**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 methodsin 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.

- The
Extra Beadtechnique uses the top most heaven bead and has a value of 5.(see Example 3: Step 3 below)

- The
Suspended Beadtechnique also uses the top most heaven bead but in this case it's placed½way down the rod so that the bead neither touches the frame above nor the bead below; it has a value of 10.(see Example 4: Steps 3 & 4 below)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## The Unit rod and the Decimal

This technique makes working with decimals and the unit rod wonderfully easy. One has only to count either the whole numbers or the trailing zeros in the multiplier then shift the unit rod accordingly.

Multiplication And Shifting the Unit Rod

If the Multiplier is....## 1.03...... one whole number - unit rod shifts 1 rod to the right.

45.003... two whole numbers - unit rod shifts 2 rods to the right.

0.75....... no whole numbers, no trailing zeros, unit rod does *not* shift.

0.0125... one trailing zero - unit rod shifts 1 rod to the left.

0.003..... two trailing zeros - unit rod shifts 2 rods to the left...etc.

Suan Pan and the Decimal- Multiplication:

Shifting

Traditional Chinese multiplication techniques differ from the Japanesemultiplicationtechniques. Particularly different is the order in which we multiply numbers. Before moving on to the the Extra Bead and the Suspended Bead techniques we'll do two simple examples that will help to illustrate how this traditional multiplication method works.

Example 1: 456 x 23 = 10,488

The Order of Multiplication:

Please refer to Fig.1 below to help with the illustration. The order of operation is as follows;## ▪ Rod H x B x A. Note: the multiplicand on rod H changes to become part of the product.

▪ Rod G x B x A. Note: the multiplicand on rod G changes to become part of the product.

▪ Rod F x B x A. Note: the multiplicand on rod F changes to become part of the product.

Step 1:Choose rod H to be the unit rod. Set multiplicand 456 on rods FGH and multiplier 23 on AB. The multiplier has two whole numbers therefore the unit rod will shift two to the right bringing us to rod J. (Fig.1)

Fig.1 Step 1A B C D E F G H I J K L M2 3 0 0 0 4 5 6 0 0 0 0 0

Multiply 6 by 3 and add 18 to rods IJ. (Fig. 2)

Step 2:

Fig.2 Step 2A B C D E F G H I J K L M2 3 0 0 0 4 5 6 0 0 0 0 0+ 1 82 3 0 0 0 4 5 6 1 8 0 0 0

Multiply 6 by 2 and add 12 on rods HI. Note that what was multiplicand 6 on H has now changed to 1. (Fig. 3)

Step 3:

Fig.3 Step 3A B C D E F G H I J K L M2 3 0 0 0 4 5 6 1 8 0 0 0+ (1)22 3 0 0 0 4 5 1 3 8 0 0 0

Multiply 5 by 3 and add 15 on to rods HI. (Fig. 4)

Step 4:

Fig.4 Step 4A B C D E F G H I J K L M2 3 0 0 0 4 5 1 3 8 0 0 0+ 1 52 3 0 0 0 4 5 2 8 8 0 0 0

Step 5:Multiply 5 by 2 and add 10 to rods GH. Note that what was multiplicand 5 on G has now changed to 1. (Fig. 5)

Fig.5 Step 5A B C D E F G H I J K L M2 3 0 0 0 4 5 2 8 8 0 0 0+ (1)02 3 0 0 0 4 1 2 8 8 0 0 0

Step 6:Multiply 4 by 3 and add 12 to rods GH. (Fig. 6)

Fig.6 Step 6A B C D E F G H I J K L M2 3 0 0 0 4 1 2 8 8 0 0 0+ 1 22 3 0 0 0 4 2 4 8 8 0 0 0

Step 7 and the answer:Multiply 4 by 2 and add 08 to rods FG. Once again note the change; what was originally multiplicand 4 on F changes to 0 and then to 1 after adding 8 to 2 on rod G. This leaves the answer 10,488 on rods F - J. (Fig. 7)

Fig.7 Step 7A B C D E F G H I J K L M2 3 0 0 0 4 2 4 8 8 0 0 0+ (0)82 3 0 0 0 1 0 4 8 8 0 0 0

Example 2: 347 x 276 = 95,772

The Order of Multiplication:

Please refer to Fig.1 below to help with the illustration. The order of operation is as follows;## ▪ Rod H x B x C x A. Note: the multiplicand on rod H changes to become part of the product.

▪ Rod G x B x C x A. Note: the multiplicand on rod G changes to become part of the product.

▪ Rod F x B x C x A. Note: the multiplicand on rod F changes to become part of the product.

Step 1:Choose rod H to be the unit rod. Set multiplicand 347 on rods FGH and multiplier 276 on rods ABC. The multiplier has three whole digits therefore the unit rod will shift three rods to the right bringing us to rod K.

