To my friend Edvaldo Siqueira of Rio de Janeiro, Brazil.
For this, the most thoughtful of gifts.
Sincere and heartfelt thanks.
This is not a manual for Lee's Abacus. Rather, this page is meant to be a demonstration illustrating a little of what this instrument has to offer. Instead of using examples from Lee's instruction book, I've chosen a few of my own. I hope that these pages may provide a little insight into the thought and ideas behind the development of the Lee Kaichen Abacus.  Totton Heffelfinger, Toronto, Ontario, Abacus: Mystery of the Bead
THE LEE KAICHEN ABACUS
Little is known about Lee Kaichen. The manuals that accompany the abacus give
no information about him or how he came to invent his abacus.
All remains a mystery. About the abacus he created, however, we do know a little more. During the middle part of the last century Lee produced a
small number of extraordinary and unusual abaci. The Lee Kaichen Abacus was made in Taiwan where it was manufactured for only a few years. Manuals show 1959 as the copyright date for the abacus and June and December 1958 as printing dates for the manual. Although the manuals that accompany his abaci seem to have remained largely unchanged, the abacus itself did go through a number of design modifications. Whether or not these modifications were a conscious effort to improve the abacus or whether it was a manufacturing decision is unclear. Perhaps the answer lies somewhere in between.

A few design modifications of note.
Three characteristics that make Lee's Abacus different from the rest:
For both addition and
subtraction, problems are solved in the
principle operation field. Problems are worked
out exactly as if using a normal Chinese suan pan or
Japanese soroban, with one possible exception. In some cases the operator
might have to make an adjustment to the vernier to ensure that the decimal point corresponds with decimal
numbers in the problem. This would be the only difference. Just to recap and to become a little more
familiar with the diagrams here are two simple examples, one each for addition and for subtraction.
Example: 76.25 + 45.33 = 121.58
The vernier: In the PO field, choose rod I to be the unit rod. Set the decimal on the vernier just to the right of rod I.
Step 1: Set 76.25 onto rods HIJK. (Fig.1)
Fig.1 
Step 1 a b c d e f g h i j k l m n o p q r 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A B C D E F G H I J K L M , , • <==== Vernier 0 0 0 0 0 0 0 7 6.2 5 0 0 Step 1 
Step 2: Solve the addition using complementary numbers where necessary just as you would on a normal suan pan. This leaves the answer 121.58 on rods GHIJK. (Fig.2)
Fig.2 
Step 2 a b c d e f g h i j k l m n o p q r 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 0 7 6.2 5 0 0 + 4 5.3 3 Step 2 0 0 0 0 0 0 1 2 1.5 8 0 0 
Example: 102.43  67.89 = 34.54
The vernier: In the PO field, choose rod I to be the unit rod. Set the decimal on the vernier just to the right of rod I.
Step 1: Set 102.43 onto rods GHIJK. (Fig.3)
Fig.3 
Step 1 a b c d e f g h i j k l m n o p q r 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A B C D E F G H I J K L M , , • <==== Vernier 0 0 0 0 0 0 1 0 2.4 3 0 0 Step 1 
Step 2: Solve the subtraction using complementary numbers where necessary just as you would on a normal suan pan. This leaves the answer 34.54 on rods HIJK. (Fig.4)
Fig.4 
Step 2 a b c d e f g h i j k l m n o p q r 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 1 0 2.4 3 0 0  6 7.8 9 Step 2 0 0 0 0 0 0 0 3 4.5 4 0 0 
Japanese multiplication techniques require setting multiplier on the left and multiplicand on the right. In setting the multiplicand on the left and the multiplier on the right, Lee does the opposite. This difference aside, Lee's technique for adding products to the frame in a rightleft manner is the same as one described by Takashi Kojima in his book, 'Advanced Abacus: Japanese Theory and Practice'. In the book Kojima illustrates several different methods for solving problems of multiplication. Of this method he says, "This arithmetic method is the oldest multiplication method here introduced. It was used extensively in Japan until around 1930, ..."
