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ABACUS
Small and Handy but a remarkably efficient Chinese Calculator
The Abacus, a sort of rectangular board with many beads -- somewhat similar to what children use at primary school to learn counting. Similar calculators (1)
exist in several other countries, including Russia as well as Japan, but with differentiating variations. There is no doubt that it is one of the most prominent articles of business equipment in Chinese business offices today. There is not a home or office in Taiwan where an Abacus is not found. Its use is compulsory in all primary and business schools. Of late, many a foreigner is becoming interested in the use of the (2)
Abacus in a increased number. Wherever a Chinese goes, be it New York, London, Bombay, Sidney or Buenos Aires, the Abacus goes with him. Chinese government officials, bankers, financiers, office clerks, junk dealers, and house wives depend largely upon it for their counting in the same manner, as a child depends on his mother. A Chinese is so dependent on the Abacus that he finds it awkward to make even simple calculations without one. (3)
Naturally a great many expressions utilizing the name of Abacus has developed out of this national habit. Today, there are more than 100 different models of the Chinese Abacus varied in size and quality of the materials used, but only one type made of brass mounting on a marble base is the best. This Chinese dependence on the Abacus is under- (4)
standable in the light of what this simple gadget can accomplish in the efficient calculations from the simplest to the most complex mathematical problems of addition, subtraction, multiplication and division. More briefly explained, the Abacus is an instrument to augment mental arithmetic. The mechanics of its use is simple, and its accuracy is assured by repetition and practice alone. (5)
Many speed and accuracy contests have been staged between the Abacus and Calculating Machines. The small, insignificant and inexpensive Abacus has usually been the victor. It must be understood, how- ever, that while the Abacus outshines its mechanical brother in addition and subtraction, it faces some disadvantages in multiplication and division problems. (6)
INTRODUCTION TO "ABACUS"
Unit points on the crosspiece are usually placed (7)
one every fifth upright. Unit points are use to indicate "ones", "thousands", "millions", "billions" and so on.
- How to manipulate your fingers.
In pushing the counters, there are three ways. 1) Use only the forefinger. 2) Using the thumb and forefinger. 3) Using the thumb, forefinger and middle finger. (8)
The third one is the method with the thumb. Of these three ways, the second one using the thumb and forefinger is recommended, it being the best and easiest. Thumb used to: "Put in" counters in the lower part. Forefinger used to: "Take off" the counters in the lower part. "Put in" a counter in the upper part. (9)
"Take off" a counter in the upper part. Note: "Put in" means pushing the counters toward the crosspiece. "Take off" means pushing the counters away from the crosspiece.
- How to manipulate counters.
the manipulation consists of the following six motions. 1) Putting in counters. (10)
2) Taking off counters. 3) Putting in five counter and taking off one-counter. 4) Putting in one-counter and taking off five counter. 5) Taking off counter of ones' place and putting in a counter of tens place. 6) Taking off a counters of tens' place and putting in counters of ones' place: Note: In order to indicate 10, a counter in the lower part on the next upright to the left is put in, (11)
i.e. a counter in tens' place is put in. In illustrating the manipulations which ar most important in calculations, the following symbols are used to facilitate one's understanding of its manipu- lation. ^ Push up the counter with the thumb or fore finger. (In the case of taking off the counter in the upper part, push up the counter with the (12)
forefinger. (subtraction) In the case of putting in the counter in the lower part, push up the counter with the thumb. (addition) Push down the counter with the forefinger. (means addition in the case of five counter) (means subtraction in the case of one counter) A counter already placed, i.e., a counter which has already been "put in" and which remains (13)
either on the top or at the bottom. (*) A counter newly "put in". (/) A counter "taken off".
