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

Traditional Multiplication techniques for Chinese Suan Pan - The "Extra Bead" and the "Suspended Bead"

The Suan Pan, or Chinese abacus, had been in use in China for two or three centuries before making its journey to Japan sometime in the late 1500's. Unlike the present day Japanese Soroban which has evolved to become an instrument with a 1/4 bead construction, the Suan Pan has retained its classic 2/5 bead construction - 2 beads above the reckoning bar, 5 beads below. The techniques described below may offer some idea as to why this might be so.

These techniques are older and differ from modern Chinese methods in that the order of multiplication and adding products to the frame is different. These techniques might even be recognizable to an older generation in Japan, many of whom grew up learning older methods. New ideas and changes to the physical makeup of the Japanese abacus early in the last century brought about changes in the way many modern day Japanese do their abacus work. In learning these Chinese techniques, not only does it enable a better understanding of abacus use, it also provides a more in-depth look at the instrument and the thought behind its development and use.

Before beginning these techniques, it would be useful to have a good understanding of both the addition and multiplication techniques described in: Abacus: Mystery of the Bead

** Please note: When setting up the problems in the following examples, it's advisable to count the number of digits in the multiplier and leave the corresponding number of rods vacant on the right hand side of the Suan Pan. In doing so the unit number in the product will fall neatly on the last rod. In the examples below there are three numbers in the multiplier, so the last three rods on the right are left vacant.**

Example 1: "The Extra Bead"

When it comes to adding products onto the abacus, most techniques require the use of complementary numbers. In the examples below we'll use this approach wherever we can. In doing so it helps to better focus on where the exclusive use of complementary numbers makes the job much more difficult.

189 x 576 = 108,864

Step 1: Set the multiplier 576 on DEF and the multiplicand 189 on HIJ. (Fig.1)

   Fig. 1

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

Step 2: Multiply the 9 on rod J by the 7 on E and add the product 63 on KL.

2a: Multiply the 9 on rod J by the 6 on F and add the product 54 on LM

2b: Multiply the same 9 on rod J by the 5 on rod D, remove the 9 on J and place the product 45 on JK . This leaves 185184 on rods HIJKLM. (Fig.2)

   Fig. 2

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

This is where we use we use the Extra Bead. At this point we're going to multiply the 8 on rod I by the 7 on E. If we were to do a carry here then we'd have have to change the 8 to 9. That would be confusing. (Especially since it's the 8 that's going to be multiplied by each of the digits in the multiplier.) Since there is an extra 5 bead available on J we'll use it to avoid confusion.

Step 3: Multiply the 8 on rod I by the 7 on E and add the product 56 on JK, leaving 1 8 "10" 7 8 4 on rods HIJKLM. (Fig.3)

   Fig. 3

   
Step 3
A B C D E F G H I J K L M
                         
0 0 0 5 7 6 0 1 8 5 1 8 4
                + 5 6      Step 3
0 0 0 5 7 6 0 1 8"10"7 8 4

* It may be interesting to note that many Japanese Soroban made between circa 1850 and 1930 have one heaven bead and five earth beads. (i.e. 1 bead above the reckoning bar and 5 beads below.)  I've always been confused as to why this might be. Learning from some of these older techniques may provide a clue. In step 3 one could, as an alternative, show the number 10 on rod J as one heaven bead down and 5 earth beads up.*

3a: From here we can continue. Multiply the 8 on rod I by the 6 on F and add the product 48 on KL

3b: Multiply the 8 on rod I by the 5 on D, remove the 8 on I and place the product 40 on rods IJ. This leaves 1 4 "11" 2 6 4 on rods HIJKLM.

3c: Next clear the 10 from rod J and carry it over to rod I changing the 4 to 5. This leaves 151264 on rods HIJKLM. (Fig.4)

   Fig. 4

   
Steps 3a-3b-3c
A B C D E F G H I J K L M
                         
0 0 0 5 7 6 0 1 8"10"7 8 4
                  + 4 8    Step 3a
             + (4)0        Step 3b
0 0 0 5 7 6 0 1 4"11"2 6 4
+ extra bead    1 0        Step 3c
0 0 0 5 7 6 0 1 5 1 2 6 4

Step 4: Now we can continue using complements in the normal way. Multiply the 1 on rod H by the 7 on rod E and add the product 7 on rod J

4a: Multiply the 1 on rod H by the 6 on rod F and add the product 6 on rod K

4b & the answer: Multiply the same 1 on rod H by the 5 on rod D, remove the 1 on rod H and add the product 5. This leaves 108,864 on rods HIJKLM, which is the answer. (Fig.5)

   Fig. 5

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

 

Example 2: "The Suspended Bead" (Xuan - zhu)

In this next example we'll use a technique where we bring down the upper most 5 bead ½ way down the rod so that it touches neither the frame above nor the bead below. This Suspended Bead has a value of 10. Once again in solving this next example, we'll use complementary numbers as much as possible. However, in two instances it will be clear that having extra beads simplifies the process.

