Vedic (SSA)

2015-2016

(For Teachers and Students)

VEDIC MATHEMATICS MANUAL

Material produced by:Resource persons/ Teachers in collaboration with IIVA

Contents

1. Introduction of Vedic Maths 1

Benefits of Vedic Maths

History of Vedic Maths

Sutras & Sub Sutras

Chapter - 1 13

2. Addition

Chapter - 2 20

3. Subtraction

(a) All from a last from 10

(b) Using Vinculum

Chapter - 3 20

4. Digit Sum (casting out 9’s, 9-check method)

Chapter - 4 24

5. Special Multiplication Methods

a) Multiplication by 11 and multiples of 11

b) Multiplication by 12 to 19

c) Multiplication by 111

d) Multiplication by 222 to 999

e) Base Method Multiplication

f) If sum of unit digit is 10 and rest place digit are same.

g) If sum of ten’s place digits is 10 and ones place digit is same.

h) Multiplication by 9

I) Multiplication by number ending with 9 i.e. 19 to 99

j) General Method

k) Algebraic Expressions

Chapter - 5 57

6. Squares and Square Roots

Chapter - 6 65

7. Division

1

Introduction of Vedic Maths

Mathematics is universally regarded as science of all sciences. In all early

civilizations, the first expression of mathematical understanding appears

in the form of counting systems. In Indian history “MATHEMATICS” had

occupied a significant place. All earlier systems of numeration had certain

drawbacks. It was Indian system that reached the western world through

the Arabs and has now been accepted universally. India has always been a

land of great mathematicians. Aryabhatta a man who knew infinity,

Srinivasa Ramanujan etc are some mathematicians who revolutionized

the world of mathematics.

In Vedic period, records of mathematical activities are mostly to be found

in Vedic texts associated with rituals.

Vedic Mathematics is a term given to this ancient

system of Mathematics which is supposed to be

formulated over many centuries by ancient sages

and rishis of India. It was rediscovered from

Vedas between 1911 and 1918 by Jagadguru

Swami Sri Bharti-Krishna Tirathji Maharaj.

The term vedic mathematics refers to sixteen

mathematical formulae or sutras and their corollaries or subsutras derived

from the vedic system. In vedic system pupil are encouraged to be

creative and use whatever method they like. All that the student has to do

is to look for certain characteristics, spot them out, identify the particular

type and apply the formulae which is applicable there to. Vedic

Mathematics is useful in preparing students for competitive examination.

2

History of Vedic Mathematics

Mathematics is as old as human race itself. A human brain understands

mathematics in a very little age. Little kids have number sense that can be

easily judged in certain practical situations. Let's take an example, if you

keep 5 toffees in one hand and 2 toffees in other hand and ask child to

choose either of the two. He/ She definitely go for hand with five toffees.

This shows the child has number sense, he/ she knows the concept of more

& less.

The evidence of mathematics was available in all early civilizations also.

During Harappa period decimal system was prevalent as weights

corresponding to ratios of 0.05, 0.1, 0.2, 0.5, 1, 2, 5 etc and scales with

decimal divisions were found.

In Indian history during Vedic period, records of mathematical activities are

mostly to be found in Vedic texts associated with ritual activities.

Arithmetic's operations (GANIT) such as additions, subtractions,

multiplications, fractions, squares, cubes and roots are enumerated in

Narad- Vishnu Purana attributed to Ved Vyas (pre 1000 B.C). Examples of

Geometric knowledge (rehka –ganit) are to be found in Sulva- Sutra of

Baudhayana (800 B.C) and Apasthamba (600 B.C) which describe

techniques for the construction of altars in use during Vedic Era.

It is considered that all the information about Vedic period was stored in a

book called VEDA. Later on it was divided into four Vedas namely RIGVEDA,

YAJURVEDA, SAMAVEDA and ATHARVAVEDA. Indian Rishis and Sages used

to teach everything in Gurukuls. Written Records of many things are not

available.

Vedic Mathematics is the name given to ancient system of mathematics

which is supposedly is a part of Sthaptya –Veda, an Upveda of

Atharvaveda.

3

It is a gift to the world and was formulated over centuries by ancient sages &

rishis of India. It was rediscovered from Vedas between 1911 & 1918 by

Jagadguru Swami Sri Bharti-Krishna Tirathji Maharaj. Swamiji was the

Shankracharya of Govardhan Math, Jagannath puri as well as Dwarka,

Gujarat (1884-1960). He was a great scholar and a brilliant student. He was

M.A in seven subjects including English, Mathematics, Science, Sanskrit,

History etc. He was a person who worked in this field and wrote 16 volumes

on the sixteen sutras of vedic mathematics. These volumes got lost and only

one volume containing all information is in print.

The term Vedic Mathematics then onwards refers to sixteen sutras on which

further derivations are being done. This work of Swamiji is considered to be

first work done in field of vedic mathematics.

The most striking features of Tirathji's work is its coherence. The whole

system is inter-related and unified. The methods are simple and help

students to give solution to various arithmetical problems mentally.

Later on, in late 1960's some mathematicians took interest in this system

and started delivering lectures on this subject. There are number of books in

print on the subject. Lots of people are working in the field of Vedic

Mathematics. In India many organizations are working in this field to make

the subject more interesting and help pupil remove the phobia of

mathematics.

Benefits of Vedic Maths

Vedic Mathematics provides pupil with the “extra something” which helps

them to be different. It help them to play with numbers and thereby

removes the phobia of Mathematics. It helps them to be more confident

about the subject. Pupil start loving Mathematics once their interest is

triggered and confidence level is increased.

Advantages:

® It helps in reducing silly mistakes.

® It helps in solving problems 10-15 times faster.

® It helps in intelligent guessing.

® Reduces Phobia of Mathematics.

® It reduces the burden (Need to learn tables only up to 9).

® It increases concentration.

Vedic Maths is a system of reasoning and mathematical working which is

fast efficient and easy to learn and use . We all are aware that world is

getting competitive day by day and in every competition there is a race

against time only those people who have ability to calculate faster will able

to win competitions as faster calculations will lead to time saving and saved

time can be utilised to solve much difficult problems.

Vedic mathematics provides answer in one line where as conventional

method requires several steps. Its techniques simplifies multiplication,

divisibility, squaring, cubing etc. It can even simplifies recurring decimals

and Auxiliary fractions. The simplicity of techniques encourages flexibility

and helps student to device newer techniques i.e. his/her own methods and

not remain limited to same rigid approach.

The simplicity of Vedic Mathematics encourage most calculations to be

carried out without the use of pen and paper. Once the mind of students

develops the understanding for system he/she begins to work closely4

5

with numbers and becomes more creative. It is very easy to understand

and practice. This process helps student to be confident about the subject

mathematics and this removes his/her phobia of mathematics.

By Applying Sutras we can save lot of time and effort in solving the

problems, compared to the formal methods presently in use . Though the

solutions appear like a magic, the application of the Sutras is perfectly

logical and rational. This course of Vedic Mathematics seeks to present an

integrated approach to learning Mathematics with interest and

inquisitiveness, avoiding the monotony of accepting theories and logical

proof of the Sutras is explained, which eliminates the misconception that

the Sutras are wrongly explained.

Application of the Sutras improves the computational skills of the learners

ensuring both speed and accuracy. The knowledge of such methods

enables the teachers to be more resourceful to enhance the students and

strengthen their talent and knowledge. To specific problems application

of sutras involves rational thinking, which, in turn, helps improve intuition

that is the bottom - line of the mastery of the mathematical geniuses of the

past and the present such as Aryabhatta, Bhaskaracharya, Srinivasa

Ramanujan etc.

This course makes the learning Mathematics at the school and college

level in a way different from what is taught at present, but strictly

embodying the principles of algebra for empirical accuracy. The

presentation and explanation of concepts are in simple language to make

it easy to understand. This course will build your confidence level as well

as help you become self dependent.

6

Sutras

1 ,dkf/kdsu iwosZ.k(Ekâdhikena Pûrve?a)

2 fuf[kya uor'pjea n'kr%(Nikhila? Navataœcarama? Daœata?)

3 Å/oZfr;ZXH;ke~ (Ûrdhva Tiryagbhyâ?

4 ijkoR;Z ;kst;sr~ (Parâvartya Yojayet)

5 'kwU;a lkE;leqPp;s (Œûnya? Sâmyasamuccaye)

6 ¼vkuq#I;s½ 'kwU;eU;r~ (Ânurûpye Œûnayamanyat)

7 ladyuO;odyukH;ke~ (Sa?kalana Vyavakalanâbhyâ? )

8 iwj.kkiwj.kkH;ke~ (Pûra?âpûra?âbhyâ? )

9 pyudyukH;ke~(Calana Kalanâbhyâ? )

10 ;konwue~ (Yâvadûna? )

11 O;f"Vlef"V% (Vya??isama??i?)

12 'ks"kk.;M~dsu pjes.k (Œe?â?ya?kena Carame?a)

13 lksikUR;};eUR;e~(Sopântyadvayamantya? )

14 ,dU;wusu iwosZ.k(Ekanyûnena Pûrvena)

15 xqf.krleqPp;% (Gu?itasamuccaya?)

16 xq.kdleqPp;% (Gu?akasamuccaya?)

By One More than the One Before.

All from 9 and the Last from 10.

Vertically and Crosswise

Transpose and Apply

If the Samuccaya is the Same it is Zero

If One is in Ratio the Other is Zero

By Addition and by Subtraction

By the Completion or Non-Completion

Differential Calculus

By the Deficiency

Specific and General

The Remainders by the Last Digit

The Ultimate and Twice the Penultimate

By One Less than the One Before

The Product of the Sum

All the Multipliers

7

vkuq:I;s.k (Ânurûpye?a)

f'k";rs 'ksÔlaK% (Œi?yate Œe?as? jña?)

vk|ek|s ukUR;eUR;su (Âdyamâdyenântyamantyena)

dsoyS% lIrda xq.;kr~(Kevalai? Saptaka? Gu?yât)

os"Vue~ (Ve??ana? )

;konwua rkonwue~ (Yâvadûna? Tâvadûna? )

;konwua rkonwuhd`R; oxZ p ;kst;sr~(Yâvadûna? Tâvadûnîk?tya Vargañca Yojayet)

vUR;;ksnZ’kds·fi (Antyayordaœake'pi)

vUR;;ksjso (Antyayoreva)

leqPp;xqf.kr% (Samuccayagu?ita?)

yksiLFkkiukH;ke~ (Lopanasthâpanâbhyâ? )

foyksdue~ (Vilokana? )

xqf.krleqPp;% leqPp;xqf.kr%(Gu?itasmuccaya? Samuccayagu?ita?)

Proportionately

The Remainder Remains Constant

The First by the First and the Last by the Last

For 7 the Multiplier is 143

By Osculation

Lesson by the Deficiency

Whatever the Deficiency, less on by that amount and set up the square of the Deficiency.

Last Totalling 10

Only the Last Terms

The sum of the Products

By Alternate Elimination and Retention

By Mere Observation

The Product of the Sum is the Sum of the Products.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

Sub-Sutras

8

Terms and Operations

(a) Ekadhika means ‘one more’

e.g : Ekadhika of 0 is 1

Ekadhika of 1 is 2

Ekadhika of 7 is 8

Ekadhika of 27 is 28

Ekadhika of 245 is 246

(b) Ekanyuna means ‘one less’

e.g : Ekanyuna of 1 is 0

Ekanyuna of 2 is 1

Ekanyuna of 24 is 23



(c) Purak means ‘ complement’

e.g: Purak of 1 is 9

Purak of 2 is 8

Purak of 3 is 7

Purak of 4 is 6

Purak of 5 is 5 and Vice Versa

(d) Rekhank means ‘a digit with a bar on its top’. In other words it is a

negative number.

e.g : A bar on 8 is written as 8. It is called Rekhank 8 or bar 8. We treat

Purak as a Rekhank.

e.g : 8 is 2 and 2 is 8

At some instances we write negative numbers also with a bar on the

top of the numbers as -2 can be shown as 2. -45 can be shown as 45.

(e) Beejank : The Sum of the digits of a number is called Beejank. If the

addition is a two digit number, then these two digits are also to be

added up to get a single digit.

e.g: Beejank of 24 is 2 + 4 = 6

Beejank of 648 is 6 + 4 + 8 = 18, further 1 + 8 = 9

Easy way of finding Beejank :

beejank is unaffected if 9 is added to or subtracted from the number.

This nature of 9 helps in finding. Beejank very quickly, by cancelling

9 or the digits adding to 9 from the number.

9

e.g. : Find the Beejank of 167432

As above we have to follow

167432---------- 1+6+7+4+3+2 ------- 23 -------- 2 + 3 ------- 5

But a quick look gives 6 & 3 ; 2&7 are to be ignored

because 6+3=9, 2+7=9. Hence remaining

1 + 4 --------- 5 is the beejank of 167432

(f) Vinculum : The numbers which by presentation contain both positive

and negative digits are called vinculum numbers.

Conversion of general numbers into vinculum numbers :

We obtain them by converting the digit which are 5 and above 5 or less

than 5 without changing the value of that number.

Consider a number say 7.(Note that it is greater than 5).

Use its complement (Purak-rekhank) from 10.

It is 3 in this case and add 1 to the left ( i.e. tens place) of 7.

Thus, 7 = 07 = 13

The number 1 contains both positive and negative digits.

i.e. 1 and 3. Here 3 is in units place, hence it is - 3 and value of 1 at tens

place is 10.

T O

1 3

3 x 1 = -3+

1 x 10 = 10

7

Conveniently we can think and write in the following way

General Number Conversion Vinculum number

8 10-2 12

96 100-4 104

387 400-13 413 etc.

10

Chapter-1

Addition is the most basic operation and adding number1 to the previous number

generates all the numbers. The Sutra “By one more than the previous one describes the

generation of numbers from unity.

0 + 1 = 1 1 + 1 = 2 2 + 1 = 3

3 + 1 = 4 4 + 1 = 5 5 + 1 = 6

6 + 1 = 7 7 + 1 = 8 8 + 1 = 9 9 + 1 = 10 ......

Completing the whole method

The VEDIC Sutra ‘By the Deficiency’ relates our natural ability to see how much something

differs from wholeness.