(A) 2 + 1 
Answer: 3 In this case, first put in 2 with the thumb, second, put in 1 with thumb. (14)
(B) 2 + 6 
Answer: 8 In this case, put in 2 with the thumb first. Then put in 5 (upper part) with the forefinger, and simultaneously put in 1 (lower part) with the thumb. (15)
(C) 2 + 4 
Answer: 6 First, put in 2 with the thu- mb; Second, to add for put in 5 of the upper part and take off 1, of the lower part with the forefinger. (16)
(D) 9 + 7 
Answer: 16 First, put in 9 with the for- efinger and thumb. Then to add 7, take off 3 with the for- efinger, and put in 1 in place with the thumb, i.e., think 7 and 3 are 10. So take off 3 and carry the 1 ten which is put in the tens' the tens' place with thumb. (17)
(E) 36 + 75 
Answer: 111 To add two place numbers and above, always commence ad- ding from the left to the right. In this case put in 7 tens i.e., add 7 tens of 75 to 3 tens of 36 in the same method as in example D. Next add 5 ones to 6 ones in the following way. (18)
Take off 5 (a five counter) with the forefinger, and simultaneously put in 1 ten in the tens' place with the thumb as in example D. [Other examples] 9 + 7 = 16 4 +27 = 31 3 + 8 = 11 7 + 5 = 12 23 +20 = 43 9 + 5 = 14 8 +13 = 21 2 + 8 = 10 21 + 8 = 29 50 +85 =135 6 + 9 = 15 76 +82 =158 6 + 7 = 13 31 + 7 = 38 8 + 4 = 14 25 +13 = 38 (19)
(F) 10 - 3 
Answer: 7 In this case there are no ones. Consequently 1 ten is borrowed by taking off 1 ten with the forefinger and 3 ones are subtra- cted from it by putting in 7 ones, the remainder, with the forefinger and thumb. (20)
(G) 12 - 6 
Answer: 6 In the same there are not enou- gh) ones to subtract from. Cons- equently, as in example (F) 1 ten must be borrowed. With the forefinger 1 ten is taken off and the remainder 4 ones are put in by putting in a five counter and taking off a one counter in the ones' place. (21)
(H) 100 - 58 
Answer: 42 In subtraction involving two place, three place num- bers and above, always com- mence from the left to the right as in the case of addi- tion. In this case subtract 5 tens of the number 58 from the tens. However, as there are no tens, 1 hundred is borrowed in a same manner as in example F. Next, in subtracting 8 (22)
ones, as there are no ones left, 1 ten must be borrowed. 8 is subtracted and 2 ones, the remainder, are put in by the same method as of example (F). [Other examples] 4 - 2 = 2 5 - 3 = 2 6 - 2 = 4 8 - 6 = 2 17 - 4 =13 11 - 6 = 5 15 - 6 = 9 12 - 6 = 6 88 -45 =43 40 -15 =25 30 -22 = 8 27 -12 =15 136 -74 =62 112 -50 =62 300 -126=174 10000-41=959 Exercise for addition and subtraction. (23)
"How to exercise?" You need not care about the unit position. (1) 1 2 3 4 5 6 7 8 9 First of all, put in (2) 2 4 6 9 1 3 5 7 8 "123,456,789" on the (3) 3 7 0 3 7 0 3 6 7 "Abacus", and then (4) 4 9 3 8 2 7 1 5 6 put in "123,456,789" (5) 6 1 7 2 8 3 9 4 5 eight times, i.e., and (6) 7 4 0 7 4 0 7 3 4 "123,456,789" to "123, (7) 8 6 4 1 9 7 5 2 1 456,789" eight times. (8) 9 8 7 6 5 4 3 1 2 ---------------------- (9) 11 1 1 1 1 1 1 0 1 (24)
The sum is "1,111,101". This is exercise for addition. For exercise of subtraction, subtract "123,456, 789" from "1,111,111,101 nine times. The remainder "0".
Determination of unit position in the product: 1) In such case where the multiplier is the integral or mixed decimal number (for example 3.56; 19.362 etc), count figures in the integral (25)
part of the multiplier. Count off equal number of uprights, from the unit position of the mult- iplicand, toward the right: this upright is the unit position for the product. 2) In case the multiplier is a decimal number (except the mixed-decimal number) and there are some 0's "Zeros" between the decimal point and the decimal-significant figure 1, 2, 3, ......9: Count a number of 0 "zero" between the decimal (26)
point and the decimal-significant figure of the multiplier. Count off an equal number of uprights, from the unit position of the multiplcand, toward the left: this upright is the unit position for the product. 3) In such case where the multiplier is the decimal number and there is no 0 "zero" between the decimal point and the decimal-significant number the unit position of the multiplicand is the unit (27)
position for the product. 67 x 2 = 134 
14 12 --- 134 Compare the last figure of the multiplicand, i.e., 7 on the upright B, with the multiplier 2. "2 x 7 is 14" therefore, after taking off 7 from the upright B, place 1, the first figure of 14, on the upright B, and (28)
Place 4, the second figure, on the upright C. Next, compare 6, another figure in the multiplicand, with the multiplier, 2. "2 x 6 is 12" therefore, after taking off 6 from the upright A, place 1, the first figure of 12, on the upright A, then add2, the second figure, to 1 on the upright B. The product is 134. (29)
9 x 7 = 63 
