slide 2: FUNDAMENTALS
APPLICATIONS OF
VEDIC MATHEMATICS
2014
State Council of Educational Research Training
Varun Marg Defence Colony New Delhi110024
slide 3: Chief Advisor
Anita Satia
Director SCERT
Guidance
Dr. Pratibha Sharma
Joint Director SCERT
Contributors
Dr. Anil Kumar Teotia Sr. Lecturer DIET Dilshad Garden
Neelam Kapoor Retired PGT Directorate of Education
Chander Kanta Chabria PGT RPVV Tyagraj Nagar Lodhi Road
Rekha Jolly TGT RPVV Vasant Kunj
Dr. Satyavir Singh Principal SNI College Pilana
Editor
Dr. Anil Kumar Teotia
Sr. Lecturer DIET Dilshad Garden
Publication Officer
Ms. Sapna Yadav
Publication Team
Navin Kumar Ms. Radha Jai Baghwan
Published by : State Council of Educational Research Training New Delhi and printed at
Educational Stores S5 Bsr. Road Ind. Area Ghaziabad U.P.
slide 4: Vedic Mathematics introduces the wonderful applications to Arithmetical computations theory of
numbers compound multiplications algebraic operations factorisations simple quadratic and higher
order equations simultaneous quadratic equations partial fractions calculus squaring cubing
square root cube root and coordinate geometry etc.
Uses of Vedic Mathematics:
It helps a person to solve mathematical problems 1015 times faster
It helps m Intelligent Guessing
It reduces burden need to learn tables up to 9 only
It is a magical tool to reduce scratch work and finger counting
It increases concentration.
It helps in reducing silly mistakes
"Vedic Mathematics" is a system of reasoning and mathematical working based on ancient Indian
teachings called Veda. It is fast efficient and easy to learn and use. Vedic mathematics which
simplifies arithmetic and algebraic operations has increasingly found acceptance the world over.
Experts suggest that it could be a handy tool for those who need to solve mathematical problems
faster by the day.
V edic Mathematics provides answer in one line where as conventional method requires several steps.
It is an ancient technique which simplifies multiplication divisibility complex numbers squaring
cubing square and cube roots. Even recurring decimals and auxiliary fractions can be handled by
Vedic Mathematics. Vedic Mathematics forms part of Jyotish Shastra which is one of the six parts
of Vedangas. The Jyotish Shastra or Astronomy is made up of three parts called Skandas. A Skanda
means the big branch of a tree shooting out of the trunk.
The basis of Vedic mathematics are the 16 sutras which attribute a set of qualities to a number or
a group of numbers. The ancient Hindu scientists Rishis of Bharat in 16 Sutras Phrases and 120
words laid down simple steps for solving all mathematical problems in easy to follow 2 or 3 steps.
V edic Mathematicsor one or two line methods can be used effectively for solving divisions reciprocals
factorisation HCF squares and square roots cubes and cube roots algebraic equations multiple
simultaneous equations quadratic equations cubic equations biquadratic equations higher degree
equations differential calculus Partial fractions Integrations Pythogorus theoram Apollonius
Theoram Analytical Conics and so on.
Preface
3
slide 5: 4
How fast your can solve a problem is very important. There is a race against time in all the
competitions. Only those people having fast calculation ability will be able to win the race. Time
saved can be used to solve more problems or used for difficult problems.
This Manual is designed for Mathematics teachers of to understand Vedic System of Mathematics.
The Chapters developed in this Manual will give teachers the depth of understanding of the Vedic
methods for doing basic operations in Arithmetic and Algebra. Some important basic devices like
Digit Sum the Vinculum are also explained along with independent Checking Methods.
All the techniques are explained with examples. Also the relevant Sutras are indicated along with
the problems. In Vedic System a manual approach is preferred. The simplicity of Vedic Mathematics
encourages most calculations to be carried out without the use of paper and pen. The content
developed in this manual will be applicable in the curriculum of VIX classes. Methods like Shudh
Method is applicable in statistics. This mental approach sharpens the mind improves memory and
concentration and also encourages innovation.