Basically Lee's procedure is this:
**Indicator**
Where both multiplicand and multiplier are whole numbers,
begin by placing the indicator directly under the unit rod.
Example 1: 78 x 23 = 1794
Step 1: In the left AO field, set the multiplicand 78 on rods hi and in the right AO field set 23 on rods lm.
The vernier: In the PO field, choose rod I to be the unit rod. Set the decimal on the vernier just to the right of rod I.
The indicator: Slide the indicator directly underneath unit rod I. **The indicator under rod I will assist in finding the correct number placement in the product.** (Fig.5)
Fig.5 
Step 1 a b c d e f g h i j k l m n o p q r 0 0 0 0 0 0 0 7 8 0 0 2 3 0 0 0 0 0 Step 1 A B C D E F G H I J K L M , , • <==== Vernier 0 0 0 0 0 0 0 0 0.0 0 0 0 ▬ <=== Indicator 
Step 2: Multiply 3 on m by 8 on rod
i. In the PO field, add the product 24 to rods
HI. (The '4' in 24 is added to the indicator rod.)
2a: Multiply the same 3 on m by 7 on rod h. In the PO field, add the product 21 to rods GH. This leaves the partial product 234 on rods GHI. (Fig.6)
Fig.6 
Step 2 a b c d e f g h i j k l m n o p q r 0 0 0 0 0 0 0 7 8 0 0 2 3 0 0 0 0 0 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 0 0 0.0 0 0 0 + 2 4 Step 2 + 2 1 Step 2a 0 0 0 0 0 0 2 3 4.0 0 0 0 ▬ <=== Indicator 
The indicator: Slide the indicator one rod to the left placing it under rod H.
Step 3: Multiply 2 on l by 8 on rod i. In the PO field, add the product 16 to rods GH. (This time the '6' in 16 is added to the indicator rod.)
3a & the answer: Multiply the same 2 on l by 7 on rod h. Add the product 14 to rods FG, leaving the answer 1,794 on rods FGHI. (Fig.7)
Fig.7 
Step 3 a b c d e f g h i j k l m n o p q r 0 0 0 0 0 0 0 7 8 0 0 2 3 0 0 0 0 0 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 2 3 4.0 0 0 0 + 1 6 Step 3 + 1 4 Step 3a 0 0 0 0 0 1,7 9 4.0 0 0 0 ▬ <=== Indicator 
When multiplying pure decimal or mixed decimal numbers there are a couple of things to keep in mind:
**The Indicator**
When setting up the problem on the abacus, for every number in the decimal
fraction in
the multiplier move the indicator one place to the right of the chosen unit rod
For the Multiplier:
17.2.......Place the indicator 1 rod to the right of the chosen unit rod. 1.23.......Place the indicator 2 rods to the right of the chosen unit rod. 0.003......Place the indicator 3 rods to the right of the chosen unit rod, and so on.
**The Vernier Decimal **
After completing the multiplication, for every decimal number in the
multiplicand move the vernier decimal one place to the left of the chosen unit rod.
For the Multiplicand:
35.4.......Move the decimal vernier 1 rod to the left of the chosen unit rod. 5.78.......Move the decimal vernier 2 rods to the left of the chosen unit rod. 0.005......Move the decimal vernier 3 rods to the left of the chosen unit rod, and so on.
Example 1: 4.5 x 7.003 = 31.5135
Step 1: In the left AO field, set the multiplicand 4.5 on rods hi. In the right AO field, set 7.003 on rods lmno.
The moveable unit markers: In both the left and right AO fields slide the unit markers to separate the whole and decimal numbers.
The vernier: In the PO field, choose rod G to be the unit rod. Set the decimal on the vernier just to the right of rod G.
The indicator: Because there are three decimal numbers in the multiplier, slide the indicator three rods to the right of unit rod G. Place it under rod J. **Once again, the indicator under rod J will assist in finding the correct number placement in the product.** (Fig.8)
Fig.8 
Step 1 a b c d e f g h.i j k l.m n o p q r 0 0 0 0 0 0 0 4.5 0 0 7.0 0 3 0 0 0 Step 1 A B C D E F G H I J K L M , , • <===== Vernier 0 0 0 0 0 0 0.0 0 0 0 0 0 ▬ <=== Indicator 
Step 2: Multiply 3 on rod o by 5 on rod
i. In the PO field, add the product 15 to rods IJ. (The '5' in 15 is added to the indicator rod.)
