set the divisor on the left and the dividend on the right.

    (Fig.1)  
Step 1

A B C D E F G H I J K L M
    .     .     .     .  
0 0 7 0 0 0 2 5 9 0 0 0 0   Step 1

Follow Kojima's rules for setting the quotient answer. Divisor 7 is larger than the dividend 2, place the first quotient number one rod to the left of the dividend.

Step 2: Chinese rule:  7 meets 2, 2 + 6.  Place quotient 2 on rod F and clear 2 from G.

    (Fig.2)  
Step 2

A B C D E F G H I J K L M
    .     .     .     .  
0 0 7 0 0 0 2 5 9 0 0 0 0   Step 2
          2
clear      -2            
0 0 7 0 0 2 0 5 9 0 0 0 0

Step 3: Add 6 to rod H leaving 11 on GH. With 11 remaining we must revise the answer.

    (Fig.3)  
Step 3

A B C D E F G H I J K L M
    .     .     .     .  
0 0 7 0 0 2 0 5 9 0 0 0 0   Step 3
             +6          
0 0 7 0 0 2 1 1 9 0 0 0 0

Step 4: Revise. Add 1 to quotient 2 and subtract at further 7 from GH leaving 49 on rods HI.

    (Fig.4)  
Step 4

A B C D E F G H I J K L M
    .     .     .     .  
0 0 7 0 0 2 1 1 9 0 0 0 0   Step 4
         +1
             -7          
0 0 7 0 0 3 0 4 9 0 0 0 0

At this stage it will be obvious to many that the next quotient number will be 7 but let's follow the rules in order to see how they work.

Step 5: Chinese rule:  7 meets 4, 5 + 5. Place quotient 5 on rod G and clear 4 from rod H.

   (Fig.5)  
Step 5

A B C D E F G H I J K L M
    .     .     .     .  
0 0 7 0 0 3 0 4 9 0 0 0 0   Step 5 
            5
clear        -4          
0 0 7 0 0 3 5 0 9 0 0 0 0

Step 6: Add 5 to rod I leaving 14 on rods HI. With 14 remaining we must revise the answer.

    (Fig.6)  
Step 6

A B C D E F G H I J K L M
    .     .     .     .  
0 0 7 0 0 3 5 0 9 0 0 0 0   Step 6
               +5        
0 0 7 0 0 3 5 1 4 0 0 0 0

Step 7: Revise. Add 2 to quotient 5 and subtract 14 from rods HI.
Step 7a: Shift the unit rod. Starting on rod H count 1 to the left. The answer is 37 on rods FG.

    (Fig.7)  
Step 7

A B C D E F G H I J K L M
    .     .     .     .  
0 0 7 0 0 3 5 1 4 0 0 0 0   Step 7
           +2
             -1 4        
0 0 7 0 0 3 7 0 0 0 0 0 0

  
Example 2: 962 13 = 74

This example follows the rules:

Step 1: Rod I will be the unit rod. Set the divisor on the left and the dividend on the right.

    (Fig.1)  
Step 1

A B C D E F G H I J K
    .     .     .    
0 1 3 0 0 0 9 6 2 0 0   Step 1

Follow Kojima's rules for setting the quotient answer. Divisor 1 is smaller than dividend 9, place the first quotient number two rods to the left of the dividend.

Step 2: Chinese rule:  1 meets 9 forward 9. Place quotient 9 on rod E and clear 9 from G. Revise because we can't continue.

    (Fig.2)  
Step 2

A B C D E F G H I J K
    .     .     .    
0 1 3 0 0 0 9 6 2 0 0   Step 2
        9
clear      -9        
0 1 3 0 9 0 0 6 2 0 0            

Step 3: Revise. Subtract 1 from quotient on rod E and add 1 to G. Revise again.

    (Fig.3)  
Step 3

A B C D E F G H I J K
    .     .     .    
0 1 3 0 9 0 0 6 2 0 0   Step 3
       -1
           +1        
0 1 3 0 8 0 1 6 2 0 0            

Step 4: Revise. Subtract 1 from quotient on rod E and add 1 to G.

    (Fig.4)  
Step 4

A B C D E F G H I J K
    .     .     .    
0 1 3 0 8 0 1 6 2 0 0   Step 4
       -1
           +1        
0 1 3 0 7 0 2 6 2 0 0            

Step 5: With 26 on rods GH we can continue. 3 * 7 = 21, subtract 21 from rods GH.

    (Fig.5)  
Step 5

A B C D E F G H I J K
    .     .     .    
0 1 3 0 7 0 2 6 2 0 0    Step 5
           -2 1      
0 1 3 0 7 0 0 5 2 0 0            

Step 6: Chinese rule: 1 meets 5 forward 5. Place quotient 5 on rod F and clear 5 from H. Revise because we can't continue.

    (Fig.6)  
Step 6

A B C D E F G H I J K
    .     .     .    
0 1 3 0 7 0 0 5 2 0 0    Step 6
          5
clear        -5      
0 1 3 0 7 5 0 0 2 0 0
 
Step 7: Revise. Subtract 1 from quotient on F and add 1 to rod H.
 
    (Fig.7)
Step 7

A B C D E F G H I J K
    .     .     .    
0 1 3 0 7 5 0 0 2 0 0    Step 7
         -1
             +1      
0 1 3 0 7 4 0 1 2 0 0

Step 8: With 12 remaining we have enough to continue. 3 * 4 = 12, subtract 12 from rods HI.
Step 8a: Shift the unit rod. Starting on rod H count 2 to the left. The answer is 74 on rods EF.

    (Fig.8)  
Step 8

A B C D E F G H I J K
    .     .     .    
0 1 3 0 7 4 0 1 2 0 0    Step 8
             -1 2    
0 1 3 0 7 4 0 0 0 0 0

A Special thanks to members of the Yahoo Soroban/Abacus newsgroup for some very stimulating discussion on this topic. In particular I'd like to thank the following;

Steve Treadwell
Hannu Hinkka
Fabrice Bouillerot

 

 

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© October, 2013
Totton Heffelfinger   Toronto Ontario  Canada
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