Since the Vedic Mathematics approach encourages flexibility the mathematics teachers encourage
their students to device his/her own method and not remain limited to the same rigid approach which
is boring as well as tedious. Once the mind of the student develops an understanding of system of
mental mathematics it begins to work more closely with the numbers and become more creative.
The students understand the numbers better. Vedic Mathematics is very flexible and creative and
appeals to all group of people. It is very easy to understand and practice.
I acknowledge a deep sense of gratitude to all the subject experts for their sincere efforts and expert
advice in developing this manual which lead to qualitative and quantitative improvement in
mathematics education and may this subject an interesting joyful and effective.
Suggestions for further improvements are welcome so that in future this manual become more useful.
—Anita Satia
slide 6: 5
Contents
Preface 0304
Introduction 0711
Chapter1 Addition and Subtraction 1224
1. Addition  Completing the whole
2. Addition from left to right
3. Addition of list of numbers  Shudh method
4. Subtraction  Base method
5. Subtraction  Completing the whole
6. Subtraction from left to right
Chapter2 Digit Sums Casting out 9s 9Check Method 2528
Chapter3 11Check method 2931
Chapter4 Special Multiplication methods 3252
1. Base Method
2. Sub Base Method
3. Vinculum
4. Multiplication of complimentary numbers
5. Multiplication by numbers consisting of all 9s
6. Multiplication by 11
7. Multiplication by twodigit numbers from right to left
8. Multiplication by three and fourdigit numbers from right to left.
Chapter5 Squaring and square Roots 5357
Squaring
1. Squaring numbers ending in 5
2. Squaring Decimals and Fraction
slide 7: 6
3. Squaring Numbers Near 50
4. Squaring numbers near a Base and Sub Base
5. General method of Squaring  from left to right
6. Number splitting to simplify Squaring Calculation
7. Algebraic Squaring
Square Roots
1. Reverse squaring to find Square Root of Numbers ending in 25
2. Square root of perfect squares
3. General method of Square Roots
Chapter6 Division 5864
1. Special methods of Division
2. Straight Division
slide 8: 7
Introduction
The “Vedic Mathematics” is called so because of its origin from Vedas. To be more specific it has
originated from “Atharva V edas” the fourth V eda. “Atharva V eda” deals with the branches like Engineering
Mathematics sculpture Medicine and all other sciences with which we are today aware of.
The Sanskrit word V eda is derived from the root Vid meaning to know without limit. The word V eda
covers all VedaSakhas known to humanity. The Veda is a repository of all knowledge fathomless ever
revealing as it is delved deeper.
Vedic mathematics which simplifies arithmetic and algebraic operations has increasingly found
acceptance the world over. Experts suggest that it could be a handy tool for those who need to solve
mathematical problems faster by the day.
It is an ancient technique which simplifies multiplication divisibility complex numbers squaring
cubing square roots and cube roots. Even recurring decimals and auxiliary fractions can be handled by
Vedic mathematics. Vedic Mathematics forms part of Jyotish Shastra which is one of the six parts of
Vedangas. The Jyotish Shastra or Astronomy is made up of three parts called Skandas. A Skanda means
the big branch of a tree shooting out of the trunk.
This subject was revived largely due to the efforts of Jagadguru Swami Bharathi Krishna Tirtha Ji
of Govardhan Peeth Puri Jaganath 18841960. Having researched the subject for years even his efforts
would have gone in vain but for the enterprise of some disciples who took down notes during his last days.
The basis of Vedic mathematics are the 16 sutras which attribute a set of qualities to a number or a group
of numbers. The ancient Hindu scientists Rishis of Bharat in 16 Sutras Phrases and 120 words laid
down simple steps for solving all mathematical problems in easy to follow 2 or 3 steps.