2a: Multiply the same 3 on rod o by 4 on rod h. In the PO field, add the product 12 to rods HI. This leaved the partial product 135 on rods HIJ. (Fig.9)
Fig.9 
Step 2 a b c d e f g h.i j k l.m n o p q r 0 0 0 0 0 0 0 4.5 0 0 7.0 0 3 0 0 0 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 0.0 0 0 0 0 0 + 1 5 Step 2 + 1 2 Step 2a 0 0 0 0 0 0 0.1 3 5 0 0 0 ▬ <=== Indicator 
The indicator: The next number to be multiplied in the right AO field is the multiplier 7 on rod l. Because it's three places to the left, slide the indicator
three rods to the left placing it under rod G.
Step 3: Multiply 7 on l by 5 on rod i. In the PO field, add the product 35 to rods FG. (The '5' in 35 is added to the indicator rod.)
3a: Multiply the same 7 on rod l by 4 on h. In the PO field, add the product 28 to rods EF.
The vernier & the answer: Because there's one decimal number in the multiplicand, adjust the vernier decimal by moving it one place to the left. This changes the unit rod from rod G to rod F and leaves the correct answer 31.5135 on rods EFGHIJ. (Fig.10)
Fig.10 
Step 3 a b c d e f g h.i j k l.m n o p q r 0 0 0 0 0 0 0 4.5 0 0 7.0 0 3 0 0 0 A B C D E F G H I J K L M , , • <==== Vernier 0 0 0 0 0 0.0 1 3 5 0 0 0 + 3 5 Step 3 + 2 8 Step 3a 0 0 0 0 3 1.5 1 3 5 0 0 0 ▬ <=== Indicator 
Similar to the technique for solving problems of multiplication, Lee works the numbers in the PO field from right to left. In this case he subtracts products from the frame in a right to left manner and uses the indicator at the bottom of the frame as a guide.
Basically the procedure is this:
**Indicator**
Where dividend and divisor are both whole numbers,
Begin by placing the indicator immediately under the rod representing
the highest order of the quotient. (Probably best explained by example.)
Example 1: 1222 ÷ 26 = 47
Step 1: In the left AO field, set divisor 26 on rods hi.
The moveable unit markers: In the right AO fields, choose rod m to carry the unit number in the quotient. Slide a unit marker between rods mn .
The vernier: In the PO field, choose rod I to be the unit rod. Set the decimal on the vernier just to the right of rod I.
1b: Set the dividend 1222 on rods FGHI.
The indicator: A quick look at the problem tells us that the first step will be dividing 26 into the 122 on rods FGH. To reflect this and to help us keep track of the quotient, slide the indicator underneath rod H. (Fig.11)
Fig.11 
Step 1 a b c d e f g h i j k l m.n o p q r 0 0 0 0 0 0 0 2 6 0 0 0 0.0 0 0 0 0 Step 1 A B C D E F G H I J K L M , , • <==== Vernier 0 0 0 0 0 1 2 2 2.0 0 0 0 Step 1b ▬ <=== Indicator 
Step 2: Divide 26 on rods hi into 122 on rods FGH. **This
is where the
indicator assists in finding the correct placement for numbers in the quotient.** The indicator is
under rod H. Rod H is one rod to the left of unit rod I. Therefore, in the right AO field, set the
quotient one rod to the left of unit rod m. Set quotient 4 on rod
l.
2a: Multiply quotient 4 on l by 6 on rod i and subtract the product 24 from rods GH in the PO field.