7close to 10

8 close to 10

9 close to 10

17, 18, 19, are close to 20

27, 28, 29 are close to 30

37, 38, 39 are close to 40

47, 48, 49 are close to 50

57, 58, 59 are close to 60

67, 68, 69 are close to 70

77, 78, 79 are close to 80

87, 88, 89 are close to 90

97, 98, 99 are close to 100 ...................

and so on

We can use this closeness to find addition and subtraction.

The Ten Point Circle

Rule : By completion non-completion

Five number pairs

1 + 9

2 + 8

3 + 7

4 + 6

5 + 5

Use these number pairs to make groups of ‘10’ when adding numbers.

Example : 37 + 33 = 30 + 7 + 30 +3 = 30+30+10

Adding a list of numbers

Below a multiple of ten Rule : By the deficiency

49 is close to 50 and is 1 short

38 is close to 40 and is 2 short.

Addition Using Compliments

5

6

7

8

9

4

3

2

1

10

11

12

13

14

15

16

17

18

19

20

5

6

7

8

9

4

3

2

1

10

11

e.g. : 49 + 4 = 49 + 1 + 3 = 50 + 3 = 53(49 is close to 50 and 1 short of 50, 49+1 = is 50)

e.g. : 28 + 14 = 28 + 2 + 12 = 30 + 12 = 42or

28 + 14 = 30 + 14 - 2 = 42 (28 is close and is 2 short too of 30) 30 + 14 - 2 So, 44 - 2 = 42

e.g. : Add 49 + 6 = ?49 is close to 50 and is 1 less then it. So we take 1 from the 6 to make up 50 and then we have 5 more to add on which gives 55 Add39 + 28 +339 + 1 + 28 + 2(As 3 = 1 + 2 and 39 + 1 = 40, 28 + 2 =30)40 + 30 = 70 [Note: we break 3 into 1 + 2 because 39 need 1 to become 40 and 28 need 2 to become 30] Add49 + 8 + 1 + 449 + 8 + 1 + 2 + 250 + 10 + 2 =62

Sum of TenThe ten point circle illustrates the pairs of numbers whose sum is 10.

Remember : There are eight unique groups of three number that sum to10 for example 1 + 2 + 7 = 10 Can you find the other seven groups of three number summing to 10 as one example given for you ?

Rule : By completion or non-completion

Look for number pairs that make a multiple of 10

7 + 6 + 3 + 4

The list can be sequentially added as follows :

7 + 6 = 13 then 13 + 3 = 16 then 16 + 4 = 20

or

you could look for number pairs that make multiples of 10.

7 + 3 is 10 and 6 + 4 is 10

hence 10 + 10 is 20

Similarly : 78 + 14 + 63 + 12

=(78 + 12) + (14 + 3 + 60)

= 90 + 77 = 167

Example 10 10

8+7+9 + 1 + 2 + 3 + 6 +5 + 4

10 10

= 10 +1 0 + 10 + 10 + 5 = 45

This method can be easily used in commutative and associative property.

12

1. 21.

2. 22.

3. 23.

4. 24.

5. 25.

6. 26.

7. 27.

8. 28.

9. 29.

10. 30.

11. 31.

12. 32.

13. 33.

14. 34.

15. 35.

16. 36.

1. 37.

18. 38.

19. 39.

20. 40.

5+23+1+7 = 23+14+67+96=

59+41+11+19 = 64+53+16+37=

43+16+27+24= 79+28+11+82=

23+45+27+25 = 14+25+45+16=

73+7+19+11 = 15+71+65+19=

24+12+8+6+13 = 28+49+32+41=

18+3+2+17 = 39+66+41+34=

27+7+33+23 = 64+71+19+16=

29+37+11+23 = 72+45+28+55=

42+51+19+18 = 64+58+46+42=

16+23+24+7= 41+9+66+44=

58+41+12+9 = 66+73+34+27=

35+14+16+25 = 70+45+40+35=

123+11+27 = 41+47+39+53=

223+130+27 = 65+48+65+22=

506+222+278= 74+36+91+89=

23+107+502+18 = 193+147+204+106=

542+143+245+104 = 417+282+198+543=

133+544+246+247= 48+714+62+536=

457+233+246+574 = 89+646+71+144=

Addition Using Compliments

Practice Sheet

13

Subtraction ( all from 9 last from 10)

Sutra applied is Nikhila? Navataœcarama? Daœata?

This kind of subtraction is done when there are number of zeros in minuend.

eg. : u1,00,000 Minuend -72,354 Subtrahend ------------ 27,646 difference -------------

We subtract unit digit of subtrahend from 10 and remaining from 9 till the highest place of subtrahend here for eg. till ten thousand place.

From the remaining digit of places of minuend subtract 1. Here for eg, remaining place is just lakhs and digit is 1 so 1-1 = 0so the difference is 27646

eg. : v50000 -433 ---------- 49567 -----------

Here subtract unit place digit of subtrahend from 10 (10-3=7) And remaining digits i.e. 4 & 3 from 9 (9-4=5; 9-3=6)

From the remaining digit of places of minuend. Here for eg. remaining places are thousand and ten thousand subtract 1 and digits are 5 and 0

So 50 -1 = 49

So the difference is 49567

Chapter-2 Subtraction

14

1. 21.

2. 22.

3. 23.

4. 24.

5. 25.

6. 26.

7. 27.

8. 28.

9. 29.

10. 30.

11. 31.

12. 32.

13. 33.

14. 34.

15. 35.

16. 36.

17. 37.

18. 38.

19. 39.

20. 40.

100 - 37= 1.000 - 0.764=

1000 - 283= 2.0 - 0.672=

100 - 81= 3.0 - 1.282=

5000 - 898= 10.00 - 0.998=

10000 - 3725= 90.000 - 1.7312=

40000 - 3888= 98.000 - 3.255=

10000 - 5328= 2.00 - 1.3628=

200 - 63= 55.00 - 14.3827=

3000 - 63= 277000 - 277=

90000 - 36= 500000 - 433=

100 - 57= 1.000 - 0.364=

1000 - 275= 2.0 - 0.652=

100 - 78= 3.0 - 1.588=

5000 - 4521= 10.- 0.561=

10000 - 8881= 80.000 - 1.7232=

40000 - 5555= 89.000 - 13.255=

10000 - 635= 2.00 - 1.2328=

200 - 163= 55.00 - 13.2427=

3000 - 1336= 277000 - 544=

80000 - 6554= 500000 - 5522=

Practice Sheet

Subtraction ( all from 9 last from 10)

15

What is vinculum ?

Vinculum : The numbers which by presentation contain both, positive and negative digits

are called vinculum numbers.

Use of Vinculum in Vedic Maths : It is used to represent negative digit in Vedic Maths.

It helps to show mixed digits i.e.+ ve in one place and-ve in the other.

Advantages of using vinculum :

(1) It gives us flexibility, we use the vinculum when it suits us.

(2) Large numbers like 6, 7, 8, 9 can be converted to smaller digits.

(3) Figures tend to cancel each other or can be made to cancel.

(4) 0 and 1 occur twice as frequently as they otherwise would.

(5) Borrowing can be avoided

Converting from positive to vinculum form :

Sutras : (Nikhila? Navataœcarama? Daœata?) & (Ekâdhikena Pûrve?a) (All from 9 and

last from 10) and (One more than the previous one).

9=1 1 (i.e. 9 -10), 8 = 12, 7 = 13, 6 = 14, 19 = 21, 29 = 31

28 = 32, 36 = 44, 38 = 42

Step to convert from positive to vinculum form :

(1) Find out the digits that are to be converted.

(2) Apply “all from 9 and last from 10” on those digits.

(3) To end the conversions “add one to the previous digit”.

(4) Repeat this as many times in the same number as necessary.

Consider a number say 7.(Note that it is greater than 5).

Use it complement (Purak-rekhank) from 10.

It is 3 in this case and add 1 to the left ( i.e. tens place) of 7.

Thus, 7 = 07 = 13

The number 13 contains both positive and negative digits i.e. 1 and 3.

Here 3 is in unit place hence it is - 3 and value of 1 at tens place is 10.

T O

1 3

3 x 1 = -3+

1 x 10 = 10

7

Conveniently we can think and write in the following way

General Number Conversion Vinculum number

8 10-2 12

96 100-4 104

387 400-13 413 etc.

Vinculum

16

Change units digit into a vinculum.X̄ means negative X, for the purpose of subtraction and division in Vedic Mathematics. If you want to vinculate unit digit of a number then subtract that number from 10 and put the bar on number obtained (i.e find its Purak-Rekhank). Then after that add 1 to tens place digit.

eg. : 58 = 62̄ (8-10 = ̄2 ) (5+1 =6)

¯) 287 = 293̄ (7-10 = 3 (28+1 = 29)

¯) ¯) 394 = 406 (4-10 = 6 (39+1 = 40)

If you want to vinculute all digit except first then subtract 10 from unit place digit and 9

from remaining except the highest place digit.Add 1 to highest place digit.

¯ ¯ ¯ ¯ ¯eg. : 38673 = 4 1 3 2 7 (3-10=7)¯(7-9=2)

¯(6-9=3)

¯(8-9=1)

¯(3+1=4 )

¯ ¯ ¯eg. : 2907 = 3 0 9 3 (7-10=3)

¯(0-9=9)

(9-9=0) or (29+1 =30)

(2+1=3)

Steps to convert from vinculum to positive form :

(1) Find out the digits that are to be converted i.e. digits with a bar on top.

(2) Apply “all from 9 and the last from 10” on those digits

(3) To end the conversion apply “one less than the previous digit”

(4) Repeat this as many times in the same number as necessary

If you want to devinculate unit place. Subtract unit place from 10 and write the number

obtained without bar and subtract 1 from remaining digits.¯eg. : 51 = 49 (10-1 = 9)

(5-1 =4)

¯ eg.: 232 = 228 (10-2 = 8)

Zero is neither positive nor negative it can be written without bar in both cases

Change units digit into a vinculum

Change all digits to vinculum except first

Devinculate

17

If you want to devinculate, all digits except the first, then subtract units digit from 10

and remaining except highest place digit from 9. Write all numbers without bars.

Now subtract 1 from the highest place digit.¯ ¯ ¯eg. : 9 3 9 8 = 8 6 0 2 (10-8=2)

(9-9=0)

(9-3=6)

(9-1=8)

If you want to devinculate a number which is without bar at tens place again a number

with bar on hundreds place and thousands place without bar. In such case subtract units

place number from 10 and subtract 1 from tens place number again subtract hundred

place number from 10 and subtract 1 from thousands place.

¯ ¯eg. : 6 3 1 2 = 5708 (10-2=8)

(1-1=0)

(10-3=7)

(6-1=5)

Do subtraction at every place and get the respective answer with sign (Minuend-

subtrahend). For every negative digit put bar on it. Thus, the answer will come in mixed

form that is positive & negative digits together. Now devinculate the answer.

eg. : 7 3 2 8 (8-1=2 )¯-1 9 3 1 (3-2=1 )¯---------- (9-3=6 )

¯ ¯6 6 1 7 (7-1=6 )

-----------¯ ¯Devinculate 6 6 1 7 = 5397 (7 as it is)

(10-1=9)

(9-6=3)

(6-1=5)

Here’s the answer 7328- 1931 = 5397

Devinculate

Subtraction using vinculum

18

1. 11.

2. 12.

3. 13.

4. 14.

5. 15.

6. 16.

7. 17.

8. 18.

8. 19.

10. 20.

39 = 37=

48 = 48=

88=

18= 99=

58= 18 =

79= 29=

58= 68 =

43= 78=

27= 98=

25= 38 =

28=

1. 11.

2. 12.

3. 13 .

4. 14.

5. 15.

6. 16.

7. 17.

8. 18.

9. 19.

10. 20.

297= 277 =

376 = 276 =

487= 387 =

598= 498 =

688= 1809=

2768= 2708=

28677= 3568 =

3768= 19677 =

2607 = 48677 =

3708 = 58687=

1. 11.

2. 12.

3. 13.

4. 14.

5. 15.

6. 16.

7. 17.

8. 18.

9. 19.

10. 20.

47= 37 =

32= 42 =

74 = 64 =

51= 41 =

21 = 24 =

73= 53 =

63 = 64 =

24 = 88 =

81 = 84 =

84 = 47=

21. 31.

22. 32.

23. 33.

24. 34.

25. 35.

26. 36.

27. 37.

28. 38.

29. 39.

30. 40.

511= 321=

632= 532=

421= 361=

742= 672=

824= 774=

613= 568=

501= 6112=

432= 6223=

578= 6312=

333= 5223=

Practice Sheet

Change unit digit into a vinculumChange all digit to vinculum except first

Devinculate

19

1. 21.

2. 22.

3. 23.

4. 24.

5. 25.

6. 26.

7. 27.

8. 28.

9. 29.

10. 30.

11. 31.

12. 32.

13. 33.

14. 34.

15. 35.

16. 36.

17. 37.

18. 38.

19. 39.

20. 40.

4121-2787= 543-171=

5432-1567= 3555-613=

6000-4872= 8775-1885=

5132-1763= 4775-1886=

33221-17682= 8625-3736=

7231-6452= 2311-1881=

8181-6282= 643-171=

4242-1353= 4513-1515=

4327-1515= 3201-1881=

3672-1881= 4346-2854=

4848-2854= 5252-2353=

4750-281= 6132-2763=

6843-181= 7000-3872=

7843-181= 4221-2787=

8684-3526= 7328-1831=

7864-2288= 8241-4341=

2354-1068= 8730-1820=

3546-1378= 7564-1289=

2454-1268= 6884-3726=

7564-1288= 5713-1515=

Subtraction using vinculum

Practice Sheet

20

Chapter-3 Digit Sum

The word digit means a single figure of a number. The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9,

0 are all digits. Big numbers can be reduced to single digit by adding the constituents.

Digit Sums

A digit sum is the sum of all the digits of a number and is found by adding all of the

digits of a number.

The digit sum of 35 is 3+5 = 8

The digit sum of 142 is 1+4+2 = 7

Note: If the sum of the digits is greater than 9, then sum the digits of the

result again, until the result is less than 10.