Vedic Mental or one or two line methods can be used effectively for solving divisions reciprocals
factorisation HCF squares and square roots cubes and cube roots algebraic equations multiple
simultaneous equations quadratic equations cubic equations biquadratic equations higher degree
equations differential calculus Partial fractions Integrations Pythogorus Theoram Apollonius Theoram
Analytical Conics and so on.
V edic scholars did not use figures for big numbers in their numerical notation. Instead they preferred
to use the Sanskrit alphabets with each alphabet constituting a number. Several mantras in fact denote
numbers that includes the famed Gayatri Mantra which adds to 108 when decoded. How fast you can
solve a problem is very important. There is a race against time in all the competitions. Only those people
having fast calculation ability will be able to win the race. Time saved can be used to solve more problems
or used for difficult problems.
Given the initial training in modern maths in today’s schools students will be able to comprehend
the logic of Vedic mathematics after they have reached the 8th standard. It will be of interest to everyone
but more so to younger students keen to make their mark in competitive entrance exams. India’s past could
well help them make it in today’s world. It is amazing how with the help of 16 Sutras and 13 subsutras
the Vedic seers were able to mentally calculate complex mathematical problems.
slide 9: 8
Sixteen Sutras
S.N. Sutras Meaning
1. dkf/kdsu iwosZ.k One more than the previous one
Ekadhikena Purvena also a corollary
2. fufkya uorpjea nkr All from 9 and last from 10
Nikhilam Navatascaramam Dasatah
3. Å/oZfrZXHke Crisscross Vertically and crosswise
Urdhvatiryagbhyam
4. ijkoRZ kstsr Transpose and adjust Transpose and apply
Paravartya Yojayet
5. kwUa lkEleqPps When the samuchchaya is the same the samuch
Sunyam Samyasamuccaye chaya is zero i.e it should be equated to zero
6. ¼vkuq:Is½ kwUeUr If one is in ratio the other one is zero
Anurupye Sunyamanyat
7. ladyuOodyukHke By addition and by subtraction
Sankalanavyavakalanabhyam
also a corollary
8. iwj.kkiwj.kkHke By the completion or noncompletion
Puranapuranabhyam
9. pyudyukHke By Calculus
CalanaKalanabhyam
10 konwue By the deficiency
Yavadunam
11. Of"Vlef"V Specific and General Use the average
Vyastisamastih
12. ks"kk.dsu pjes.k The remainders by the last digit
Sesanyankena Caramena
13. lksikUReURe The ultimate twice the penultimate
Sopantyadvayamantyam
14. dUwusu iwosZ.k By one less than the previous one
Ekanyunena Purvena
15. xqf.krleqPp The product of the sum of coefficients in the factors
Gunitasamuccdyah The whole product
16. xq.kdleqPp Set of Multipliers
Gunakasamuccayah
slide 10: 9
Thirteen SubSutras
S.N. Sutras Meaning
1. vkuq:is.k
Anurupyena
2. fk"rs ks"klaK
Sityate Sesasanfitah
3. vkeksukUReURsu
Adyamadyenantyainantyena
4. dsoyS lIrda xq.kr
Kevalalh Saptakan Gunyat
5. os"Vue
Vestanam
6. konwra rkonwue
Yavadunam Tavadunam
7. konwua rkonwuhd`RoxZ p kstsr
Yavadunam Taradunikrtya Varganca Yojayet
8 vRksnZkds·fi
Antyayordasakept
9 vURksjsn
Antyayoteva
10 leqPpxqf.kr
Samuccayaguaitah
11 yksiLFkkiukHke
Lopanasthapandbhyam
12 foyksdue
Vilokanam
13 xqf.krleqPp leqPpxqf.kr
Gunitasamuccayah Samuccayagunitah
Proportionately
The remainder remains constant
The first by the first and last by the last
In case of 7 our multiplicand should be
143
By osculation
Lessen by the Deficiency
Whatever the extent of its deficiency lessen
it still to that very extent and also set up
the square of that deficiency.
