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 theYahoo Soroban/Abacusnews 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 advantageis easily adapted for use on a 1:4 bead soroban. Let's see how how it works.

Example 1: 259 ÷ 7 = 37This example follows the rules:

- 7 meets 2, 2 + 6
- 7 meets 4, 5 + 5

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

(Fig.1) Step 1A 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 1Follow Kojima's rules for setting the

quotient answer. Divisor 7 is larger than the dividend 2, place the first quotient numberone 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 2A 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 2clear -20 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 3A 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+60 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 4A 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 +1-70 0 7 0 0 3 0 4 9 0 0 0 0At 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 5A 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 5clear -40 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 6A 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+50 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 7A 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 40 0 7 0 0 3 7 0 0 0 0 0 0

Example 2: 962 ÷ 13 = 74This example follows the rules:

- 1 meets 9, forward 9
- 1 meets 5, forward 5

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

(Fig.1) Step 1A B C D E F G H I J K . . .0 1 3 0 0 0 9 6 2 0 0 Step 1Follow Kojima's rules for setting the

quotient answer. Divisor 1 is smaller than dividend 9, place the first quotient numbertwo 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 2A B C D E F G H I J K . . .0 1 3 0 0 0 9 6 2 0 0 Step 2 9clear -90 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 3A B C D E F G H I J K . . .0 1 3 0 9 0 0 6 2 0 0 Step 3 -1+10 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 4A B C D E F G H I J K . . .0 1 3 0 8 0 1 6 2 0 0 Step 4 -1+10 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 5A B C D E F G H I J K . . .0 1 3 0 7 0 2 6 2 0 0 Step 5-2 10 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 6A B C D E F G H I J K . . .0 1 3 0 7 0 0 5 2 0 0 Step 6 5clear -50 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 7A B C D E F G H I J K . . .0 1 3 0 7 5 0 0 2 0 0 Step 7 -1+10 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 8A B C D E F G H I J K . . .0 1 3 0 7 4 0 1 2 0 0 Step 8-1 20 1 3 0 7 4 0 0 0 0 0

A Special thanks to members of the YahooSoroban/Abacus newsgroupfor 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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Abacus: Mystery of the Bead

▪Advanced Abacus Techniques## © October, 2013

Totton Heffelfinger Toronto Ontario Canada

totton[at]idirect[dot]com