The digit sum of 57 is 5 + 7 = 12 à 1 + 2 = 3 greater than 9, so need to add again

Hence, the digit sum of 57 is 3

The digit sum of 687 is 6+8+7 = 21 à 2+1 = 3

Hence, the digit sum of 687 is 3

?Keep finding the digit sum of the result, until it is less than 10

?0 and 9 are equivalent

Look and understand some more examples:

To find the digit sum of 18, add 1 and 8, i. e. 1+8 = 9. So the digit sum of 18 is 9. And

the digit sum of 234 is 9 because 2+3+4 = 9

Following table shows the digit sum of the following numbers:

15 6

12 3

42 6

17 8

21 3

45 9

300 3

1412 8

23 5

22 4

Sometimes, two steps are needed to find the digit sum.

So, for the digit sum of 29, we add 2+9 = 11. Since 11 is 2 digit number, we again find

digit sum, i. e. add 1+1 = 2

So, for the digit sum of 29, we can write

29 = 2+9 = 11 = 1+1 = 2

Similarly digit sum of 49 = 4+9 = 13 = 1+3 = 4

So the digit sum of 49 is 4.

Number 14 Digit sum 1+4 = 5 Single digit sum is 5

19 1+9 = 10 1

39 3+9 = 12 3

58 5+8 = 13 4

407 4+0+7 = 11 2

Digit sums, casting out 9's and 9' check method

21

Casting Out Nine

27

18

9

26

17

8

25

16 7

24

15

6

5

14

23

4

13

22

3

12

21

2 11 20

19

10

1

Adding 9 to a number does not affect its digit sum

So, 5, 59, 95, 959 all have same digit sum of 5.

For example, to find out the digit sum of 4939, we can cast out 9s and just add 3 and

4. So the digit sum is 7 or using 4+9+3+9 = 25 2+5 = 7

There is another way of casting out the 9s from number when you are finding its digit

sum.

Casting out 9s and digit totaling 9 comes under the Sutra “ When samuccaya is the

same, it is zero.”

Observe in 465, 4 & 5 addition is 9, they are cast out and the digit sum becomes 6.

When the total is the same (as 9) it is zero (can be cast out), cancelling a common

factor in a fraction is another example.

Number Digit Sum

1326 3

25271 8

9643 4

23674 4

128541 3

1275 6

6317892 9 or 0

Nine Point Circle

Number at each point on the circle have the same digit sum. By casting out 9s, finding

a digit sum can be done more quickly and mentally.

Check Method

Digit sum can be used to check whether the answers are correct.

Example: Solve 23 + 21 and check the answer using the digit sums

23 = digit sum of 23 is 2+3 = 5

+21 = digit sum of 21 is 2+1 = 3

44 = digit sum of 44 is 4+4 = 8

If the sum has been done correctly, the digit sum of the question and answer should

also be same, i. e. 8.

22

Digit sum of 44 = 8, so according to this check method, the answer is correct.

There are 4 steps to use digit sum to check the answers:

1. Do the question.

2. Write down the digit sums of the numbers in the question.

3. Add the digit sum

4. Check whether the two answers are same in digit sums.

If the digit sums are same, then the answer is correct.

Add 278 and 119 and check the answer.

278

+119

397

1. We get 397 for the answer.

2. We find the digit sum of 278 and 119 which are 8 & 2 respectively.

3. Adding 8 and 2 gives 10, digits sum of 10 = 1+0 = 1

4. Digit sum of 397 is

3+9+7 = 19 1+9 = 10 = 1+0 = 1

This confirms that the answer is correct.

CAUTION:

Check the following sum:

279 2+7+9=18 1+8=9 9

+121 1+2+1=4 +4

490 4+9+0=13 1+3=4 13

Here, an estimation can help you, to find the result more accurate. If by mistake, you

write 400 in place of 490, then also it will show that the result is correct.

The check is 9+4 = 13 = 4, which is same as the digit sum of the answer which

confirms the answer.

However, if we check the addition of the original number, we will find that it is correct.

This shows that the digit sum does not always find errors. It usually works but not

always.

Note : The difference of 9 and its multiples in the answer make errors.

So, keep in mind a rough estimation.

23

Digit sum puzzles

1. The digit sums of a two digit number is 8 and figures are the same, what is the

number?

2. The digit sums of a two digit number is 9 and the first figure is twice the second,

what is it?

3. Give three 2 digit numbers that have a digit sum of 3.

4. A two digit number has a digit sum of 5 and the figures are the same. What is

the number?

5. Use casting out 9s to find the digit sum of the following numbers:

Numbers

465

274

3456

7819

86753

4017

59

6. Add the following and check your answer using digit sum check:

(1) 66+77 = (2) 94+89 =

(3) 787+132 = (4) 5131+5432 =

(5) 57+34 = (6) 304+233 =

(7) 389+414 = (8) 456+654 =

Practice Problems

24

Multiplication by 11 and multiples of 11

Sub Sutra applied is Ânurûpye?a

Application of sutra is very easy, we just have to add digits in pairs. Only addition is required and no need to learn the table of 11. We call this rule as Naught Sandwich. Naught means zero.

I) Step-1 Make sandwich of multiplicand between single zeroe.

0 2314 0 x11

----------------Step-2 Add the numbers in pairs from right to left

u02314+0 (0+4=4) àv 0231+40 (1+4=5) x11 x11 ----------- -----------

4 54

w023+140 (1+3=4) à x 02+3140 (2+3=5)

x11 x11 ----------- ----------- 454 5454

�0+23140(0+2=2)

x11 ----------------

25454 Here’s your answer 2314 x 11 =25454

ii) When sum of any two digits pair of multiplicand is a two digit number.

Step 1 & step 2 are same as above but when the sum of two digits of multiplicand add up to two digits, then carry the ten's place number and add to the next digit

u 02824+0 (0+4=4) àv 0282+40 (2+4=6)

x11 x11 ----------- ----------- 4 64

w028+240 (8+2=10) àx 02+8240 (8+2+1=11)

x11 x11 ----------- ----------- 064 10341 1

y0+28240(0+2+1=3)

x11 ----------------

31064 Here’s your answer 2824 x 11 = 31064

Chapter-4 Multiplication

25

For two digit number

Example 1 : 43 x 11

Short cut method for two digit number43

x11

uWrite ones place of digit on ones place. v Tens place digit on hundreds place 43 43

x11 x11 3 4 3

w Now add the two digit of multiplicand & write the sum of digits at tens place 43 (4 +3 = 7)

x11 473

So, 43 x 11 = 473

b. Multiplication by 22 to 99 ( Multiples of 11)

Multiples mean numbers appearing in table of 11

Step-1 Find out the rank of multiple of 11 (eg. 22 is 2nd multiple or 88 is 8th multiple) and multiply the multiplicand with the number of rank, then apply the rule for 11

Example- 45 x 22u 45 (45x2=90) v 090+0 (0+0 = 0)

x 22 x11

45x 2 = 90 0

w 0 9+00 (9+0 = 9) x 0 + 900 (0+9 = 9) x11 x11

90 990 Ans. 990

Example 45 x 33u 45 (45x3=135) v 0135+0 (5+0 = 0)

x 33 x11

45x 3 = 135 5

w 0 13+50 (3+5 = 8) x 0 1+350 (1+3 = 4) x11 x11 85 485

� 0 +1350 (0+1 = 1) x11

1485

Ans. 1485

26

1. 18. 35.

2. 19. 36.

3. 20. 37.

4. 21. 38.

5. 22. 39.

6. 23. 40.

7. 24. 41.

8. 25. 42.

9. 26. 43.

10. 27. 44.

11. 28. 45.

12. 29. 46.

13. 30. 47.

14. 31. 48.

15. 32. 49.

16. 33. 50.

17. 34.

45 x 11 = 92 x 11 = 919 x 11 =

99 x 11 = 84 x 11 = 411 x 11 =

67 x 11 = 75 x 11 = 624 x 11 =

84 x 11 = 33 x 11 = 458 x 11 =

82 x 11 = 86 x 11 = 325 x 11 =

72 x 11 = 79 x 11 = 420 x 11 =

95 x 11 = 69 x 11 = 165 x 11 =

63 x 11 = 38 x 11 = 198 x 11 =

28 x 11 = 234 x 11 = 385 x 11 =

59 x 11 = 432 x 11 = 658 x 11 =

39 x 11 = 562 x 11 = 628 x 11 =

48 x 11 = 298 x 11 = 255 x 11 =

23 x 11 = 325 x 11 = 866 x 11 =

68 x 11 = 628 x 11 = 719 x 11 =

79 x 11 = 728 x 11 = 451 x 11 =

62 x 11 = 955 x 11 = 824 x 11 =

81 x 11 = 826 x 11 =

1. 18. 35.

2. 19. 36.

3. 20. 37.

4. 21. 38.

5. 22. 39.

6. 23. 40.

7. 24. 41.

8. 25. 42.

9. 26. 43.

10. 27. 44.

11. 28. 45.

12. 29. 46.

13. 30. 47.

14. 31. 48.

15. 32. 49.

16. 33. 50.

17. 34.

35 x 22 = 64 x 33 = 345 x 77 =

68 x 44 = 78 x 66 = 245 x 88 =

25 x 55 = 86 x 77 = 632 x 22 =

54 x 77 = 94 x 22 = 163 x 66 =

42 x 33 = 67 x 88 = 226 x 77 =

56 x 99 = 91 x 55 = 125 x 66 =

61 x 66 = 85 x 99 = 232 x 77 =

58 x 77 = 48 x 77 = 642 x 44 =

89 x 44 = 672 x 44 = 628 x 88 =

76 x 88 = 581 x 66 = 921 x 99 =

52 x 22 = 114 x 88 = 324 x 55 =

65 x 44 = 324 x 22 = 652 x 44 =

96 x 55 = 546 x 33 = 351 x 33 =

28 x 99 = 153 x 55 = 457 x 77 =

35 x 66 = 424 x 66 = 894 x 22 =

42 x 44 = 561 x 44 = 586 x 55 =

35 x 55 = 632 x 22 =

Practice Sheet

Multiplication by 11

Multiplication by multiples of 11

27

Sutra applied is Sopântyadvayamantya?

Using the sutra which says that the ultimate and twice the penultimate, we can solve the multiplication by 12. Using same pattern, we can solve the multiplication by of 13, 14........19.

Step-1 Sandwich the multiplicand between single zero0 1235 0 x12 ----------------

Step-2 Multiply every number of multiplicand by 2 from right to left and add it to the number following it.

u 0 1 2 3 5+0 (2x5+0=10) à v 0 1 2 3+5 0 (2x3=6+5=11+1=12)

x1 2 x1 2 ----------- ----------- 0 201 1

w 0 1 2+3 5 0 (2x2 =4+3=7+1=8) àx 0 1+2 3 5 0 (2x1=2+2=4)

x1 2 x1 2 ----------- -------------- 820 4820

y0+1235 0 (0x2=0 +1=1)

x1 2 ---------------- 14820 Here’s your answer 1235 x 12 = 14820

Note : If the answer after product and addition, in any step is 2 digit number, then 10s place digit is carry forward to next higher place digit

For any number between 12 to 19 same procedure will apply: In step 2 multiplication of number will change. The given multiplicand would be multiplied by 3 in case of 13 ; 4 is case of 14 and so on till 19.

eg. 1235x 15

u 0 1 2 3 5+0 5x5+0=25 à v 0 1 2 3+50 (5x3=15+5=20)

x15 x15 (20+2) ----------- ----------- 5 252 2

w0 1 2+35 0 (5x2=10+3=13) à x 0 1+2 3 5 0 (5x1=5+2=7)

x15 (13+2) x15 (7+1) ----------- ----------- 525 85251

y0+1 2 3 5 0 (5x0=0+1=1)

x15 ---------------- 18525 Here’s your answer 1235 x 15 =18525

x

x

x

x

x

x

x

x

x

x

Multiplication by 12 to 19

28

1. 20. 39.

2. 21. 40.

3. 22. 41.

4. 23. 42.

5. 24. 43.

6. 25. 44.

7. 26. 45.

8. 27. 46.

9. 28. 47.

10. 29. 48.

11. 30. 49.

12. 31. 50.

13. 32.

14. 33.

15. 34.

16. 35.

17. 36.

18. 37.

19. 38.

23 x 12 = 55 x 15 = 290 x 15 =

32 x 13 = 82 x 16 = 817 x 17 =

52 x 14 = 38 x 17 = 728 x 16 =

46 x 15 = 49 x 18 = 179 x 19 =

44 x 16= 58 x 19 = 348 x 18 =

56 x 17 = 79 x 12 = 235 x 12 =

19 x 18 = 52 x 14 = 584 x 13 =

39 x 19 = 86 x 13 = 643 x 14 =

79 x 12 = 99 x 15 = 223 x 12 =

86 x 13 = 97 x 16 = 437 x 16 =

74 x14 = 89 x 17 = 217 x 18 =

83 x 15 = 65 x 18 = 856 x 13 =

29 x 16 = 74 x 19 =

37 x 17 = 44 x 16 =

54 x 18 = 76 x 12 =

73 x 19 = 168 x 13 =

28 x 12 = 124 x 15 =

45 x 13 = 217 x 17 =

64 x 14 = 661 x 14 =

Practice Sheet


Concept : Double Naught Sandwich

Step-I Sandwich the mulitlplicand between two zeroes 1348 x 111 =00 1348 00

x111 -----------------

Step-II Add three digits in groups from right to left.

u 001348+0+0 (0+0+8=8) à v 00134+8+00 (0+8+4=12) x111 x111 ----------- ----------- 8 281

w0013+4+800 (8+4+3=15) àx 001+3+4800 (1+3+4=8)

x111 (15+1=16 x111 (8+1) ----------- ----------- 628 96281

y00+1+34800 (0+1+3=4) z 0+0+134800 (0+0+1=1) x111 x111 ---------------- ---------------- 49628 149628

Here’s your answer 1348 x 111 = 149628

Step-1 Find out the rank of multiple of 111 and multiply it with the multiplicand

(eg. 222 is 2nd multiple & 444 is 4th multiple)

1348 1348x 2 = 2696x222

----------------2696

Step-2 Now apply the concept of 111 on it. Add three digits of multiplicand in groups.

u 002696+0+0 (0+0+6=6) à v00269+6+00 (0+6+9=15) x111 x111 ----------- ----------- 6 561

w0026+9+600 (6+9+6=21) àx 002+6+9600 (9+6+2=17)

x111 (21+1=22) x111 (17+2-19) ----------- ----------- 256 92562 1

y00+2+69600 (6+2+0=8) z0+0+269600 (0+0+2=2) x111 (8+1) x111 ---------------- ---------------- 99256 299256

--------------

Here’s your answer 1348 x 222 = 299256 29



30

1. 18. 35.