Whose last digits together total 10 and
whose previous part is exactly the same
Only the last terms
The sum of the coefficients in the product
By alternate elimination and retention
By observation
The product of sum of the coefficients in
the factors is equal to the sum of the
coefficients in the product.
slide 11: 10
In the text the words Sutra aphorism formula is used synonymously. So are also the words Upa
sutra Subsutra Subformula corollary used.
The Sutras apply to and cover almost every branch of Mathematics. They apply even to complex
problems involving a large number of mathematical operations. Application of the Sutras saves a lot of
time and effort in solving the problems compared to the formal methods presently in vogue. Though the
solutions appear like magic the application of the Sutras is perfectly logical and rational. The computation
made on the computers follows in a way the principles underlying the Sutras. The Sutras provide not only
methods of calculation but also ways of thinking for their application.
This course on Vedic Mathematics seeks to present an integrated approach to learning Mathematics
with keenness of observation and inquisitiveness avoiding the monotony of accepting theories and
working from them mechanically. The explanations offered make the processes clear to the learners. The
logical proof of the Sutras is detailed which eliminates the misconception that the Sutras are a jugglery.
Application of the Sutras improves the computational skills of the learners in a wide area of problems
ensuring both speed and accuracy strictly based on rational and logical reasoning. The knowledge of such
methods enables the teachers to be more resourceful to mould the students and improve their talent and
creativity. Application of the Sutras to specific problems involves rational thinking which in the process
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 use of the Sutras and SubSutras stated above for presentation of their application
for learning Mathematics at the secondary school level in a way different from what is taught at present
but strictly embodying the principles of algebra for empirical accuracy. The innovation in the presentation
is the algebraic proof for every elucidation of the Sutra or the SubSutra concerned.
Terms and Operations
a Ekadhika means ‘one more’
e.g: Ekadhika of 0 is 1 Ekadhika of 1 is 2
Ekadhika of 8 is 9 Ekadhika of 23 is 24
Ekadhika of 364 is 365
b Ekanyuna means ‘one less’
e.g: Ekanyuna of 1 2 3 ..... 8 ..... 14 .....69 ......is 0 1 2 ..... 7 ......13 .... 68 ......
c Purak means ‘complement’ e.g: Purak of 1 2 3 ..... 8 9 from 10 is 9 8 7..... 2 1
d Rekhank means ‘a digit with a bar on its top’. In other words it is a negative number.
e.g: A bar on 7 is written as 7. It is called Rekhank 7 or bar 7. We treat Purak as a Rekhank.
e.g: 7 is 3 and 3 is 7
At some instances we write negative numbers also with a bar on the top of the numbers as
–4 can be shown as
4
.
–21 can be shown as
21
.
slide 12: 11
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 27 is 2 + 7 9.
Beejank of 348 is 3 + 4 + 8 15 further 1 + 5 6. i.e. 6 is Beejank.
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. 1: Find the Beejank of 632174.
As above we have to follow
632174 → 6 + 3 + 2 + 1 + 7 + 4 → 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 632174.
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 digits which are 5 and above 5 or less than 5 without changing the value
of that number.
Consider a number say 8. Note that it is greater than 5. Use it complement purak rekhank from
10. It is 2 in this case and add 1 to the left i.e. tens place of 8.
Thus 8 08 12
The number 1 contains both positive and negative digits.
i.e.1 and 2. Here 2 is in unit place hence it is –2 and value of 1 at tens place is 10.
Thus 12 10 – 2 8
Conveniently we can think and write in the following way
General Number Conversion Vinculum number
6 10 – 4 14
97 100 – 3 103
289 300 – 11
311
etc.
slide 13: 12
Chapter 1 Addition and subtraction
Addition is the most basic operation and adding number 1 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.
7 close to 10 The ten Point Circle
8 close to 10
9 close to 10
171819 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 easily say that 9 is close to 10
19 is close to 20 etc.
We can use this closeness to find addition and subtraction.