2a: Multiply the same quotient 4 by 2 on h and subtract the product 08 from rods FG in the PO field. This leaves the partial quotient 4 on rod l and the remainder 182 on rods GHI. (Fig.12)
Fig.12 
Step 2 a b c d e f g h i j k l m.n o p q r 0 0 0 0 0 0 0 2 6 0 0 4 0.0 0 0 0 0 Step 2 A B C D E F G H I J K L M , , • 0 0 0 0 0 1 2 2 2.0 0 0 0  2 4 Step 2a 0 0 0 0 0 0 9 8 2.0 0 0 0  0 8 Step 2b 0 0 0 0 0 0 1 8 2.0 0 0 0 ▬ <=== Indicator 
The indicator: Slide the indicator one rod to the right placing it under rod I.
Step 3: Divide 26 on rods hi into 182 on rods FGH. Set the quotient 7 on rod m in the right AO field.
3a: Multiply quotient 7 by 6 on rod i and subtract the product 42 from rods HI in the PO field.
3b & the answer: Multiply the same quotient 7 by 2 on h and subtract the product 14 from rods GH in the PO field. This finishes the problem and leaves quotient 47 on rods lm in the right AO field.
Fig.13 
Step 3 a b c d e f g h i j k l m.n o p q r 0 0 0 0 0 0 0 2 6 0 0 4 7.0 0 0 0 0 Step 3 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 1 8 2.0 0 0 0  4 2 Step 3a 0 0 0 0 0 0 1 4 0.0 0 0 0  1 4 Step 3b 0 0 0 0 0 0 0 0 0.0 0 0 0 ▬ <=== Indicator 
Lee's Abacus makes dividing decimals very easy. The only possible adjustment one might have to make is to the moveable vernier decimal. This only occurs when there are decimal numbers in the divisor.
**Decimals in the Divisor**
When setting up problems that have decimal numbers in the divisor:
for every decimal in the divisor move the vernier decimal one place to the right of unit rod.
In this first example we'll divide a decimal number by a whole number. Because the divisor is a whole number, no adjustment to the vernier will be necessary.
Example 1: 0.0016 ÷ 8 = 0.0002
The moveable unit markers: In the right AO field, choose rod k to be the unit rod for quotient. Slide a unit marker between rods kl.
Step 1: In the left AO field, set divisor 8 on rod i.
The vernier: In the PO field, choose rod G to be the unit rod. Set the decimal on the vernier just to the right of rod G.
1b: Set the dividend 0.0016 on rods GHIJK.
The indicator: Simplify the problem and think of it as dividing 8 into the 16. Slide the indicator underneath rod K. (Fig.14)
Fig.14 
Step 1 a b c d e f g h i j k.l m n o p q r 0 0 0 0 0 0 0 0 8 0 0.0 0 0 0 0 0 0 Step 1 A B C D E F G H I J K L M , , • <==== Vernier 0 0 0 0 0 0 0.0 0 1 6 0 0 Step 1b ▬ <=== Indicator 
Step 2: Divide 8 on rod i into 16 on rods JK. **This is where the
indicator assists in finding the correct decimal place in the quotient.** The indicator is under rod K.
Rod K is four rods to the right of unit rod G. Therefore, in the right AO field, we must follow suit and
set the quotient number four rods to the right of unit rod
k. Set quotient 2 on rod
o.
2a & the answer: Multiply quotient 2 on o by 8 on rod i and subtract the product 16 from rods JK in the PO field. This leaves the answer 0.0002 on rods klmno in the right AO field .
Fig.15 
Step 2 a b c d e f g h i j k.l m n o p q r 0 0 0 0 0 0 0 0 8 0 0.0 0 0 2 0 0 0 Step 2 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 0.0 0 1 6 0 0  1 6 Step 2a 0 0 0 0 0 0 0.0 0 0 0 0 0 ▬ <=== Indicator 
Example 2: 0.0000384 ÷ 0.0032 = 0.012
The moveable unit markers: In the left AO field, choose rod e to be the unit rod for the divisor Slide a marker between rods ef. In the right AO field choose rod K to be the unit rod for the quotient. Slide a marker between rods kl.
Step 1: In the left AO field, set divisor 0.0032 on rods efghi.
The vernier: In the PO field, choose rod D to be the unit rod. Set the decimal on the vernier just to the right of rod D.