2. 19. 36.

3. 20. 37.

4. 21. 38.

5. 22. 39.

6. 23. 40.

7. 24. 41.

8. 25. 42.

9. 26. 43.

10. 27. 44.

11. 28. 45.

12. 29. 46.

13. 30. 47.

14. 31. 48.

15. 32. 49.

16. 33. 50.

17. 34.

41 x 111 = 47 x 111 = 929 x 111 =

92 x 111 = 84 x 111 = 566 x 111 =

67 x 111 = 75 x 111 = 627 x 111 =

82 x 111 = 43 x 111 = 258 x 111 =

85 x 111 = 86 x 111 = 385 x 111 =

72 x 111 = 99 x 111 = 865 x 111 =

45 x 111 = 65 x 111 = 365 x 111 =

63 x 111 = 98 x 111 = 188 x 111 =

58 x 111 = 254 x 111 = 175 x 111 =

89 x 111 = 437 x 111 = 898 x 111 =

34 x 111 = 568 x 111 = 428 x 111 =

68 x 111 = 398 x 111 = 253 x 111 =

21 x 111 = 825 x 111 = 867 x 111 =

74 x 111 = 668 x 111 = 794 x 111 =

56 x 111 = 328 x 111 = 478 x 111 =

62 x 111 = 965 x 111 = 825 x 111 =

86 x 111 = 826 x 111 =

1. 18. 35.

2. 19. 36.

3. 20. 37.

4. 21. 38.

5. 22. 39.

6. 23. 40.

7. 24. 41.

8. 25. 42.

9. 26. 43.

10. 27. 44.

11. 28. 45.

12. 29. 46.

13. 30. 47.

14. 31. 48.

15. 32. 49.

16. 33. 50.

17. 34.

37 x 222 = 64 x 333 = 345 x 777 =

67 x 444 = 78 x 666 = 245 x 888 =

28 x 555 = 86 x 777 = 632 x 222 =

34 x 777 = 94 x 222 = 163 x 666 =

47 x 333 = 67 x 888 = 226 x 777 =

86 x 999 = 91 x 555 = 125 x 666 =

62 x 666 = 85 x 999 = 232 x 777 =

58 x 777 = 48 x 777 = 642 x 444 =

84 x 444 = 672 x 444 = 628 x 888 =

96 x 888 = 581 x 666 = 921 x 999 =

52 x 222 = 114 x 888 = 432 x 555 =

65 x 444 = 324 x 222 = 652 x 444 =

96 x 555 = 546 x 333 = 351 x 333 =

28 x 999 = 153 x 555 = 457 x 777 =

35 x 666 = 424 x 666 = 894 x 222 =

42 x 444 = 561 x 444 = 586 x 555 =

35 x 555 = 632 x 222 =

Practice Sheet


Multiplication multiples of 111

31

Base Method Multiplication

Sutra applied is Nikhila? Navataœcarama? Daœata? and Sub Sutra Ânurûpye?a

A Base is a number followed by zeros (0) eg: 10, 20, 100, 500, 1000, 10000

A) Below Base-10

Step-1 Write down the number with the negative sign

9 -1 (1 below base i.e.1 deficiency)

x 9 -1 (1 less than 10)

------

Step-2 Multiply the deficiencies and write the product at one’s place

9 -1 (-1 x -1=1)

x 9 -1

------

1

Step-3 Add the numbers diagonally or cross addition and write the sum on ten’s place

(both addition will give same answer.)

9 -1 9+ (-1) =8

x9 -1

------

81 Here’s your answer 9x9= 81

------

In case If the number obtained in step 2 is double digit then we carry forward theE subscript and add it to next place digit in next step after cross addition.

7 -3 (-3 x -4= 12)

x6 -4

------

21Cross addition.

7 -3 [7+(-4)+1 =4]

x6 -4 [6+(-3)+1=4]

------

42 Here’s your answer 7 x 6= 4

------

B) Below Base 20-90

Step 1 Write down the numbers with negative sign after subtracting the given number

from 20,30, 40 & 50..... till 90 according to the base Add the numbers diagonally

and write the answer on tens and hundreds places.

eg. u19 -1 (1 below base 20 i.e.1 deficiency)

x18 -2 (2 below base 20 i.e. 2 deficiency)

-------

32

Step 2 Multiply the deficiencies and write the product on one’s place.

eg. u 19 -1 (-1 x -2 =2)

x 18 -2

--------

2

Step 3 : Add the number diagonally or cross addition (both additions will give the same

number). Now multiply the sum with the first digit of base number i.e. by 2 for 20

base, by 3 for 30 base by 4 for 40 base and so on.

eg.u 19 -1 [{19+ (-2)} x 2]

x 18 -2 [{18+ (-1)} x 2]

------- (=34)

342


In case of step 2 if product obtained is 2 digit number then write ten’s place digit Eas subscript and later add it, in step 3 after cross addition.

eg: 17 x 16

Step-2 17 -3 Step-3 17 -3 [(17 + (-4)] x2

16 - 4 x16 -4 = (17-4) x 2

---------- ------- = 13 x 2 = 26

2 272 = 26+1 =271

------

Here’s your answer 17 x 16= 272

C) Below Base-100 Step-1 Write down the deficiency with -ve sign .

99 -01 (01 below base 100 i.e.1 deficiency)

x 98 -02 (02 below base 100 i.e.1 deficiency)

----------

Step-2 Multiply the deficiencies and write down the product on one’s and ten’s place.

99 -01

x 98 -02

----------

02

Step-3 Add the number diagonally or cross addition and write the sum on hundred’s

and thousand’s places (Both additions will give same result)

99 -1 [99+(-2)= 97]

x 98 -2 [98+ (-1)=97]

----------

9702

-----------


33

EIn case the product obtained in step 2 is three digit number then write the hundred’s

place number as subscript. . After addition in step 3, add the number written as

subscript to obtain the answer.

eg. 90 x 98

Step-2 90 -10 Step-3 90 -10 i) -10 x(-12) =120

X 98 -12 X 98 -12 ii) 90+(-12) = 78

------------ ------------ [78+1= 79]

20 7920 1

------------ ------------


D) Below Base 200-900

Step-1 Write down the deficiency with -ve sign .

u298 -02 v596 -04 (600-596)

X 295 -05 x 594 -06 (600-594)

--------- -------------

Step-2 Multiply the deficiencies numbers and write the product on one’s & ten’s place

as base is of 100

u298 -02 (-2x-5)=10 v596 -04 (-4 x-6) =24

X 295 -05 x 594 -06 ----------- ------------

10 24 ------------ -----------

Step-3 Add the numbers diagonally or cross addition. Multiply the sum with base

highest place digits 2 for 200, 3 for 300 etc and write the answer on hundred

thousand and ten thousand place.

u 298 -02 [298+ (-5) =293] v596 -04 [596+ (-6)=590]

X 295 -05 (293 x 3) Base of 300 X 594 -06 (590 x 6)B ase of 600 -------------- --------------

87910 354024 -------------- --------------

Here’s your answer 298 X 295 = 87910

596 X 594 = 354024

EIn Case in step 2 if the product is 3 digit number then write the hundred’s place

digit as subscript and after performing step-3 add the number. Now write it on

hundreds, thousand and & ten thousands place.

eg. 290 x 289 Step-2 290 -10 Step-3 290 -10 (I) [( -10) x (-11) = 110)]

x289 -11 x289 -11 (ii) 289+(-10) = 279 -------- ---------- (iii) 279x3 = 837

1 0 83810 (iv) 837 + 1 = 838 1

-------- ----------Here’s your answer 290 x 289 = 83810

34

E) Above Base 10 Step 1 Write down the surplus with + ve sign aftereg. : 11 + 1 (1 above base 10 i.e. 1 surplus)

x12 + 2 (2 above base 10 i.e. 2 surplus) ------

Step-2 Multiply the surplus numbers and write the product on one’s place. 11 + 1 (1 x 2)

x12 + 2 -------- 2

Step-3 Add the numbers diagonally or cross addition and write the answer on tens and hundreds place.

11 + 1 11+ 2 = 13x12 + 2 12 +1 = 13

-----------132 Here’s your answer 11 x 12 = 132

EIn case number obtained in step 2 is two digit number then write the ten’s place digit

as subscript. After performing step 3 add the subscript digit and write the answer

on tens and hundreds place. eg. : 14 x 14

Step-2 14 + 4 Step-3 14 + 4 (I) 4x4 = 16

14 + 4 x14 + 4 (ii) 14+4 = 18 ----------- ------------ iii) 18+1 = 19

6 1961

------------Here’s your answer 14 x 14 = 196

F) Above Base 20-90

Step-1 Write down the surplus with +ve sign.

eg: u21 +1 (I above base 20 i.e. 1 surplus ) v 42 + 2 x 23 +3 (3 above base 20 i.e. 3 surplus ) x 42 + 2 --------- --------Step-2 Multiply the surplus numbers and write down the product at one’s place.

eg: u 21 +1 (1X 3=3 ) v 42 + 2 (2X2 =4) x 23 +3 x 42 + 2 --------- ---------

3 4 --------- ---------

Step-3 Add the number diagonally or cross addition, then multiply the sum by bases(2, 3, 4 etc) and write the answer at tens and hundreds place.

u 21 +1 (23+1=24) v 42 + 2 (42+2=44) x 23 +3 (24x2=48) X 42 + 2 (44x4=176) --------- ---------- 483 1764 -------- ----------Here’s your answer (a) 21 x 23 = 483 (b) 42 x 44 = 1764

35

EIn case the product obtained in step 2 is two digit number then write the ten’s place

number as subscript. After Step 3 Add the subscript digit to answer and write then

on ten’s & hundreds place.

eg. 25 x 24 25 +5 (i) 5x4 = 20

x 24 +4 (ii) 25 +4 = 29 -------- (iii) 29 x 2 = 58

600 (iv) 58 +2 = 60 —------ Here’s your answer 25 x 24 = 600

G) Above Base 100

Step-1 Write down the surplus [in two digit] with +ve sign.eg: 101 +01 (I above base 100 i.e. 1 surplus )

x103 +03 (3 above base 100 i.e. 3 surplus ) ----------

Step-2 Multiply the surplus numbers and write down the product at one’s and ten’s place as base is of 100.

101 +01 (01x03= 03) x 103 +03 ----------

03

Step-3 Add the numbers diagonally cross addition and write down answer at hundred, thousand and tens thousand place.

101 +01 (101+3 = 104) x103 +03 ----------- 10403


EIn case the product obtained in step 2 is three digit number then write the hundred place digit as subscript and after obtaining answer in step 3 add that subscript digit. eg. 110 x 111 110 + 10 (I) 10 x11 = 110

x111 + 11 (ii) 110+11 = 121

----------- (iii) 121+1 = 122

12210


36

H) Above Base 200-900Step-1 Write down the surplus with +ve sign.

eg. u 205 +05 (205-200=05) v 507 + 07 (507-500=07)

x206 +06 (206-200=06) x508 + 08 (508-500=08)

Step-2 Multiply the surplus numbers and write the product on ones and tens place.

u 205 + 05 (5x6 =30) v 507 + 07 (7x8 =56)

x 206 + 06 x 508 + 08

-------- --------

30 56

Step-3 Add the number diagonally or cross addition and multiply the sum with respective

bases (2,3,4,.....9)

u 205 +05 (i) 205 +6= 211 v507 +07 (i) (507+8=515)

x 206 +06 (ii) 211 x2= 422 x 508 +08 (ii) (515x5=2575)

------------ ------------

42230 257556

------------ ------------

Here’s your answer 205 x 206 = 42230 507 x 508 = 257556

EIn case the answer obtained in step-2 is 3 digit number then add hundred’s place

digit to answer obtained in step-3.

eg. 310 +10 (i) 10 x 12 =120

x 312 +12 (ii) 310 + 12 = 322

----------- (iii) 322 x 3 = 966

96720 (iv) 966 + 1 = 967

-----------

Here is your answer 310 x 312 = 96720

EIn case of Base above or below 1000 same procedure will follow except after

multiplication in step 2 answer will be written at three places since in thousand

three zeros are there. Similarly for bases above or below 10,000 answer of

multiplication will be written at four places.

eg. : 999 -001 i) (-001) x (-002) = 002

x 998 -002 ii) 999 + (-002) = 997

-----------

997002

-------------

1005 + 005 i) 005 x 102 = 510

x 1102 + 102 ii) 1005+102 = 1107

--------------

1107510

---------------

37

I) Base method when one number is above & other is below the same base

Step-1 Write the number with the + ve sign which is surplus and write the number with -ve sign which is deficit.

eg. : 105 +05 (05 above base 100 then 100 i.e. 5 surplus )x95 -05 ((05 below base 100 then 100 i.e. 5 deficiency) )

--------

Step-2 Multiply the surplus and deficit. Since the result of multiplication is -ve number. Write the product using vinculum

eg.: 105 +05 [(+5) x (-5) = 25)]x95 -05

-------- 25

as base is 100 so we write product at two places.

Step-3 Add the number diagonally or cross addition and write the answer at ones & tens places.

eg.: 105 +05 (105+(-05) = 100)x95 -05 (95 + 05 =100)

-------- 10025 --------

Step-3 Devinculate the number 10025 (by using all from 9 and last from 10) = 9975


Example : 103 x 98

Step 1 : Here, Base is 100

103 + 03 (03 above base i.e. 3 surplus) (i) 03 x -02 = 06

97 - 02 (02 below base i.e. 2 deficiency) (ii) 103 + (-02) = 101

-------------

10106

-------------

Now Devinculate 10106 = 10094

38

J) Numbers near different base (When both the number are above base)

eg 1 : 104 x 11

Step 1 Vertical Multiplication (multiplication of surplus)

for 104, base is 100 & surplus is 4

for 11, base is 10 & surplus is 1

The digits in the lower base is one, so digits permitted in second part of answer [i.e.

product of surplus] is only one.

