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

Short Division Techniques - Chinese Suan Pan

This is a Chinese technique for doing short division. This technique works best with problems where divisors are made up of whole numbers and have a value of less than 10. Before going on to try traditional Chinese long division techniques, it's a good idea to completely understand how this short division method works. After a little practice many find this technique enhances their ability to solve problems of long division using division techniques as taught by Takashi Kojima.

Below is a list of traditional rules. Understanding these rules is integral to understanding this method. They may look a little complicated at first glance but they are easily understood. As such there should be no need to commit them to memory. (See below for a brief explanation.)

TRADITIONAL RULES

1/1 = forward 1

1/2 = 5
2/2 = forward 1

1/3 = 3 plus 1
2/3 = 6 plus 2
3/3 = forward 1

1/4 = 2 plus 2
2/4 = 5
3/4 = 7 plus 2
4/4 = forward 1

1/5 = 2
2/5 = 4
3/5 = 6
4/5 = 8
5/5 = forward 1

 1/6 = 1 plus 4
2/6 = 3 plus 2
3/6 = 5
4/6 = 6 plus 4
5/6 = 8 plus 2
6/6 = forward 1


1/7 = 1 plus 3
2/7 = 2 plus 6
3/7 = 4 plus 2
4/7 = 5 plus 5
5/7 = 7 plus 1
6/7 = 8 plus 4
7/7 = forward 1

1/8 = 1 plus 2
2/8 = 2 plus 4
3/8 = 3 plus 6
4/8 = 5
5/8 = 6 plus 2
6/8 = 7 plus 4
7/8 = 8 plus 6
8/8 = forward 1


1/9 = 1 plus 1
2/9 = 2 plus 2
3/9 = 3 plus 3
4/9 = 4 plus 4
5/9 = 5 plus 5
6/9 = 6 plus 6
7/9 = 7 plus 7
8/9 = 8 plus 8
9/9 = forward 1

    In the above set of rules
• "plus 1" means adding 1 to the column immediately to the right, "plus 2" means adding 2 and so on...
• "forward 1" means adding 1 to the column immediately to the left.


Explanation of Rules:

Using a few random examples, this is how I think of it;

a) 2 /3 = 6 plus 2  ==> 20 ÷ 3 = 6 with remainder 2. Place 2 one column to the right.
b) 6 /7 = 8 plus 4  ==> 60 ÷ 7 = 8 with remainder 4. Place 4 one column to the right.
c) 4 /7 = 5 plus 5  ==> 40 ÷ 7 = 5 with remainder 5. Place 5 one column to the right.
d) 5 /5 = forward 1 ==>  5 ÷ 5 = 1, place 1 one column to the left... and so on.

All is best explained using examples.

Example 1:  128 ÷ 2 = 64

 Step 1: Set the dividend 128 on FGH  and the divisor 2 on C.  (Fig.1)

   Fig. 1

   
Step 1
A B C D E F G H I J K L M
                         
0 0 2 0 0 1 2 8 0 0 0 0 0

 Step 2: Following   rule 1 /2 = 5    ===>  divide 2 on rod C into 1 on F. (Think of it in terms of 10 ÷ 2 = 5). Remove the 1 and replace it with the quotient 5 on rod F, leaving 528 on rods FGH. (Fig.2)
 

   Fig. 2

   
Step 2
A B C D E F G H I J K L M
                         
0 0 2 0 0 1 2 8 0 0 0 0 0
         (5)               Step 2  
0 0 2 0 0 5 2 8 0 0 0 0 0

 Step 3: Following   rule 2 /2 = forward 1    ===>  this means clear the 2 on rod G and add 1 to the rod on the left.  Therefore, 5 on rod F becomes 6, leaving 608 on rods FGH. (Fig.3)

   Fig. 3

   
Step 3
A B C D E F G H I J K L M
                         
0 0 2 0 0 5 2 8 0 0 0 0 0
clear      (2)
forward   1                Step 3
0 0 2 0 0 6 0 8 0 0 0 0 0

Step 4: Following   ** rule 8 / 2 = forward 4**     ===>  this means clear the 8 on rod H and add 4 to the rod on the left. Therefore, 0 on rod G becomes 4, leaving 64 on rods FG, which is the answer. (Fig.4)
** This is just   rule 2 /2 = forward 1   that's been applied four times **

   Fig. 4

   
Step 3
A B C D E F G H I J K L M
                         
0 0 2 0 0 6 0 8 0 0 0 0 0
clear        (8)
forward     4              Step 4
0 0 2 0 0 6 4 0 0 0 0 0 0

 

Example 2:  259 ÷ 7 = 37

 Step 1: Set the dividend 259 on FGH  and the divisor 7 on C.  (Fig.5)

   Fig. 5

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

 Step 2:  Step 2: Following   rule 2 /7 = 2 plus 6    ===>  divide 7 on rod C into 2 on F. (Think of it in terms of 20 ÷ 7 = 2 with a remainder of 6). Place the quotient 2 on rod F and place the remainder 6 on G, leaving 2 "11" 9 on rods FGH. (Fig.6)

   Fig. 6

   
Step 2
A B C D E F G H I J K L M
                         
0 0 7 0 0 2 5 9 0 0 0 0 0
         (2)
plus      + 6              Step 2   
0 0 7 0 0 2"11"9 0 0 0 0 0
 

Step 3: The 11 on rod G is a number greater than 7. So take 7 away from rod G and  *forward 1*  onto rod F changing the 2 to a 3. This leaves 349 on rods FGH. (Fig.7)

   Fig. 7

   
Step 3
A B C D E F G H I J K L M
                         
0 0 7 0 0 2"11"9 0 0 0 0 0
          - 7
forward   1                Step 3
0 0 7 0 0 3 4 9 0 0 0 0 0

Step 4:  Following   rule 4 /7 = 5  plus 5   ===>  (Think of it in terms of 40 ÷ 7 = 5 with a remainder of 5). Replace the 4 on G with the quotient 5. Place the remainder 5 onto rod H, leaving 3 5 "14" on rods FGH. (Fig.8)

   Fig. 8

   
Step 4
A B C D E F G H I J K L M
                         
0 0 7 0 0 3 4 9 0 0 0 0 0
           (5)
plus          5            Step 4
0 0 7 0 0 3 5"14"0 0 0 0 0

Step 5: Following   ** rule 14 /7 = forward 2 **     ===>  this means clear the 14 on rod H and add 2 to the rod on the left. Therefore, 5 on rod G becomes 7, leaving 37 on rods FG, which is the answer. (Fig.9)
** This is just    rule 7 /7 = forward 1    that's been applied two times **

   Fig. 9

   
Step 5
A B C D E F G H I J K L M
                         
0 0 7 0 0 3 5"14"0 0 0 0 0
clear       ("14")
forward     2              Step 5
0 0 7 0 0 3 7 0 0 0 0 0 0

It's evident that some of the above steps could have been done more efficiently. For instance in  Example 2:  Step 3, instead of dividing 7 into the 4 on rod G, we could have looked ahead and noticed 49 on GH. At this point dividing 7 into 49 would have allowed us to come to our answer of 37 that much more quickly. I took the longer approach only because it may be of help to see some of the thinking behind the process.

 

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