The ten Point Circle
Rule : By completion noncompletion
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.
10
1
2
3
4
5
6
7
9
8
19
18
17
16
15
14
13
12
11
20
10
1
2
3
4
5
6
7
9
8
slide 14: 13
Example : 24 + 26 20 + 4 + 20 + 6 20 + 20 + 10 50
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.
Example : 59 + 4 59 + 1 + 3 60 + 3 63 59 is close to 60 and 1 short 50 59 + 4 is 60
Example : 38 + 24 38 + 2 + 22 40 + 22 62
or
38 + 24 40 + 24 – 2 64 – 2 62 38 is close and is 2 sheet so 38 + 24 is 2
short from 40 + 24 hence 38 + 24 40 + 24
– 2 64 – 2 62
Example
Add 39 + 6
39 is close to 40 and is 1 less then it.
So we take 1 from the 6 to make up 40 and then we have 5 more to add on which gives 45
Add
29 + 18 + 3
29 + 18 + 1 + 2 As 3 1 + 2 and 29 + 1 30 18 + 2 20
30 + 20 50 Note we break 3 into 1 + 2 because 29 need 1 to become 30 and 18 need
2 become 20
Add
39 + 8 + 1 + 4
39 + 8 + 1 + 2 + 2
40 + 10 + 2 52
Sum of Ten
The ten point circle illustrates the pairs of numbers whose sum is 10.
Remember : There are eight unique groups of three number that sum to 10 for example 1 + 2 + 7 10
1 + 2 + 7 10
Can you find the other seven groups of three number summing to 10 as one example given for you
2 + 3 + 5 10
slide 15: 14
Adding a list of numbers
Rule : By completion or noncompletion
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.
Similarily : 48 + 16 + 61 + 32
48 + 32 + 16 + 1 + 60
80 + 77 157
10 10 10
or 7 + 8 + 9 + 2 + 3 + 5 + 3 + 1 + 2 + 3 + 7 + 9
10 10
10 + 10 + 10 + 10 + 10 + 9 59
PRACTICE PROBLEMS
Add by using completing the whole method
1. 39 + 8 + 1 + 5 2. 18 + 3 +2 + 17
3. 9 + 41 + 11 +2 4. 47 + 7 + 33 23
5. 23 + 26 + 27 + 34 6. 22 + 36 + 44 + 18
7. 33 + 35 + 27 + 25 8. 18 + 13 + 14 + 23
9. 3 + 9 + 8 + 5 + 7 + 1+ 2 10. 37 + 25 + 33
11. 43 + 8 + 19 + 11 12. 42 + 15 + 8 +4
13. 24 + 7 + 8 + 6 +13 14. 16 +43 + 14 +7
15. 13 + 38 +27
ADDITION
Completing the whole method class VI commutative associative property
1. 39 + 17 + 11 + 13 2. 16 + 23 + 24 + 7
3. 12 + 51 + 9 + 18 4. 35 + 12 +55
slide 16: 15
5. 123 + 118 + 27 6. 35 + 15 + 16 + 25
7. 58 + 41 + 12 + 9 8. 223 + 112 + 27
9. 24 + 106 + 508 + 12 10. 506 + 222 + 278
Adding from left to right
The conventional methods of mathematics teachers use to do calculation from right and working towards
the left.
In Vedic mathematics we can do addition from left to right which is more useful easier and
sometimes quicker.