989 x 898 = 888, 122

Step 1: Set the multiplier 898 on DEF and the multiplicand 989 on HIJ. (Fig.6)

   Fig. 6

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

Step 2: Multiply the 9 on rod J by the 9 on E and add the product 81 on KL.

2a: Multiply the 9 on rod J by the 8 on F and add the product 72 on LM

2b: Multiply the same 9 on rod J by the 8 on rod D, remove the 9 on J and place product 72 on JK . This leaves 988082 on rods HIJKLM. (Fig.7)

   Fig. 7

   
Step 2
A B C D E F G H I J K L M
                         
0 0 0 8 9 8 0 9 8 9 0 0 0
                  + 8 1    Step 2
                    + 7 2  Step 2a
                 (7)2      Step 2b
0 0 0 8 9 8 0 9 8 8 0 8 2

This is where we use we use the Suspended Bead. At this point we're going to multiply the 8 on rod I by the 9 on E. If we were to do a carry here then we'd have have to change the 8 to 9. That would be confusing. (Especially since it's the 8 that's going to be multiplied by each of the digits in the multiplier.) We might loose track. Instead we use the "Suspended Bead" technique.

Step 3: Multiply the 8 on rod I by the 9 on E and add the product 72 on JK. In bringing down the suspended bead to signify 10 and subtracting 3 we've added 7 to rod J giving the rod a total value of 15.

3a: Multiply the 8 on rod I by the 8 on F and add the product 64 on KL. This leaves a value of 9 8 "15" 9 2 2 on rods HIJKLM. (Fig.8)

   Fig. 8

   
Step 3-3a
A B C D E F G H I J K L M
                         
0 0 0 8 9 8 0 9 8 8 0 8 2
                + 7 2      Step 3
                  + 6 4    Step 3a
0 0 0 8 9 8 0 9 8"15"9 2 2

3b: Multiply the same 8 on rod I by the 8 on rod D, remove the 8 on I and place the product 64 on IJ . This leaves 9 6 "19" 9 2 2 on rods on rods HIJKLM.

3c: Next clear the "Suspended" 10 from rod J and carry it over to rod I changing the 6 to 7. This leaves 979922 on rods HIJKLM. (Fig.9)

   Fig. 9

   
Step 3b-3c
A B C D E F G H I J K L M
                         
0 0 0 8 9 8 0 9 8"15"9 2 2
               (6)4        Step 3b   
0 0 0 8 9 8 0 9 6"19"9 2 2
+ suspended     1 0        Step 3c
0 0 0 8 9 8 0 9 7 9 9 2 2

Step 4: Multiply the 9 on H by the 9 on E and add the product 81 on IJ using the Suspended Bead technique.

4a: Multiply the 9 on H by the 8 on F and add the product 72 on JK. This leaves 9 "16" 8 1 2 2 on HIJKLM. (Fig.10)

   Fig. 10

   
Steps 4-4a
A B C D E F G H I J K L M
                         
0 0 0 8 9 8 0 9 7 9 9 2 2
              + 8 1        Step 4
                + 7 2      Step 4a
0 0 0 8 9 8 0 9"16"8 1 2 2

4b: Multiply the 9 on H by the 8 on D, remove the 9 on H and place the product 72 on HI. This leaves 7 "18" 8 1 2 2 on rods HIJKLM.

4c & the answer: Next clear the "Suspended" 10 from rod I and carry it over to rod H changing the 7 to 8. This leaves 888122 on rods HIJKLM, which is the answer. (Fig.11)

   Fig. 11

   
Step 4b-4c
A B C D E F G H I J K L M
                         
0 0 0 8 9 8 0 9"16"8 1 2 2
             (7)2          Step 4b
0 0 0 8 9 8 0 7"18"1 2 2
+ suspended   1 0          Step 4c
0 0 0 8 9 8 0 8 8 8 1 2 2

 

Over the centuries, both the Japanese and the Chinese have developed and used many different techniques for solving problems of multiplication on the abacus. One widely used method is the one illustrated in the 1954 book, "The Japanese Abacus, its use and Theory" by Takashi Kojima. When introducing his technique, even Kojima says, "There are several methods of multiplication on the abacus. The one introduced in the following pages is a recent method..."

Even though it's possible to solve multiplication problems while using these Chinese techniques on a 1/4 bead Soroban, the above examples show the answers come more easily with the help of extra beads.

 

▪ Abacus: Mystery of the Bead
▪ Advanced Abacus Techniques

 

 

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