104 + 04

x 11 + 1

---------

4

Step 2 : Do cross - addition [only of higher base number and surplus of lower base

number].

104

+10 (one zero is attached with surplus as number of digits in higher base is one more)

114 (again take care to line the numbers properly, so as to get 114 i.e. First

part of answer)

104 + 04

X11 + 1

-----------

1444

Checking :

104 = 1 + 4 = 5, 11 = 1+1 = 2

LHS = 5 x 2 = 10 = 1 + 0 = 1

RHS = 1144 = 1+1+4+4 = 10 = 1+0 = 1

: LHS = RHS, so answer is correct

eg 2 10006 x 1002

Now doing the question directly

10006 + 0006 = (base 10000)

x 1002 + 002 = (base 1000)

012 (Three digits are permitted as per our method for second part)

For first part of the answer we will do cross addition

10006 +0020

---------- 10026 So the required Answer to question 10006x 1002=10026012

--------

Checking = 10006 = 1+ 6 = 7, 1002 = 1+2 = 3

LHS = 7 x 3 = 21 = 2 + 1 = 3

RHS = 10026012 = 1+ 2 + 6 + 1 + 2 = 12 = 1+ 2 = 3

AS LHS = RHS so answer is correct

39

k) Numbers near different base (Both Numbers below base)

eg. 97 x 9

97 - Here base is 100 and deficiency is 03

9 - Base is 10 and deficiency is 1

97 - 03

9 - 1

Step 1 : Multiply the deficiencies and write the answer in ones place as the number of

digits permitted after multiplication is equal to number of digits in the lower

base.

97 - 03

9 - 1

3

Step 2 : Cross subtraction from higher base number

97

- 10

---------

87

The subtraction is done after placing the number of extra zeros which are in higher base

deficiency starting from ones place, while writing subtrahend. Here in this case 1 becomes

10 after writing extra zero

so the answer to question 97 x 9= 873

eg. 998 x 98



998 - 002

x98 - 02

-----------------

Step1: Multiply the deficiencies and write the answer on ones and tens place as number

of digits will be equal to the lower base digits

998 - 002

x 98 - 02

-------------

04

Step 2: Cross subtraction will be done from higher base number after placing a zeroes

equal to the higher base

998 - 20 = 978

998 - 002

x 98 - 02

---------------

978 / 04 so answer to question 998 x 98 = 97804

----------------

Checking : - (through the check method)

992 = 2, 98 = 8

LHS = 2 x8 = 16 =1+6 = 7

RHS = 97216 = 1+6 = 7

As LHS = RHS, so answer is correct

40

1.2.3.4.5.6.7.8.9.10.11.1213.14.15.16.

29 X 23=28 X 25=27 X 26=28 X 28=29 X 24=28 X 23=27 X 22=28 X 26=29 X 22=27 X 27=28 X 26=

. 26 X 24=28 X 23=27 X 25=25 X 24=26 X 23=

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.

39 X 31=38 X 35=37 X 34=38 X 38=39 X 34=38 X 33=37 X 32=38 X 36=39 X 32=37 X 37=38 X 36=36 X 36=38 X 37=37 X 36=35 X 31=36 X 32=

1.2.3.

5.6.7.8.9.10.11.12.13.14.15.16.


4.

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


Below Base -40

Below Base -10+

Below Base -50

Below Base -20

Below Base -60

Below Base -30

1. 2.3.4.5.6.

8.9.10.11.12.13.14.15.16.


7.

1.2.3.4.5.67.8.9.10.11.12.13.14.15.16.

7 X 6=9 X 5=6 X 5=8 X 8=9 X 4=

. 8 X 3=7 X 2=8 X 6=9 X 2=7 X 7=8 X 6=9 X 4=8 X 3=7 X 5=5 X 4=6 X 3=


Practice Sheet

41

Below Base -100

Below Base -70

Below Base -200

Below Base -80

Below Base -300

Below Base -90

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.



Practice Sheet

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


42

Below Base-700

Below Base -400

Below Base -800

Below Base-500

Below Base -900

Below Base -600

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


12345678910111213141516


12345678910111213141516

597 X 592=599 X 595=597 X 595=597 X 595=596 X 590=598 X 592=595 X 595=591 X 581=597 X 597=598 X 585=597 X 579=598 X 595=597 X 595=598 X 596=

X 596 X 585=587 596=

12345678910111213141516


12345678910111213141516


12345678910111213141516


Practice Sheet

Base Meltthod Muiplication

43

Above Base -40

Above Base -10

Above Base -50

Above Base -20

Above Base -60

Above Base -30

1.2.3.4.5.6.78.9.10.11.12.13.14.15.16.

12 X 13=15 X 14=16 X 15=19 X 16=18 X 18=13 X x13=12 X 19=14 X 12=11 X 11=12 X 14=15 X 18=17 X 17=12 X 16=13 X 16=14 X 16=15 X 19=

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


Practice Sheet


44

Above Base-100

Above Base -70

Above Base-200

Above Base -80

Above Base-300

Above Base -90

1.2.3.45.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12. =13.14.15.16.

82 X 87=88 X 85=81 X 88=88 X 81=84 X 83=87 X 86=84 X 89=81 X 81=83 X 84=87 X 82=89 X 86=86 X 8881 X 84=85 X 83=82 X 89=88 X 88=

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


Practice Sheet


45

Above Base-700

Above Base-400

Above Base -800

Above Base-500

Above Bas -900

Above Base-600

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.


Practice Sheet


46

12 X 9=

11 X 8=

15 X 7=

19 X 9=

16 X 6=

21 X 19=

32 X 28=

51 X 49=

59 X 62=

89 X 91=

108 X 95=

101 X 99=

108 X 92=

102 X 92=

103 X 95=

98 X 109=

201 X 199=

303 X 296=

405 X 399=

298 X 302=

699 X 501=

1005 X 995=

10002 X 9998=

10001 X 9994=

10009 X 9999=

104 X 11=

105 X 12=

108 X 14=

1005 X 13=

1009 X 15=

1008 X 11=

1002 X 102=

1004 X 101=

1006 X 105=

1007 X 102=

10001 X 11=

10002 X 13=

10005 X 12=

10002 X 15=

10006 X 14=

10001 X 1002=

10003 X 1005=

10004 X 1002=

10005 X 1001=

10008 X 1002=

10001 X 103=

10002 X 102=

10009 X 101=

10004 X 108=

10002 X 105=

97 X 9=

98 X 8=

96 X 7=

999 X 9=

998 X 7=

996 X 8=

998 X 98=

999 X 97=

996 X 95=

996 X 92=

9995 X 9=

9996 X 8=

9994 X 7=

9998 X 6=

9999 X 99=

9997 X 96=

9995 X 94=

9992 X 97=

9993 X 98=

9997 X 95=

9999 X 998=

9999 X 999=

9998 X 995=

9996 X 992=

9999 X 992=

WHERE ONE NUMBER IS

ABOVE BASE AND ANOTHER

NUMBER IS BELOW BASE

NUMBERS NEAR DIFFERENT

BASE BOTH THE NUMBERS ARE

ABOVE BASE

NUMBERS NEAR DIFFERENT

BASE BOTH NUMBERS ARE

BELOW BASE

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

Practice Sheet


47

Sub sutra applied is Antyayordaœake'pi

Step-1 Multiply the units digit and write the answer on one’s and ten’s place

eg: u 67 7x3= 21 eg.v 91 1x 9 = 09 x63 x 99

----- ------ 21 09 Here in this case product of 1 & 9

is 9 so we write it as 09

Step-2 Multiply the tens place number with its successor

eg. u 67 6x7= 42 eg.v 91 9x 10 = 90 x63 x 99

-------- ------ 4221 9009 -------- ------

Here are the answer 67 x 63 = 4221 91 x 99 = 9009

Note : In case of triple digit number, when unit place add to 10 and tens and hundreds place digits are same, then tens and hundred’s place number are together multiplied by its successor.eg. : 106 (i) 6 x 4 = 24

x104 (ii) 10 x 11 = 110---------

11024--------

Here,s is the answer 106 x 104 = 110024

Step1. Multiply the units digits and write the answer on one’s and ten’s place.

eg. : u 22 (2x2 = 04) eg. v 77 (7x7 = 49)x82 x37------- -------- 04 49

Step2. Multiply the tens place digits and add units place digit to it and write the answer on hundred and thousand place.

eg. : u 22 (8x2 +2) eg. v 77 7x3+7 = 21x82 16+2 =18 x37 21 +7 = 28------- --------1804 2849------- --------

22 x 82 = 1804 77 x 37 = 2849

If the sum of units digits is 10 and rest place digits are same

If the sum of tens place digits is 10 and ones place digit is same

48

1. 19. 37.2. 20. 38.3. 21. 39.4. 22. 40.5. 23. 41.6. 24. 42.7. 25. 43.8. 26. 44.9. 27. 45.10. 28. 46.11. 29. 47.12. 30. 48.13. 31. 49.14. 32. 50.15. 33.16. 34.17. 35.18. 36.

25 x 85 = 52 x 52 = 73 x 33 =19 x 99 = 63 x 43 = 34 x 74 =82 x 22 = 51 x 51 = 53 x 53 =73 x 33 = 44 x 64 = 64 x 44 =56 x 56 = 32 x 72 = 55 x 55 =77 x 37 = 21 x 81 = 45 x 65 =84 x 24 = 11 x 91 = 22 x 82 =35 x 75 = 34 x 74 = 12 x 92 = 45 x 65 = 43 x 63 = 36 x 76 =81 x 21 = 32 x 72 = 42 x 62 =86 x 26 = 24 x 84 = 47 x 67 =89 x 29 = 25 x 85 = 35 x 75 =41 x 61 = 83 x 23 = 15 x 95 =66 x 46 = 41 x 61 = 87 x 27 =55 x 55 = 57 x 57 =62 x 42 = 68 x 48 =76 x 36 = 78 x 38 =71 x 31 = 62 x 42 =

1. 18. 35.2. 19. 36.3. 20. 37.4. 21. 38.5. 22. 39.6. 23. 40.7. 24. 41.8. 25. 42.9. 26. 43.10. 27. 44.11. 28. 45.12. 29. 46.13. 30. 47.14. 31. 48.15. 32. 49.16. 33. 50.17. 34.

67 x 63 = 62 x 68 = 63 x 67 =91 x 99 = 34 x 36 = 92 x 98 =28 x 22 = 77 x 73 = 207 x 203 =43 x 47 = 93 x 97 = 134 x 136 =65 x 65 = 21 x 29 = 124 x 126 =48 x 42 = 53 x 57 = 104 x 106 =88 x 82 = 75 x 75 = 112 x 118 =66 x 64 = 88 x 82 = 155 x 155 =39 x 31 = 68 x 62 = 124 x 126 =27 x 23 = 76 x 74 = 139 x 131 =45 x 45 = 14 x 16 = 117 x 113 =81 x 89 = 42 x 48 = 125 x 125=44 x 46 = 39 x 31 = 102 x 108 =52 x 58 = 51 x 59 = 111 x 119 =29 x 21 = 65 x 65 = 125 x 125 =81 x 89 = 79 x 71 = 116 x 114 =44 x 46 = 84 x 86 =

Practice Sheet

If the sum of unit digit is 10 and rest place digits are same

If the sum of ten� s place digit is 10 and one� s place digit is same

49

Step-1 Multiply the number by 10

eg. : 53 53x10 = 530x9

----- Step-2 Subtract the multiplicand from the answer obtained in step-1

53 530-53 = 477x9

-----477


Step-1 : Multiply the multiplicand by 20,30,..... as the case may be.

eg. : u 53 1060 ( 53 x 20) v 53 2650 ( 53x 50) x19 x49 ------- -------

Step-2 :Subtract the multiplicand from the answer obtained in step-1

eg. : u 53 (1060 - 53) v 53 (2650 - 53)x19 x49------- -------1007 2597------ ------

53 x 19 = 1007 53 x 49 = 2597

Multiplication by 9

Multiplication of number ending with 9 i.e. 19-99

50

1. 37. 73. =2. 38. 74.3. 39. 75.4. 40. 76.5. 41. 77.6. 42. 78.7. 43. 79.8. 44. 80.9. 45. 81.10. 46. 82.11. 47. 83.12. 48. 84.13. 49. 85.14. 50. 86.15. 51. 87.16. 52. 88.17. 53. 89.18. 54. 90.19. 55. 91.20. 56. 92.21. 57. 93.22. 58. 94.23. 59. 95.24. 60. 96.25. 61. 97.26. 62. 98.27. 63. 99.28. 64. 100

29. 65.30. 66.31. 67.32. 68.33. 69.34. 70.35. 71.36. 72.

52 X 9 = 49 X 19 = 357 X 29 15 X 9 = 95 X 29 = 25 X 39 =54 X 9 = 39 X 39 = 82 X 49 =45 X 9 = 71 X 49 = 18 X 59 =28 X 9 = 92 X 59 = 78 X 69 =77 X 9 = 86 X 69 = 17 X 79 =78 X 9 = 89 X 79 = 73 X 89 =165 X 9 = 36 X 89 = 83 X 99 =177 X 9 = 48 X 99 = 57 X 9 =58 X 9 = 49 X 19 = 85 X 19 =62 X 9 = 32 X 19 = 22 X 29 =99 X 9 = 12 X 29 = 42 X 39 =72 X 9 = 121 X 19 = 35 X 49 =218 X 9 = 101 X 39 = 87 X 59 =146 X 9 = 52 X 9 = 49 X 69 =157 X 19 = 15 X 19 = 95 X 79 =47 X 29 = 54 X 29 = 39 X 89 =69 X 39 = 45 X 39 = 71 X 99 =21 X 9 = 28 X 49 = 92 X 9 =46 X 9 = 77 X 59 = 86 X 19 =226 X49 = 78 X 69 = 89 X 29 =37 X 59 = 165 X 79 = 36 X 39 =357 X 69 = 177 X 89 = 48 X 49 =25 X 79 = 58 X 99 = 49 X 59 =82 X 89 = 62 X 89 = 32 X 69 =18 X 99 = 99 X 79 = 12 X 79 =78 X 19 = 72 X 69 = 121 X 89 =17 X 29 = 218 X 59 = 101 X 99 =73 X 39 = 146 X 49 =83 X 49 = 157 X 39 =57 X 59 = 47 X 29 =85 X 69 = 69 X 19 =22 X 79 = 21 X 9 =42 X 89 = 46 X 99 =35 X 99 = 226 X 9 =87 X 19 = 37 X 19 =

Practice Sheet

Multiplication by 9 and Numbers Ending with 9

51

Sutra applied is Ûrdhva Tiryagbhyâ?