63 Compare the multiplicand, i. e., 8, with the multiplier 7. "9 x 7 is 63", ---therefore, take off 9 and place 6, the first figure of 63, on the upright A, and place 3, the second figure on the upright. B, after taking off 9 from the upright A. The pro- duct is 63 (30)
26 x 14 = 364 
0 6 2 4 0 2 0 8 ------------- 3 6 4 (31)
In illustration No. 1, we see that the multiplier is a two place integral. Therefore, count off two uprights, from the unit position of the multiplicand, i. e., from the upright B, toward the right. The upright D is the unit position for product. Process of multiplication: 1) Compare the last figure of the multiplicand, i. e., 6 on the upright B, with the first figure of the multiplier i.e. 1. (32)
"1 x 6 is 6" --- therefore, after taking off 6 from the upright B, place 6 on the upright C. 2) Next, compare the same 6 with the second figure of the multiplier. 4, "4 x 6 is 24" --- therefore, add 2, the first figure of 24, to 6 on the upright C, and place 4, the second figure, on the upright D. 3) Compare 2, the first figure of the multiplcand, with 1, the first figure of the multiplier. "1 x 2" (33)
is 2", therefore, place 2 on the upright B, after taking off 2 from the upright A. Finally, compare the same 2, the first figure of the multiplicand, with 2, the second figure of the multiplier, "4 x 2 is 8" --- therefore add 8 to 8 on the upright C. See illustration No. 3. The product is 364 as shown on the "Abacus" now. (34)
678 x 345 = 233910 
2 4 (1) 3 2 (2) 4 0 (3) 2 1 (4) 2 8 (5) 3 5 (6) 1 8 (7) 2 4 (8) 3 0 (9) --------------- Required answer... 2 3 3 9 1 0 (35)
Determination of the unit ---------------------- position for the product: In multiplicand 394 this case, the multiplier is a ---------------------- three place integral, therefore, multiplier | product count off three uprights, from ---------------------- the unit position of the mul- 2 | tiplicand, i.e., from the up- 4 | right C, toward the right: the 6 | upright F is in the unit position 12 | 19 | 20 | 123 | 456 | ---------------------- (36)
for the product. The diagram above shows the diff- erent steps on "Abacus" in the course of calculation of "678 x 345". A B C D E F G Note: The encircled --------------- figures indicate the 4 0 7 6........multiplicand steps in calculation. 1 2......(1) (2 x 6) 4.076 x 0.028 = 0.11428 4 8....(2) (8 x 6) multiplier 1 4........(3) (2 x 7) 0.028 5 6......(4) (8 x 7) 8............(5) (2 x 4) 3 2..........(6) (8 x 4) --------------- 0 1 1 4 1 2 8 (37)
Note: The encircled figures indicate the steps in calculation. Determination of the unit position for the product: In the above example, we see that the multiplier has one 0 "zero" between the decimal point and the decimal-significant figure. Therefore, count off one upright from the unit position of the multiplicand, i. e., from the upright B, towards the left: Then the upright A is the unit position for the product. (38)
Note: At first, calculate by placing both "multipli- cand" and "multiplier" on the "Abacus". As you practice on the "Abacus", calculate by placing either the "multiplicand" or "multiplier" on the "Abacus". If you practice further and become proficient in calculation on the "Abacus", calculation by placing neither of them is possible. Calculation can be done more rapidly this way. (39)
Determination of the unit position for the quotient: 1) Count the figures in the integral part of the divisor. Count off an equal number of uprights, from the unit position of the dividend towards the left: count off one more: and the upright is the unit position for the quotient. 2) In such case as there are some 0's (zero's) between the decimal point and the decimal-significant (40)
figure of the divisor. (A) Count the number of 0's ("zero"s) between the decimal point an the decimal-significant figure of the divisor. Count off an equal number of upright, from the unit position of the dividend towards the right: count off one less: and the last upright is the unit position for the quotient. (B) In such case as there is one "0" ("zero") (41)
between the decimal point and the decimal- significant figure of the divisor, the unit position of the dividend is the unit position for the quotient. 3) In such case as there is no "0" ("zero") between the decimal point and the decimal-significant figure, next upright from the unit position of the dividend towards the left is the unit position for the quotient. (42)
93 / 3 = 31 
divisor dividend 3 A B C D 9 3 3 9 --------- 3 0 0 3 1 -3 --------- 3 1 (43)
In the above illustration, we see that the divisor is a one-place integra therefore, count off one upright, from the unit position of the dividend, i.e., from the upright D, towards the left: count off one more: then the last upright, i.e., the upright B is the unit position for the quotient. Process of division: 1) Compare the first figure of the dividend, i.e., 9 on the upright C, with the divisor 3. Find how (44)