Add from left to right
1. 23 2. 2 3 4
+ 15 + 5 2 4
38 7 5 8
3. 1 5 4. 2 3 5
3 8 5 2 6
4 3 7 5 1
Add 1 Add 1
53 761
The method: This is easy enough to do mentally we add the first column and increase this by 1 if
there is carry coming over from the second column. Then we tag the last figure of the second column onto
this
Mental math
Add from left to right
1 6 6 2 5 4 6 3 5 3 4 4 1 4 5 7
+ 5 5 + 6 7 1 + 7 1 7 + 2 8 5 7
5 4 5 6 3 1 2 4 6 5 7 7 4 5 8 1 4 3 2
+ 7 6 + 7 6 1 2 4 6 + 2 7 + 8 6 6 8
9 8 5 10 5 3 7 11 4 5 6 12 2 6 4 8
+ 2 3 + 7 1 8 + 1 2 7 + 8 3 6 5
13 1 3 4 5 14 5 4 6 15 7 8 8 5 16 3 7 8
+ 5 8 3 6 + 4 5 6 1 + 1 5 4 3 + 4 8
17 3 5 6 7 1 18 2 4 6 8
+ 1 2 3 4 5 + 1 2 3
slide 17: 16
Shudh method for a list of number
Shudh means pure. The pure numbers are the single digit numbers i.e. 0 1 2 3…9. In Shudh method of
addition we drop the 1 at the tens place and carry only the single digit forward.
Example: Find 2 + 7 + 8 + 9 + 6 + 4
2
• 7
• 8
9
• 6
4
36
We start adding from bottom to top because that is how our eyes naturally move but it is not
necessary we can start from top to bottom. As soon as we come across a twodigit number we put a dot
instead of one and carry only the single digit forward for further addition. We put down the single digit
6 in this case that we get in the end. For the first digit we add all the dots 3 in this case and write it.
Adding two or three digit numbers list
. 23.4 We start from the bottom of the right most columns and get a single digit 6 at the unit
6.5.8 place. There are two dots so we add two to the first number 4 of
.81.8 the second column and proceed as before. The one dot of this
46 column is added to the next and in the end we just put 1 down
1756 for one dot as the first digit of the answer.
Shudh method
• 52 6
• 9 • 4•5
4 3 4
• 6 • 8 1
7 5 2
• 8 23 8
4
43
Add the following by Shudh method
1. 5 2. 37 3. 345
7 64 367
6 89 289
826 + 167
4 + 71
+ 9
slide 18: 17
4. 3126 5. 468 6. 235
1245 937 579
4682 386 864
+ 5193 654 + 179
7. 59 8. 49 9. 98
63 63 83
75 78 78
82 85 62
+ 91 + 97 + 44
10. 37 11. 2461 12. 9721
79 4685 2135
52 6203 5678
88 1234 207
+ 91 + 5432 + 1237
Number Spliting Method
Quick mental calculations can be performed more easily if the numbers are split into more manageable
parts.
For example : Split into two more manageable sums
+ 3642 36 42 Note : The split allows us to add 36 + 24
2439 + 24 39 and 42 + 39 both of which can be done
60 81 mentally
Remember : Think about where to place the split line. Its often best to avoid number carries over the
line.
For example : 342 3 42 34 2
+ 587 5 87 58 7
2 29 92 9
carry 1 No carry is required
A carry of 1 over the line is required
slide 19: 18
SUBTRACTION
Sutra: All from 9 and the Last from 10
The Concept of Base
Numbers made up of only 1’s and 0’s are known as a Base.
Examples of a Base are
10 100 1000 1 .01….etc
The base method is used for subtracting multiplying or dividing numbers. Like 98 898 78999 etc
that are close to base.
Applying the formula “All form 9 and Last form 10” to any number especially the big one’s reduces
it to its smaller Counterpart that can be easily used for calculations involving the big digits like 7 8 and 9.
Applying the formula “All from 9 and the last from 10”
Example: Apply ‘All from 9 Last from 10’ to
Subtract 789 from 1000
7 8 9
↓ ↓ ↓ Here all from 9 last from 10 means subtract 78 8 from 9 and 9 from 10 so weget 211
2 1 1
We get 211 because we take 7 and 8 from 9 and 9 from 10.
from 10000 from 100 from 100 from 100000
2772 54 97 10804
↓↓↓↓ ↓↓ ↓↓ ↓↓↓↓↓
7228 46 03 89196
If you look carefully at the pairs of numbers in the above numbers you may notice that in every case
the total of two numbers is a base number 10 100 1000 etc.