Step 1: Divide the answer block in three parts (Blocks are always 1 less than total no. of digits in a question)

Step 2: Multiply the two digits in right column and write the answer in last block i.e. units place digit with units place digit.

Step 3: Cross multiply the unit place digit with the tens place digits to obtain two products. Add the two products and write the answer in the middle block.

Step 4: Again vertically multiply the digits in left column i.e. tens place digit with tens place digit. and write the answer in left block.

eg.: 1 1 x 1 4 ------------- 1 5 4 --------------

= 154 1) In this first we multiply 4 x1=4 and write the answer in the last block.

2) We multiply cross wise 4 x1=4, 1x1=1 and add 4 and 1 and write answer 5 in middle block.

3) Multiply vertically the tens place digits 1x1=1 and write in the first block.

4) If each block has only one digit then these digits make the answer.

Eg. 72 x 43

7 2

x 4 3

28 21 06

+ 2 +8

30 29

+0

29 Ans. : 3096

Same steps as explained in above example will be followed. If number of digit in each block is more than one, then the higher place digits will be carried and added in next block.

General Method (2 digit x 2 digit)

52

Beejank : The Sum of the digits of a number is called Beejank. If the addition is a two digit number, then these two digits are also to be added up to get a single digit.

e.g: Beejank of 24 is 2 + 4 = 6

Beejank of 648 is 6 + 4 + 8 = 18, further 1 + 8 = 9

Easy way of finding Beejank :

Beejank is unaffected, if 9 is added to or subtracted from the number. This nature of 9 helps in finding, Beejank very quickly, by cancelling 9 or the digits adding to 9 from the number.

e.g. : Find the Beejank of 167432

As above we have to follow

167432 1+6+7+4+3+2 23 2 + 3 5

But a quick look gives 6 & 3 ; 2&7 are to be ignored. [This method is called casting out 9s] because 6+3=9, 2+7=9.

Hence remaining 1 + 4 --------- 5 is the beejank of 167432.

eg. : 24 x 32

24

x32

6 4 8

+1 +12

7 16

=768

Cross Checking

24 x 32 = 768

(2+4) x (3+2) = 7+6+8

6 x 5 = 21

30 = 2+1

3+0 = 3

3 = 3

LHS= RHS Hence the product is correct.

This digit cross check method can be used in any kind of calculation to cross check your answer. This is also explained earlier.

53

1. 21. 41.

2. 22. 42.

3. 23. 43.

4. 24. 44.

5. 25. 45.

6. 26. 46.

7. 27. 47.

8. 28. 48.

9. 29. 49.

10. 30. 50.

11. 31.

12. 32.

13. 33.

14. 34.

15. 35.

16. 36.

17. 37.

18. 38.

19. 39.

20. 40.

54 X 27 = 26 X 34 = 98 X 59 =

27 X 58 = 37 X 19 = 86 X 64 =

58 X 89 = 35 X 29 = 89 X 87 =

48 X 34 = 25 X 63 = 36 X 39 =

28 X 46 = 82 X 49 = 48 X 42 =

78 X 56 = 18 X 87 = 49 X 57 =

89 X 69 = 78 X 96 = 35 X 63 =

15 X 79 = 17 X 56 = 12 X 78 =

17 X 84 = 73 X 89 = 37 X 89 =

28 X 29 = 83 X 55 = 17 X 59 =

62 X 82 = 57 X 58 =

94 X 73 = 85 X 19 =

72 X 56 = 22 X 96 =

21 X 59 = 42 X 32 =

46 X 48 = 35 X 75 =

57 X 35 = 87 X 58 =

47 X 28 = 49 X 62 =

69 X 19 = 56 X 74 =

21 X 91 = 78 X 82 =

26 X 96 = 91 X 98 =

Practice Sheet

General Method (2 digit x 2 digit)

54

Multiplication (Algebraic Expressions)

I) Binomial x Binomial (2x+3y) (3x+4y)

Step 1: Divide the answer block in three parts (Blocks are always 1 less than total no. of digits in a question)

Step 2: Multiply the two digits in right column and write the answer in last block

Step 3: Cross multiply the first digit column with the second digit of second column, and second digit of the first column with the first digit of the second column. Then add the two products. Write the answer in second block.

Step 4: Again vertically multiply the digits in left column and write the answer in starting block.

Step 5: The answer in each block will br written separately & like term will be added (if any).

eg:- (3x+2y) (4x+5y)

3x + 2y x4x + 5y

--------------------------------------------- 3x X 4x 3x X 5y = 15xy 2y x 5y

2 2 =12x 4x X 2y = 8xy = 10y =23xy

2 2Ans. 12x +23xy+10y

2 2II) Trinomial x Trinomial (2x +2x+5) (3x +2x+1)

I) Step1: In 3 digit * 3 digit multiplication, we make 5 blocks (Here there are 6 digits in all, so 1 less than 6 is 5 blocks).

Step2: Vertically multiply the right hand side digits and write the answer in the first block from right.

Step3: Cross multiply the digits of the two columns from right. Add both the products and write the answer in second block starting from right to left.

55

Step4: Cross multiply the digits in three columns i.e., multiply extreme right digits with extreme left of different blocks and multiply vertically the middle digits. Now add the three products and write the answer in third block.

Step5 : Multiply the digits of two columns from the left. Add the two products and write the answer in the fourth block.

Step6 : Vertically multiply the digits of the column of the left side and write the answer in th5 block.

Step7 : The answers in each block are written separately & like terms are added (if any).

eg.

2 2 (2x +2x+5) (3x +2x+1)

22x +2x+5

23x +2x+1

2 2 2 3 2 2 2x +3x 2x X2x=4x 2x X1= 2x 2x +1x =2x 5x1

2 3 2 2 3x X2x= 6x 3x X 5= 15x 2xX 5 = 10x

2 2x X 2x= 4x

4 3 2 =6x =10x = 21x = 12x =05

4 3 2 =6x +10x +21x +12x+5

2 (7x -3x-2) (3x+2) (for Trinomial x Binomial multiplication put 0 in Binomial to make it trinomial)

27x -3x-2

0+ 3x+2 2 2 2

7x X 3x 7x X 2 = 14x -3x X 2 -2

0X-3x 0 X 2 = 0 3x X 2 x 2

3 2 =21x -3x X 3x = -9x -6x+(-6x) -4 = -12x

2 5x

3 2 Ans. 21x + 5x -12x-4

56

Practice Sheet

Multiplication (Algebraic Expressions)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

(x+5) (x-1)

(3p-2) (9p+7)

2 2(x +1) (x -1)

(5k-4) (4k+6)

(9m-4n) (m+n)

(2ab-3c) (7ab+c)

(2x-3y) (3x-9y)

(3a-7b) (5a+9b)

(9y-4z) (7y+8z)

(7x-2y) (8x+3y)

(x+6) (x+8)

(9p-6) (8p+4)

2 2(x +5) (x -5)

(8x+9) (3x-4)

(6m-7n) (m+n)

(8ab-9c) (4ab+2c)

(5x-7y) (3x-4y)

(2a-9b) (6a+7b)

(y-7z) (3y+4z)

(8x-3y) (4x+y)

2 2(3x +x+5) (x +2x-1)

(3p+6q-2) (9p-4q+7)

2 2 2 2(x +y -1) (4x -3y +7)

(7m-3n-4) (2m+2n+8)

(12a+3b-1) (2a-2b+4)

4 2 4 2(2x +3x -7) (5x -5x -7)

2 2(2x -3x+5) (8x -9x+8)

(3a-7b+7) (5a+9b-7)

(2x+9y-4z) (-5x+7y+8z)

(8x-5y+3) (5x+3y-2)

2 2(x +3x-5) (6x -9x-1)

(p-3q+2) (6p-7q-37)

2 2(x -7x+1) (x -4)

(7m+-4n-4) (6n+6)

(9a-4b+3c) (7b+3c)

2 2 2(7a -3b +3) (6b -4)

(2x-35y) (3x-2y)

2 2(7a -7b) (6a +9b)

(3x-5y-4z) (8x-9y+2z)

(2x-3y) (5x+y)

57

Chapter-5 Squares and Squares Roots

Square of number ending with 5

For square of number ending with 5, same procedure discussed earlier, where the sum

of the ones place digits is 10 and tens place digits are same is followed

2 25 = 625

Ist Multiply 5 with 5 to get 25 (second half)

2nd Multiply 2 by 3 ( successor) to get 6 (first half)

Square of number starting with 5

For square of number starting with 5, same procedure as discussed earlier where the

sum of tens place digits is 10 and ones place digits are same is followed

252 = 2704

Step 1 Multiply 2 by 2 and write the answer as 04 [ 2 x2 = 04] (second half)

Step-2 Multiply 5 by 5 and add the common number 2 [5 x 5 +2 =27] (first half)

Squaring Base Method

Above Base (10-90)

I) If we are to find square of 11. Determine the nearest base for 11 (i. e. 10) and find its

difference from the base i. e. +1.The answer comes in two parts. For first part, add the

difference from base to the given number (11+1=12). For second part, square the

difference and put answer 1.

2 211 = 11+1/1 = 12/1=121

2 214 =14+4/4 = 18/16= 19/6=196

In the above example, the square of 4, (16) is in double digit, so put 1 as subscript and

add it to 18 and put 6 in second part.

II) For finding squares where base is between 20 to 90, the whole procedure is same.

Lastly, we will multiply the respective base rank with the answer of first part.(2 for

20, 3 for 30 …..9 for 90)

2 2e. g. 63 =6(63+3)/3 = 66x6/9=396/9=3969

+

58

Above Base (100-900)

I) If we are to find square of 101. Same procedure for above base is applied. Only the

placement in second part changes, answer 01 is written, as the base has two zeroes.

2 2101 = 101+1/1 = 102/01=10201

II) In case the base is between 200 to 900, the whole procedure remains same. Just

multiply the respective base rank with the answer of first part.(2 for 200, 3 for 300

…..9 for 900)

2 2e. g. 603 =6(603+3)/3 = 606x6/09=3636/09=363609

Below Base (10-90)

I) If we are to find square of 9. Same method as for above base is applied, the only

difference is the deficiency of base is now subtracted from the number and answer is

written in the first part. Now square the deficiency 9-10= -1. For second part, square

the difference from of number and the base. i. e. -1

2 29 = 9-1/(-1) = 81

II) In case where base is between 20 to 90, the whole procedure is same. Lastly, we

will multiply the respective base rank with the answer of first part. (2 for 20, 3 for

30 …..9 for 90)

2 2e. g. 48 =5(48-2)/(-2) = 230/4= 2304 (Here base will be 50)

Below Base (100-900)

I) If we are to find square of 98. Same procedure discussed above is applied. Now the

placement in second part changes, as the base has two zeroes.

2 298 = 98-2/(-2) = 96/04=9604

II) In case the base is between 200 to 900, the whole procedure remains same. Just

multiply the respective base rank with the answer of first part.(2 for 200, 3 for 300

…..9 for 900)

2 2e. g. 399 =4(399-1)/(-1) = 398x4/01= 1592/01= 159201

59

1. 11.

2. 12.

3. 13.

4. 14.

5. 15.

6. 16.

7. 17.

8. 18.

9. 19.

10. 20.

2 215 145 =

2 2 25 = 155 =

2 275 = 205 =

2 245 = 35 =

2 285 = 165 =

2 265 = 125 =

2 295 = 135 =

2 2105 = 55 =

2 2125 = 215 =

2 2115 = 305 =

=

1. 11.

2. 12.

3. 13.

4. 14.

5. 15.

6. 16.

7. 17.

8. 18.

9. 19.

10. 20.

2 251 502 =

2 255 = 506 =

2 252 = 509 =

2 259 = 508 =

2 257 = 511 =

2 2501 = 510 =

2 2504 = 56 =

2 2505 = 58 =

2 2503 = 512 =

2 2507 = 504 =

=

Practice Sheet

Square of number starting with 5

Square of number ending with 5

60

General method of squaring

The Duplex

Sub Sutra: “ ”. We will use the term Duplex, ‘D' as

follows:2

For Single digit, its duplex will be its square, eg. D (4) = 4 = 16

For Double digit, its duplex will be twice of their product. For eg. D (35) = 2X3X5 = 302

For Three digit, D (341), duplex will be : 4 + 2X3X1 = 16 + 30 = 46

Question: Find the square of 123

For finding the square, we will find the duplex: 2

D(3) = 3 = 9

D(23) = 2 x 2 x 3 = 122D(123) = 2 + 2 x 1 x 3 = 10

D(12) = 2 x 1 x 2 = 42

D(1) = 1 = 1

So the answer becomes 15129

Algebraic Squaring:

Above method is applicable for squaring algebraic expressions also:-2

Example: (x+5)2 D(5) = 5 = 25

D(x, 5) = 2 X x X 5 = 10x2

D(x) = x2

So square of x+5 is x + 10x +252Example: (x – 3y)

2 2D(-3y) = (-3y) = 9y

D(x, -3y) = 2 X x X -3y = -6xy2

D(x) = x2 2So square of x-3y is x - 6xy +9y

BASE-10 BASE-60 BASE-200 BASE-7002 2 2 213 = 63 = 204 = 701 =2 2 2 2 17 = 67 = 212 = 753 =

BASE-20 BASE-70 BASE-300 BASE-8002 2 2 223 = 73 = 303 = 824 =2 2 2 2

25 = 75 = 313 = 803 =

BASE-30 BASE-80 BASE-400 BASE-9002 2 2 235 = 81 = 411 = 946 =2 2 2

2 39 = 84 = 462 = 986 =

BASE-40 BASE-90 BASE-5002 2 2

42 = 91 = 514 =2 2 2

48 = 97 = 552 =

BASE-50 BASE-100 BASE-6002 2 253 = 101 = 663 =

2 2 2 52 = 109 = 672 =

1. 1. 1. 1.2. 2. 2. 2.