many 3's there are in 9 --- "9 / 3 is ?". There are three 3's in 9, therefore, place 3 on the upright A, as the quotient figure. 2) Multiply the divisor 3, by the quotient figure, 3 on the upright A. "3 x 3 is 9", therefore subtract 9 from the upright C. 3) Compare the remainder, 3 of the dividend on the upright D, with the divisor 3. Find how many 3's there are in 3, --- "3 / 3 is ?". (45)
There is one 3 in 3, therefore, place on the upright B as the quotient figure. 4) Multiply the divisor by the second quotient figure of 1. "3 x 1 is 3" --- therefore, take off 3 from the upright D. The quotient is 31. (46)
1476 / 12 = 123 
divisor dividend 12 A B C D E F 1 4 7 6 1 -1 -2 --------------------- partial........2 7 6 dividend 2 -2 -4 --------------------- partial..........3 6 dividend 3 -3 -6 --------------------- quotient 1 2 3 0 (47)
In the left illustration, we see that the divisor is a two-place integral; therefore, count off two uprig- hts, from the unit position of the dividend, i.e., from the upright F towards the left: count off one more; then the last one the upright C, is the unit position for the quotient. Process of division: 1) Compare the first figure of the dividend, i.e., 1 on the upright C with the first figure of the (48)
divisor of 12, i.e., 1. Find how many 1's there are in 1 --- "1 / 1 is ?" There is one 1 in 1, there- fore, place 1 on the upright A as the quotient figure. 2) Multiply 1, the first figure of the divisor, by 1, the first quotient figure on the upright A. "1 x 1 is 1" --- therefore, subtract 1 from the upright C. 3) Multiply 2, the second figure of the divisor, by 1, the first quotient figure "1 x 2 is 2"; --- therefore, (49)
subtract 2 from the upright D. Now you will get 276 which is the partial dividend. 4) Compare the first figure of the partial dividend i.e., 2, with the first figure of the divisor of 12 i.e., 1. Find how many 1's there are in 2 --- "2 / 1 is ?". There are two 1's in 2, therefore, place 2 on the upright B as the quotient figure. 5) Multiply 1, the first figure of the divisor by 2, the second quotient figure on the upright B. "1 (50)
x 2 is 2" --- therefore, subtract 2 from the upri- ght D. 6) Multiply 2, the second figure of the divisor, by 2, the second quotient figure. "2 x 2 is 4" --- therefore, subtract 4 from the upright E. Now you will get 36 as the partial dividend. 7) Compare the first figure of the partial dividend, i.e., 3 with the first figure of the divisor of 12, i.e., 1. (51)
Find how many 1's there are in 3 --- "3 / 1 is ?". There are three 1's in 3 therefore, place 3 on the upright C as the quotient figure. 8) Multiply 1, the first figure of the divisor of 12, by 3, the last quotient figure on the upright C. "1 x 3 is 3" --- therefore, subtract 3 from the upri- ght E. 9) Multiply 2, the second figure of the divisor by 3, the last quotient figure "2 x 3 is 6" -- therefore, (52)
638 / 22 = 29 subtract 6 from the divisor dividend upright F. The 22 A B C D E quotient is 123. 6 3 8 From the foregoing 3 explanations you can 6 see that the unit -6 position for the quotient ----------- is the upright B. 3 0 0 3 8 1) Compare the first -1 +2 ----------- 2 0 2 3 8 -4 ----------- 2 0 1 6 8 9 -1 8 -1 8 ----------- 2 9 (53)
figure of the dividend, 6, with the figure of the divisor 2. Find how many 2's there are in 6, --- "6 / 2 is ?". There are three 2's in 6. Therefore, place 3 on the upright A. 2) Multiply 2, the first figure of the divisor, by 3 the quotient figure on the upright A. "2 x 3 is 6" --- therefore subtract 6 from the upright C. 3) Multiply 2, the second figure of divisor, by 3 (54)
the quotient figure "2 x 3 is 6" --- therefore, sub- tract 6 from the upright D. But in this case, 6 cannot be subtracted from 3 on the upright D. So, take 1 from 3, the quotient figure on the upright A and replace 2, on the upright C (i.e., this 2, the first figure of divisor which had been multiplied by 3. Then, multiply 2, the correct quotient figure thus obtained on the upright A, by 2, the second figure of divisor. "2 x 2 is 4" --- (55)
therefore, subtract 4 from the upright D. 4) Divide 198, the partial dividend, by 22, the divisor. In this case, find how many 2's, the first figure of the divisor, there are in 19, the first and second figure of the partial dividend. "19 / 2 is ?". There are nine 2's in 19. Therefore, place 9 on the upright B as the last quotient figure. 5) Multiply 2, the first figure of the divisor, by 9, (56)
the last quotient figure. "2 x 19 is 18" --- therefore, subtract 18 from the upright C and D. Finally, multiply 2, the second figure of the divisor, by 9 the last quotient figure "2 x 9 is 18" --- therefore, subtract 18 from the upright D and E. The quotient is 29. -- E N D -- (57)