This gives us an easy way to subtract from base numbers like 10 100 1000…….
Subtracting from a Base
Example:  1000 – 784 216
Just apply ‘All from 9 and the Last from 10’ to 784 difference of 7 from 9 is 2 8 from 9 is 1 4 from
10 is 6 so we get 216 after subtraction.
When subtracting a number from a power of 10 subtract all digits from 9 and last from 10.
1000 from 9 from 10
– 276
724 276
↓↓↓
724
slide 20: 19
Subtracting from a Multiple of a Base
Sutra: ‘All from 9 and the last from 10’
and
‘One less than the one before’
Example: 600 – 87
We have 600 instead of 100. The 6 is reduced by one to 5 and the All from 9 and last from 10 is applied
to 87 to give 13. Infact 87 will come from one of those six hundred so that 500 will be left.
∴ 600 – 87 513 Note : First subtract form 100 then add 500 as 500 + 13 513
Example: Find 5000 – 234
5 is reduced to one to get 4 and the formula converts 234 to 766
∴ 50002344766
Example: 1000 – 408 592
Example:100 – 89 11
Example:1000 – 470 530 Remember apply the formula just to 47 here.
If the number ends in zero use the last nonzero number nonzero number as the last number for
example.
10000 from 9 from 10
– 4250
5750 4250
↓↓↓
5750
Hence 1000 – 4250 5750
Adding Zeroes
In all the above sums you may have noticed that the number of zeros in the first number is the same as
the numbers of digits in the number being subtracted.
Example: 1000 – 53 here 1000 has 3 zeros and 53 has two digits.
We can solve this by writing
1000
– 053
947
We put on the extra zero in front of 53 and then apply the formula to 053.
Example: 10000 – 68 Here we need to add two zeros.
10000 – 0068 9932
slide 21: 20
Practice Problems
Subtract from left to right
1 86 – 27 2 71 – 34
3 93 – 36 4 55 – 37
5 874 – 567 6 804 – 438
7 793 – 627 8 5495 – 3887
9 9275 – 1627 10 874– 579
11 926 – 624 12 854– 57
13 8476 – 6278 14 9436 – 3438
Subtract the following mentally
1 55 – 29 2 82 – 558
3 1000 – 909 4 10000 – 9987
5 10000 – 72 6 50000 – 5445
7 70000 – 9023 8 30000 – 387
9 46678 – 22939 10 555 – 294
11 8118 – 1771 12 61016 – 27896
Example: Find 9000 – 5432
Sutra: ‘One more than the previous one’ and ‘all from 9 and the Last from the 10’
Considering the thousands 9 will be reduced by 6 one more than 5 because we are taking more than
5 thousand away
‘All from 9 and the last from 10’ is than applied to 432 to give 568
9000 – 5432 3568
Similary—7000 – 3884
3116 3 7 – 4 4 is one more than 3 and 116 4000 – 3884 by all from a and the last from 10
If the number is less digits then append zero the start :
1000 from 9 from 10
– 425
9575 042 5
↓↓↓ ↓
957 5
slide 22: 21
When subtracting form a multiple ofa power of 10 just decrement the first digit by 1 then subtract
remaining digits :
4000 from 9 from 10
– 257
3743 257
↓↓↓
4 – 1 → 3 753
Look at one more example :
Money: A great application of "all from 9 and last from 10" is money. Change can be calculated by
applying this sutra mentally for example :
10.00 from 9 from 10
– 4.25
5.75 4.25
5.75
This is helpful because most our rupee notes are multiple of 10s.