1. 1. 1. 1.2. 2. 2. 2.

1. 1. 1. 1.2. 2. 2. 2.

1. 1. 1.2. 2. 2.

1. 1. 1. 2. 2. 2.

BASE-10 BASE-60 BASE-200 BASE-7002 2 2 2

9 = 58 = 179 = 698 =2 2 2 2

7 = 56 = 191 = 668 =

BASE-20 BASE-70 BASE-300 BASE-8002 2 2 218 = 69 = 222 = 779 =2 2 2 219 = 62 = 297 = 795 =

BASE-80BASE-30 BASE-400 BASE-90022 2 2 76 =29 = 359 = 875 =22 2 279 =28 = 399 = 892 =

BASE-90BASE-40 BASE-50022 2 85 =39 = 445 =22 289 =35 = 489

BASE-100BASE-50 BASE-60022 294 =47 = 545 =22 298 =46 = 598 =

1. 1. 1.2. 2. 2. 2.

1. 1. 1. 1.2. 2. 2. 2.

1.1. 1. 2.2. 2. 2.

1.1. 1. 2.2. 2.

1. 1. 1. 2.2. 2.

1.

1.

Practice Sheet

Square Above Base

Square Below base

61

62

Practice Sheet

Squares (Using duplex Method)

1. 21. 41.

2. 22. 42.

3. 23. 43.

4. 24. 44.

5. 25. 45.

6. 26. 46.

7. 27. 47.

8. 28. 48.

9. 29. 49.

10. 30.

50.11. 31.

12. 32.

13. 33.

14. 34.

15. 35.

16. 36.

17. 37.

18. 38.

19. 39.

20. 40.

2 2 2 21 = 125 = (x+1)

2 2 243 = 136 = (2a+b)

2 2 272 = 231 = (3x+4y)

2 2 226 = 214 = (8c+4d)

2 2232 = 511 = (3x+2y)

2 2243 = 116 = (7a+3b)

2 251 = 215 = 2(9c+4d)2 262 = 172 =

2(m+3)2 234 = 139 =

2(2m+n)2 2

23 = 142 =2(5z+8k)2 2

61 = 236 =

2 227 = 189 =

2 253 = 215 =

2 283 = 312 =

2 224 = 415 =

2 284 = 251 =

2 291 = 932 =

2 241 = 192 =

2 254 = 163 =

2 275 = 342 =

63

Square Roots :

General Method:2 2 2 2 2 2 2 2 2As 1 = 1, 2 = 4, 3 = 9, 4 = 1(6), 5 = 2(5), 6 = 3(6), 7 = 4(9), 8 = 6(4), 9 =

8(1), i. e., square numbers only have digits 1, 4, 5, 6, 9, 0 at the units place

(or at the end)

Also in 16, digit sum = 1+6 = 7, 25 = 2+5 = 7, 36 = 3+6 = 9, 49 = 4+9 = 13 = 4,

64 = 6+4 = 10 = 1, 81 = 8+1 = 9, i. e. square number only have digit sums of

1, 4, 7 & 9.

This means that square numbers cannot have certain digit sums and they cannot end

with certain figures (or digits). Using above information state which of the following

are not square numbers:

(1) 4539 (2) 6889 (3) 104976 (4) 27478 (5) 12345

Note: If a number has a valid digit sum and a valid last figure that does not mean that

it is a square number. If 75379 is not a perfect square inspite of the fact that its digit

sum is 4 and last figure is 9.

Square Root of Perfect Squares:

Example 1 : 5184

Step 1: Pair the numbers in two, from right to left i. e. 51 84

Therefore, answer will come in 2 digits

Step 2: For the first digit, look where 51 lies in squares table, and consider

the lower number. Here, 51 lies in between 49 and 64. So, we will

consider 7 at tens place, as 49 is square of 7.2 2

Step 3: Look at the last digit of number, 4. 4 lies in 2 = 04 and 8 = 64. So, the

ones place digit is either 2 or 8

We can say the answer is either 72 or 78.

Step 4: Now find the square of 75 and compare with the number (given in

the question). If the number is lesser, then consider the lower value

and if number is more than the square, then consider the higher value.

Here 5184<5625, therefore we will consider the lower value that is 72

Example 2 : 9216

Step 1: Pair the numbers in two, from right to left i. e. 92 16

Therefore, answer will come in 2 digits2 2

Step 2: The first digit is 9, as 92 lies between 9 and 10 , and we will take

lower value, 9 as 81 is square of 9.2 2

Step 3: The last digit of number, 6. 6 lies in 4 = 16 and 6 = 36. So, the ones

place digit is either 4 or 6

The answer is either 94 or 96.

Step 4: Now, find the square of 95 and compare with the number

(given in the question).

Here 9216>9025, therefore we will consider the higher value

that is 96

64

Practice Sheet

Find the Square Roots of following

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

676

1521

576

1444

361

1369

6724

4761

1225

7396

1156

8649

1764

1024

2209

2809

4356

3136

4489

6084

7056

3025

2916

2025

625

1849

2704

9801

2401

1681

1296

961

3721

4489

9409

9604

2116

1936

784

441

65

For Division by 9 we need to know the table of 9 but in vedic mathematics we can easily calculate Quotient and Remainder with help of just addition Write first place number as it is then add the number diagonally to the second number and repeat this addition till the unit place of dividend you will get the desired answer.

If sum is 2311÷ 9 it will be written as 9 231 1 (leaving unit place separately to show the remainder.

a) The initial 2 is brought straight down as the first digit of quotient

b) This 2 is then added to the 3 in 2311 and 5 is next quotient digit.

c) This 5 is then added to the 1 in the 2311 and 6 is next quotient digit

d) This 6 is then added to 1 to give the remainder, 7Here the answer 2311÷ 9 = 256 (Q) & 7

Type-II

(I) If the digit of dividend add upto two digit number then tens place number is carried and added to the next number

eg. : 9 2 8 4 31

2 0

If we add 2 + 8 then result obtained is 10 then we will write tens place no. of 10 i.e.1 below as a subscript as shown above and second digit i.e. zero just along with 2.

(ii) Now add 10 to the next number i.e. 4. 10+4 = 14. Write 1 as subscript and four in answer row.

(iii) Now add 14 to next number 3. Write 17 after the slash as 3 is the last number of dividend.

9 2 8 4 3

2 0 4 17

(iv) Add the carry forward in the quotient

9 2 8 4 3

2 0 4 173 1 4 ( 2+1=3) (0+1 =1)

9 2 3 1 1 2

9 2 3 1 1 2 5

9 2 3 1 1 2 5 6

9 2 3 1 1 2 5 6 7

a)

b)

c)

d)

1

1 1

1 1

Chapter-6 Division Division by 9

66

(v) Since remainder cannot be more than the divisior. So remainder 17 will be divided with help of line and 1 will remain as it is on the left and 1 is added in right side to form 8 (1+7)

9 2 8 4 3

3 1 4 1/1+7

First part i.e. 1 will be added in the Quotient and 8 will be the remainder.

9 2 8 4 3

3 1 4 1/8 + 1

3 1 5 8Here’s your answer 2843 ÷ 315 (Q) and 8

I) Actually in Division of 9 we use base method i.e.(Sutra Nikhila? Navataœcarama? Daœata?)Type-1 if a number is to be divided by 9 then since it is 1 less than base 10 we will take 1 under 9 to act as multiplier

eg. : 9 1 0 1 1 2 1ii) Since it is near base 10 i.e. having one zero we will leave unit place of dividend and

put a line Whatever is obtained on left hand side will be quotient and whatever obtained on right hand side will be remainder.

9 1 0 1 1 2 Quotient Remainder

iii) Write initial digit of dividend down as first digit of quotient. Then multiply this digit by 1 and write the answer under second digits.

9 1 0 1 1 2

1 1

1iv) Add second digit numbers and write the answer.

9 1 0 1 1 2

1 1

1 1 v) Multiply the obtained second digit with 1 and write the answer under third digit then

add third place digits. Now repeat the procedure till the unit place of dividend.9 1 0 1 1 21

1 1 2 3

1 1 2 3 5

Q = 1123 R = 5 Here’s your answer 10112 ÷ 9 = 1123 (Q) and 5

67


If a number is to be divided by 8 then since it is 2 less than base 10 we will write 2 below 8 to act as multiplier

e.g. : 8 1 2 3 12

ii) Since it is near base 10. i.e. having one zero we will leave unit place of dividend and

put a line whatever obtained on left hand side will be the a Quotient and whatever obtained on right hand side of line will be the remainder.

8 1 2 3 12 Quotient Remainder

iii) Put initial digit of dividend i.e. 1 down as first quotient digit multiply this digit by 2. The product obtained i.e. 2 is written under second digit i.e. 2 and added.

8 1 2 3 12

21 4

iv) The answer obtained after adding second digit i.e. 4 again multiplied by 2 and written under third digit. The product obtained i.e. ( 4 x 2= 8) is added to next digit 3 and answer obtained i.e. 11 will be written down as shown.

8 1 2 3 12

81 4 1

v) Multiply 11 by 2 and add to next number i.e. 18 1 2 3 12

8 221 4 1 23

vi) Add carry overs in quotient and divide the remainder by 8 since 23 is more than 8. After converting to mix fraction ( 23 7 ) add whole no. i.e. 2 to quotient

8 8and 7 will be the remainder

8 1 2 3 12

8 221 4 1 23

1 5 1 23 7 8 8

151 +2-----

(Q) 153 7 Here’s the answer 1231 ÷ 8 = 153 (Q) & 7

1

1

1

= 2

= 2

Division by 8

68

If a number is to be divided by 11 then since it is one more than the base 10 then we will write 1 below 11 to act as a multiplier

eg. : 11 3 0 0 2 1

1

ii) Since base is 10 unit place is separated for showing remainder11 3 0 0 2 1

1 Quotient Remainder

iii) Put initial digit i.e. 3 down as first digit of quotient. Multiply 3 by 1 and write the answer i.e. 3 under second digit i.e. 0. After adding the second place digit you will obtain 3 as second quotient digit.

11 3 0 0 2 11 3

3 3

iv) Multiply 3 by 1 you will get 3. Now add this three to next digit i.e. zero. Sum 3 is written as shown below.

11 3 0 0 2 11 3

3 3 3

v) Multiply 3 by 1. You will get 3. Add 3 to next digit 2 you will get 1 as an answer

11 3 0 0 2 11 3

3 3 3 1

vi) Multiply 1 by 1 you will get 1 as an answer add 1 to next digit 1 to get 2 as remainder

11 3 0 0 2 11 1

3 3 3 1 2

vii) Now devinculate the digits obtained in quotient.

11 3 0 0 2 1

1 1

3 3 3 1 2 Devinculate 3 3 3 1= 2729

(Q ) 2729 (R) 2

Here’s the answer 30021÷11= 2729 (Q) & 2

Division by 11

69

Possible cases of remainder in cases of 11, 12 or Numbers above Base 100, 1000 etc.

1) In case of 11the remainder that appear at the end can be either positive or negative.

2) If remainder is positive and less than 11 then it will be shown as remainder for that particular sum.

3) If remainder is positive and more than 11 than it will be converted to the mixed fractions as discussed in earlier cases of 8 and 9 and whole no. will be added in quotient and numerator will act as remainder.

4) If remainder is negative and absolute value is less than 11 then 11 will be added to remainder and whatever positive answer we obtain would be remainder and 1 will be subtracted from the Quotient since we have used divisor 11 once.

5) If remainder is negative and absolute value is more than 11 then add the nearest multiple of 11 to that particular negative number so that answer obtained will be positive and less than 11. This particular answer is our remainder. What ever multiple we used in adding that particular rank of multiple will be subtracted from the quotient eg:- suppose we obtained 18 as our remainder so we will add nearest multiple i.e 22 (2nd rank or 11 x 2) to 18 and obtained 4 as remainder. Now subtract 2 from the quotient as it is the second multiple of 11

6) The same procedure would be followed in other cases of 12 or above 100 base according to the remainder & divisor.

Sutra applied is Nikhila? Navataœcarama? Daœata?If a number is to be divided by 12 same procedure as 11 will be followed except the change of multipler. In case of 12 multiplier would be 2 and rest everything will be similar to 11.

12 3 0 0 2 1

2 6 12 24 44

3 6 2 2 45

3 5 0 2 Add the carryover 6 + 1 = 5

2 + 2 = 0

Devinculate 3 5 0 2 = 2 4 9 8 45 9 12 12

2 4 9 8

+ 3 --------- 2 5 0 1

---------Here is your answer 30021 ÷12 = 2501 (Q) 9

(

(

=3

1 2

(Q)

Division by 12

70


Eg: 1230÷ 99 it will be written as 99 123099

There is no need for table of 99

i) If a number is to be divided by 99 then since it is one less than the base number 100 we will write 01 under 99 to act as multiplier

99 1 2 3 0 101

ii) Since it is near base 100 i.e. having two zeros we will put a line before ten’s place digit to differentiate between quotient and remainder.

99 1 2 3 0 101

Quotient Remainder

iii) Put initial digit i.e. 1 down as first digit of the Quotient now multiply this digit with multiplier 01 and write the answer as two digit number i.e. 01 under next two digit which are 2 and 3.

99 1 2 3 0 101

0 1

1

iv) Add the numbers at second place only i.e. 2 & 0 and write the answer i.e. 2. Now multiply 2 by 01 and write the answer 02 under third and fourth digit i.e. under 3 and 0.

99 1 2 3 0 101

0 1 0 2

1 2

v) Add the numbers at third place i.e. 3+1+0 and write the answer i.e. 4. Now multiply 4 by 01 and write answer under next two digits i.e. under 0 and 1. Now no more digit are left in dividend . Add the digit s till the unit place of dividend.