1a: Set the dividend 0.0000384 on rods DEFGHIJK. (Fig.16)
Fig.16 
Step 1 a b c d e.f g h i j k.l m n o p q r 0 0 0 0 0.0 0 3 2 0 0.0 0 0 0 0 0 0 Step 1 A B C D E F G H I J K L M , • <==== Vernier 0 0 0 0.0 0 0 0 3 8 4 0 0 Step 1a ▬ <=== Indicator 
Step 2:
The vernier: Because the divisor has four decimal numbers, move the decimal on the vernier four rods to the right. This places it just to the right of rod H. Rod H has become the unit rod.
The indicator: Simplify this first part of the problem and think of it as dividing 32 on rods hi into 38 on rods IJ. Slide the indicator underneath rod J. (Fig.17)
Fig.17 
Step 2 a b c d e.f g h i j k.l m n o p q r 0 0 0 0 0.0 0 3 2 0 0.0 0 0 0 0 0 0 A B C D E F G H I J K L M , , • <==== Vernier 0 0 0 0.0 0 0 0 3 8 4 0 0 ▬ <=== Indicator 
Step 3: Divide 32 on rods hi into 38 on rods IJ. **The indicator under rod J will assist in finding the correct decimal place in the quotient.** The indicator is under rod J. Rod J is two rods to the right on unit rod H. Therefore, in the right AO field, set the quotient two rods to the right of unit rod K. Set quotient 1 on rod m.
3a: Multiply the quotient 1 on rod m by 2 on rod i and subtract the product 02 from rods IJ in the PO field.
3b Multiply the same quotient 1 on rod m by 3 on rod h and subtract the product 03 from rods HI in the PO field. This leaves the partial quotient 1 on rod m and the remainder 64 on rods JK. (Fig.18)
Fig.18 
Step 3 a b c d e.f g h i j k.l m n o p q r 0 0 0 0 0.0 0 3 2 0 0.0 1 0 0 0 0 0 Step 3 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 0 0.3 8 4 0 0  0 2 Step 3a  0 3 Step 3b 0 0 0 0 0 0 0 0.0 6 4 0 0 ▬ <=== Indicator 
The indicator: Slide the indicator one rod to the right placing it under rod K.
Step 4: Divide 32 on rods hi into 64 on rods JK. Set quotient 2 on rod n.
4a: Multiply the quotient 2 on rod n by 2 on i and subtract the product 04 from rods JK.
4b & the answer: Multiply the same quotient 2 on rod n by 3 on h and subtract the product 06 from rods IJ. This finishes the problem and leaves the quotient 0.012 on rods klmn.
Fig.19 
Step 4 a b c d e.f g h i j k.l m n o p q r 0 0 0 0 0.0 0 3 2 0 0.0 1 2 0 0 0 0 Step 4 A B C D E F G H I J K L M , , • 0 0 0 0 0 0 0 0.0 6 4 0 0  0 4 Step 4a  0 6 Step 4b 0 0 0 0 0 0 0 0.0 0 0 0 0 ▬ <=== Indicator 
Lee's abacus is a lot of fun to operate. Unlike it's more basic soroban/suan pan cousins, it has lots of little moving parts which
make things very interesting.
The techniques are easy to follow and fun to learn. An abacus of this type might be a good instrument
for school children. For teaching place value it has clear advantages. Because Lee's abacus consists of 3
separate frames this would be a very handy tool for doing more advanced calculations or for storing numbers
interim. Having said this, when it comes solving basic arithmetical problems, in the hands of a good operator a simple soroban or suan pan will always be
much faster and much more efficient.  T.P.H March, 2005
Readers often email me asking questions about buying a Lee Kaichen abacus. Since they are no longer manufactured those looking to find one are faced with a difficult task. Even though they rarely come up for auction doing an Ebay search is probably the best way to locate one. Many of Lee's abaci are in the hands of collectors.
REFERENCES:
Lee Kaichen Kojima, Takashi. Heffelfinger, Totton & Flom, Gary. Abacus: Mystery
of the bead
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© March, 2005
Totton Heffelfinger
Toronto Ontario Canada
Email
totton[at]idirect[dot]com