Fig.1 Step 1A B C D E F G H I J K L M2 7 6 0 0 3 4 7 0 0 0 0 0

Step 2:Multiply 7 by 7 and add 49 to rods IJ.

2a:Multiply 7 by 6 and add 42 to rods JK.

2b:Multiply 7 by 2 and add 14 to rods HI. Note that what was multiplicand 7 on H has now changed to 1. (Fig. 2)

Fig.2 Step 2A B C D E F G H I J K L M2 7 6 0 0 3 4 7 0 0 0 0 0 + 4 9 + 4 2+ (1)42 7 6 0 0 3 4 1 9 3 2 0 0

Step 3:Multiply 4 by 7 and add 28 to rods HI.Multiply 4 by 6 and add 24 to rods IJ.

3a:Multiply 4 by 2 and add 08 to rods GH. Note that what was originally multiplicand 4 on G changes to 0 and then to 1 after adding 8 to rod H. (Fig. 2)

3b:

Fig.3 Step 3A B C D E F G H I J K L M2 7 6 0 0 3 4 1 9 3 2 0 0 + 2 8 + 2 4+ (0)82 7 6 0 0 3 1 2 9 7 2 0 0

Step 4:Multiply 3 by 7 and add 21 to rods GH.Multiply 3 by 6 and add 18 to rods HI.

4a:

4b and the answer:Multiply 3 by 2 and add 06 to rods FG. Once again note the change; what was originally multiplicand 3 on F changes to 0. This leaves the answer 95,772 on rods G - K. (Fig. 4)

Fig.4 Step 4A B C D E F G H I J K L M2 7 6 0 0 3 1 2 9 7 2 0 0 + 2 1 + 1 8+ (0)62 7 6 0 0 0 9 5 7 7 2 0 0

Example 3: "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. However in Step 3 (see below) using the complementary number might be confusing.

189 x 576 = 108,864

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

Fig. 1Step 1A B C D E F G H I J K L M0 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 H - M. (Fig.2)

Fig. 2Step 2A B C D E F G H I J K L M0 0 0 5 7 6 0 1 8 9 0 0 0 + 6 3 Step 2 + 5 4 Step 2a(4)5Step 2b0 0 0 5 7 6 0 1 8 5 1 8 4

This is where we usewe use theExtra 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.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 H - M. (Fig.3)

Step 3:

Fig. 3Step 3A B C D E F G H I J K L M0 0 0 5 7 6 0 1 8 5 1 8 4+ 5 6Step 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 H - M.

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 H - M. (Fig.4)

Fig. 4Steps 3a-3b-3cA B C D E F G H I J K L M0 0 0 5 7 6 0 1 8"10"7 8 4 + 4 8 Step 3a+ (4)0Step 3b 0 0 0 5 7 6 0 1 4"11"2 6 4+ extra bead 1 0Step 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 H - M, which is the answer. (Fig.5)

Fig. 5Step 4A B C D E F G H I J K L M0 0 0 5 7 6 0 1 5 1 2 6 4 + 0 7 Step 4 + 0 6 Step 4a+ 0 5Step 4b 0 0 0 5 7 6 0 1 0 8 8 6 4

Example 4: "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 Beadhas 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.It should be noted that while it's fun to do a problem like this it's a fairly extreme example and instances where one needs the suspended bead are quite rare.

989 x 898 = 888, 122

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

Fig. 6Step 1A B C D E F G H I J K L M0 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 H - M. Fig.7)

Fig. 7Step 2A B C D E F G H I J K L M0 0 0 8 9 8 0 9 8 9 0 0 0 + 8 1 Step 2 + 7 2 Step 2a(7)2Step 2b 0 0 0 8 9 8 0 9 8 8 0 8 2

This is where we usewe use theSuspended 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 H - M. (Fig.8)

Fig. 8Step 3-3aA B C D E F G H I J K L M0 0 0 8 9 8 0 9 8 8 0 8 2 + 7 2 Step 3+ 6 4Step 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 H - M.

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 H - M. (Fig.9)

Fig. 9Step 3b-3cA B C D E F G H I J K L M0 0 0 8 9 8 0 9 8"15"9 2 2(6)4Step 3b 0 0 0 8 9 8 0 9 6"19"9 2 2+ suspended 1 0Step 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 theSuspended Beadtechnique.

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. 10Steps 4-4aA B C D E F G H I J K L M0 0 0 8 9 8 0 9 7 9 9 2 2 + 8 1 Step 4+ 7 2Step 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 H - M.

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 H - M, which is the answer. (Fig.11)

Fig. 11Step 4b-4cA B C D E F G H I J K L M0 0 0 8 9 8 0 9"16"8 1 2 2(7)2Step 4b 0 0 0 8 9 8 0 7"18"1 2 2+ suspended 1 0Step 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 isthe one illustratedin 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

totton[at]idirect[dot]com