99 1 2 3 0 101

0 10 2 0 4

(Q) 1 2 4 2 5 (R)

Here’s your answer 12301 ÷ 99 = 124 (Q) and 25 (R)

Note : In case remainder is more than 99 then same process of conversion in mixed fraction is followed. Whole number is added to Quotient and numerator will act as Remainder.

Division by 99

71


Eg. 1242 ÷ 102 it will be written as 102 1242

i) If a number is to be divided by number above base 100 for example by 102 then we will write 02 under 102 to act as multiplier.

e.g. 102 1 2 4 2 102

ii) Put a line before 10’ s place digit to differentiate between Quotient and Remainder.

102 1 2 4 2 102

Quotient Remainder

iii) Put the initial digit i.e. 1 down as first digit of Quotient Now multiply it with multiplier 02 and write the answer 02 under two digit 2 and 4

102 1 2 4 2 102

0 2 1

iv) Add second place digit i.e. 2 & 0 and write 2 as answer. Multiply 2 by multiplier 02 and write the answer 04 under 4 and 2

102 1 2 4 2 102

0 2 0 4

1 2v) Add third place digits i.e. 4 + (2) + 0 and write the answer i.e.2 Multiply 2 by 02 and

write 0 4 under 2 and 1. Since no digits are left in dividend add remaining digits to obtain the answer.

102 1 2 4 2 102

0 2 0 4 0

0 4 1 2 2 2 3

( here 2 + 4 =2 1 + 4 =3)

vi) Since remainder is a negative number and smaller than 102 so add 102 to 23 = 79 will be the remainder and subtract 1 from the quotient i.e. 122-1 =121 121 will be the quotient.

102 1 2 4 2 102

0 2

0 4

0 4

1 2 2 2 3 + 102 - 1 = 79 (R)

----------(Q) 1 2 1

Note :1) If quotient obtained has negative

numbers then devinculate the digits.2) In case remainder is -ve then add the

divisor to -ve remainder to make it positive subtract 1 from the quotient.

3) In case -ve remainder has greater absolute value than divisor then to obtain positive remainder add the multiple of divisor to the remainder.

Division by number above base 100

72


Type-I (20-90)

(I) Let us say 2563 is to be divided by 72. Since 72 is 2 more than base 70, we write 2 under 72, to act as multiplier. Start dividing the number by rank of multiple of 10, i.e.7.

7 2 2 5 6 32

÷7ii) Since the base has one zero put the line before units place to differentiate between

quotient and remainder

7 2 2 5 6 32

÷7 Quotient Remainder

iii) Divide the first digit by first digit of base i.e 2 by 7, since 2 is less than 7, we will take two digits together i.e. 25 ÷7. Quotient obtained is 3 (since 7 x 3 =21), write 3 as first digit of quotient and subtract 21 from 25 and write the difference, 4 to the next digit i.e. 6 as a subscript.

72 2 5 6 32

2 1 ÷7 3

iv) Multiply the quotient i.e. 3 by multiplier 2 and write the product 6 under next digit i.e. 46 and add the two, 46 + 6 and write the answer, 40 below as new dividend.

72 2 5 6 32

6

40 ÷7 3

(v) Now repeat the steps and divide 40 by 7. Put quotient 5 to the right of 3 as next Quotient digit. Now multiply 7x 5. Subtract 35 from 40. The difference obtained 5 will be attached to next digit i.e. 3. Now multiply 5 by 2 and add the answer obtained i.e. 10 + 53

72 2 5 6 3 2 6

40 10 -35

÷7 3 5 4 3

(Q) (R) Here’s your answer 2563 ÷ 72 = 35 (Q) & 43( R)

4 5

Note 1) In case after addition of product

(multiplication of quotient and multiplier) the answer obtained is -ve number, then go back to earlier step and lessen the Quotient by 1, then result will always be positive.

4

4

Division Base Method (Above Base)

73

Type-II (200-900)

eg. : 3 2 4 3 1÷ 512

(I) Let us say 3 2 4 3 1÷ 512. Since 512 is 12 more than base 500, we will write 12 under 512 and divide the number by 5, since base is 500. Put a line before tens place digit to differentiate between quotient and remainder

512 3 2 4 3 11 2

÷5

ii) Divide first digit by 5, since first digit is less than 5, we will divide 32 by 5. The first digit of the quotient is 6. Now subtract 30 from 32 and attach the answer, 2 to next digit 4. 512 3 2 4 3 1

1 2 3 0

÷5 6

iii) Multiply the quotient with multiplier, 6 x 12 = 72. Put 72 under next two digits i.e. 4 and 3 but add only first place digit i.e. 24 and 7. Write the answer 17 below as new dividend. 512 3 2 4 3 1

1 2 7 2

17

÷5 6

iv) Repeat the steps of division. Quotient, 3 will be placed to the right of 6. Multiply 3 and 12. Subtract 15 from 17 attach 2 to 31. Now, multiply 3 and 12. Write 36 under 231 ( 3 & 6 under digit is 23 and 1 resp. Add the digits to get the remainder.

512 3 2 4 3 11 2

7 2 17 3 6

15

÷5 6 3 225 Here’s your answer 2431 ÷512

= 63 (Q) and 175 ( R)

Note In case result after addition is negative number then go back to earlier step and lessen the quotient by 1 and repeat till you get the positive remainder.

2

2

2 2 R =225 i. (10-5 =5) = 175 ii. (9-2 = 7)

iii. (2-1 = 1)

74

Type -1 (20-90)

(I) Let us say 3861 ÷ 49, as 49 is 1 less than base 50, we write 1 under 50, to act as a

multiplier and start dividing the number by rank of multiple of that base, here division is to be performed by 5. Since base has one zero, put line before units place.

49 3 8 6 1 1

÷5 Quotient Remainder

ii) Since first digit i.e. 3 is smaller than 5, we take two digits i.e. 38 and divide by 5 to obtain the Quotient 7. Write 7 as first digit of the quotient and subtract 35 from 38. Now write answer, 3 to the left of digit 6 as subscript.

49 3 8 6 1 1

3 5

÷5 7iii) Multiply the quotient 7 by multiplier 1 and add the answer i.e. 7 to next digits which is

36. Write the answer 43 below as new dividend. 49 3 8 6 1

1 7

43

÷5 7iv) Repeat the steps and write 8 as second digit of quotient (5 x 8 = 40) and

subtract 40 from 43. Write the answer, 3 to the left of 1 as subscript. 49 3 8 6 1

1 7 43 -40

÷5 7 8v) Multiply the quotient 8 by multiplier 1 and write it under next digit i.e. 31 add and

obtain the remainder i.e. 39

49 3 8 6 11 7 8

43 -40

÷5 7 8 39 (Q) (R)

Here’s the answer 3861 ÷ 49 = (Q) 78 39

3

3

3

3

3

3

Division Base Method (Below Base)

75

Type-II (200-900)

(I) Another example, 35364 ÷ 894. 894 is 6 less than base 900, we will write 06 under

894, to act as multiplier and divide the number by rank of multiple of base here it is 9. Put a line before 10s place digit to differentiate between Quotient and Remainder.

894 3 5 3 6 406

÷ 9 Quotient Remainder

ii) Divide first digit by 9 since 3<9, we divide 35 by 9 to get the Quotient 3 ( 9 X 3= 27) Subtract 27 from 35. The answer obtained, 8 is attached to next number 3 as shown below.

894 3 5 3 6 406

2 7

÷ 9 3iii) Multiply 3 by 06 and write answer 18 under two digit 3 and 6. Add only Ist place

digit i.e. 83 and 1. 894 3 5 3 6 4

06 1 8 84

÷ 9 3

iv) Repeat the steps of division. Divide 84 by 9 . Put 9 as Quotient, subtract 81 from 84 and attach 3 to next digit i.e 64. Multiply 9 with 06 and add 54 to obtain the remainder.

894 3 5 3 6 406

1 8 0 84

81 5 4

÷ 9 39 498 (Q) (R)

Here’s your answer 35364 ÷ 894 = 39 (Q) and 498

8

8

3

76

1. 15. 29.

2. 16. 30.

3. 17. 31.

4. 18. 32.

5. 19. 33.

6. 20. 34.

7. 21. 35.

8. 22. 36.

9. 23. 37.

10. 24. 38.

11. 25. 39.

12. 26. 40.

13. 27.

14. 28.

24÷8= 235÷8= 2584÷8=

35÷8= 427÷8= 3044÷8=

15÷8= 3002÷8= 24007÷8=

82÷8= 154÷8= 26184÷8=

76÷8= 569÷8= 16018÷8=

32÷8= 290÷8= 16417÷8=

47÷8= 8624÷8= 31786÷8=

39÷8= 7178÷8= 14526÷8=

10÷8= 1233÷8= 24789÷8=

22÷8= 2860÷8= 13798÷8=

146÷8= 9932÷8= 24381÷8=

248÷8= 1481÷8= 26590÷8=

562÷8= 8428÷8=

123÷8= 5511÷8=

Practice Sheet

Division by 9

1. 15. 29.

2. 16. 30.

3. 17. 31.

4. 18 32.

5. 19. 33.

6. 20. 34.

7. 21. 35.

8. 22. 36.

9. 23. 37.

10. 24. 38.

11. 25. 39.

12. 26. 40.

13. 27.

14. 28.

21÷9= 421÷9= 5487÷9=

32÷9= 201÷9= 2567÷9=

43÷9= 121÷9= 23417÷9=

71÷9= 110÷9= 14155÷9=

52÷9= 111÷9= 71076÷9=

23÷9= 244÷9= 32411÷9=

12÷9= 1121÷9= 132761÷9=

42÷9= 7777÷9= 34512÷9=

51÷9= 1827÷9= 44231÷9=

10÷9= 2301÷9= 32248÷9=

122÷9= 8282÷9= 64324÷9=

102÷9= 4568÷9= 88532÷9=

214÷9= 1234÷9=

314÷9= 6542÷9=

Division by 8

77

Practice Sheet

Division by 11

Division by 12

23÷11= 654÷11= 900÷11=

93÷11= 138÷11= 349÷11=

76÷11= 693.÷11= 263÷11=

19÷11= 247÷11= 326÷11=

49÷11= 197÷11= 266÷11=

53÷11= 387÷11= 797÷11=

82÷11= 794÷11= 587÷11=

341÷11= 568÷11= 255÷11=

888÷11= 685÷11= 65÷11=

848÷11= 119÷11= 826÷11=

651÷11= 222÷11= 355÷11=

741÷11= 221÷11= 255÷11=

233÷11= 442÷11=

242÷11= 458÷11=

15.

16.

17.

18.

19.

20

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.38.

39.

40.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

18÷12= 796÷12= 1201÷12=

28÷12= 381÷12= 1001÷12=

39÷12= 324÷12= 2917÷12=

304÷12= 808÷12= 1468÷12=

606÷12= 400÷12= 21234÷12=

326÷12= 672÷12= 3876÷12=

120÷12= 928÷12= 6434÷12=

550÷12= 496÷12= 1121÷12=

345÷12= 213÷12= 2121÷12=

117÷12= 996÷12= 4321÷12=

256÷12= 128÷12= 3452÷12=

514÷12= 101÷12= 1224÷12=

360÷12= 234÷12=

315÷12= 1021÷12=

15.

16.

17.

18.

19.

20

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.38.

39.

40.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

78

Practice Sheet

Division by 99

Division by number above base 100

502÷99= 167÷99= 4182÷99=

121÷99= 4144÷99= 1698÷99=

431÷99= 2009÷99= 1005÷99=

617÷99= 1122÷99= 12312÷99=

111÷99= 1212÷99= 12345÷99=

112÷99= 1111÷99= 10301÷99=

107÷99= 1234÷99= 11210÷99=

234÷99= 2122÷99= 10013÷99=

141÷99= 3001÷99= 30111÷99=

114÷99= 2211÷99= 10010÷99=

221÷99= 4756÷99= 23101÷99=

198÷99= 1582÷99= 13401÷99=

144÷99= 4608÷99=

213÷99= 6580÷99=

15.

16.

17.

18.

19.

20

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.38.

39.

40.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

246÷112= 1659÷103= 9228÷113=

8919÷113= 9228÷113= 13905÷113=

1116÷112= 6012÷106= 15542÷110=

2728÷102= 9268÷112= 10623÷105=

8958÷120= 4734÷121= 25953÷123=

1542÷112= 5114÷113= 31274÷103=

7878÷101= 1296÷113= 14678÷121=

8996÷104= 1234÷112= 15921÷106=

1970÷113= 2688÷102= 16453÷102=

1776÷105= 1241÷112= 98112÷111=

2824÷111= 1486÷112= 24557÷101=

1698÷123= 1389÷113= 312749÷104=

9215÷121= 6012÷106=

3174÷112= 9268÷112=

15.

16.

17.

18.

19.

20

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.38.

39.

40.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

79

Practice Sheet

Division Above Base

Division Below base

386÷72= 4250÷51= 1771÷514=

416÷53= 3768÷84= 4951÷603=

396÷34= 6236÷73= 2229÷611=

543÷81= 2926÷62=

511÷33= 1223÷71= 4132÷402=

732÷51= 6481÷41= 5214÷601=

581÷44= 8136÷58= 3412÷710=

348÷23= 4972÷63= 2563÷415=

711÷62= 5768÷34= 1871÷502=

743÷81= 7812÷63= 4891÷502=

963÷84= 1012÷402= 1084÷201=

245÷41= 1488÷611= 10764÷713=

9101÷62= 1386÷704=

1601÷32= 2482÷221=

3694÷333=

15.

16.

17.

18.

19.

20

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.38.

39.

40.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

521÷68= 654÷87= 3536÷894=

343÷64= 384÷69= 3914÷996=

983÷89= 396÷78= 2558÷197=

769÷89= 395÷79= 2564÷296=

241÷38= 314÷58= 1197÷399=

487÷38= 388÷98= 2664÷296=

845÷39= 2485÷58= 1297÷399=

396÷49= 1341÷87= 3736÷894=

722÷49= 4208÷69= 6128÷895=

468÷49= 7443÷89= 5404÷386=

572÷68= 4784÷57= 5354÷595=

772÷68= 5016÷39= 5355÷595=

921÷57= 3861÷49=

117÷16= 2758÷197=

15.

16.

17.

18.

19.

20

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37

38.

39.

40.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Vedic (SSA)

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