PRACTICE PROBLEMS
Subtract base method
1 1000 – 666 2 10000 – 3632
3 100 – 54 4 100000 – 16134
5 1000000 – 123456 6 1000 – 840
7 1000 – 88 8 10000 – 568
9 1000 – 61 10 100000 – 5542
11 10000 – 561 12 10000 – 670
Subtract multiple of base
1 600 – 72 2 90000 – 8479
3 9000 – 758 4 4000 – 2543
5 7000 – 89 6 300000 – 239
7 1 – 0.6081 8 5 – 0.99
Subtracting Near a base
Rule : By completion or non completion.
when subtracting a number close to a multiple of 10. Just subtract from the multiple of 10 and correct
the answer accordingly.
slide 23: 22
Example : 53 – 29
29 is just close to 30 just 1 short so subtaract 30 from 53 making 23 then add 1 to make 24.
53 – 29 53 – 30 + 1
23 + 1
24
Similarily
45 – 18
45 – 20 + 2
25 + 2
27 18 is near to 20 just 2 short
Use the base method of calculating
To find balance
Q. Suppose you buy a vegetable for Rs. 8.53 and you buy with a Rs. 10 note. How much change would
you expect to get
Ans. You just apply “All from 9 and the last from 10” to 853 to get 1.47.
Q. What change would expect from Rs. 20 when paying Rs. 2.56
Ans.The change you expect to get is Rs. 17.44 because Rs. 2.56 from Rs.10 is Rs. 7.44 and there is Rs.
10 to add to this.
Practice Problem
Q1. Rs. 10 – Rs. 3.45
Q2. Rs. 10 – Rs. 7.61
Q3. Rs. 1000 – Rs. 436.82
Q4. Rs. 100 – Rs. 39.08
Subtracting number just below the base
Example: find 55 – 29
Subtraction of numbers using "complete the whole"
Step 1: 20 is the sub base close to 19
19 is 1 below 20
Step 2: take 20 from 55 to get 35
Step 3: Add 1 back on 55 – 19 36
Example
61 – 38
38 is near to 40 40 – 38 2
61 – 40 21
61 – 38 21 + 2 23
slide 24: 23
Example
44 – 19
19 + 1 20
44 – 20 24
44 – 19 24 – 1 23
Example 88 – 49
49+150
88 – 50 38
88 – 49 38 + 1 39
Example
55 – 17
17 + 3 20
55 – 20 35
55 – 17 35 + 3 38
Number spliting Method
As you have use this method in addtion the same can be done for subtraction also :
+ 3642 36 42 Note : The split allows on to add 36 – 24
2439 + 24 39 and 42 – 39 both of which can be done
12 03 mentally
General Method of subtraction
Subtraction from left to right
In this section we show a very easy method of subtracting numbers from left to right that we have probably
not seen before. We start from the left subtract and write it down if the subtraction in the next column
can be done. If it cannot be done you put down one less and carry 1 and then subtract in the second
column.
Subtraction from left to right.
Example: Find Find
83 – 37 78 – 56
8 3 7 8
– 3 7 – 5 6
4 6 2 2
slide 25: 24
Left to right
3 4 5
5
1
13
1
2
1
1 3 0
1
1
– 4 9 – 2 8 9 –2 0 4
0 2 0 3 2 1 9 7
6 3 0 1 7 3 5
1
5 6
1
7
–2 0 1 –1 1 8 2 8
1 0 0 2 3 7 3 9
Starting from the left we subtract in each column 312 but before we put 2 down we check that in
next column the top number is larger. In this case 5 is larger than 1 so we put 2 down
In the next column we have 514 but looking in the third column we see the top number is not larger
than the bottom 5 is less than 8 so instead putting 4 down we put 3 and the other 1 is placed as the flag
as shown so that 5 becomes 15 so now we have 1587. Checking in the next column we can put this
down because 6 is greater than 2. In the fourth column we have 624 but looking at the next column
7 is smaller than 8 we put down only 3 and put the other flag with 7 as shown finally in the last column
1789.
slide 26: 25
Chapter 2 Digit sums casting out 9’s and 9’ check method
The word digit means a single figure 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 of 57 is 5 + 7 12 → 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 → 2 + 4 3
Hence the digit sum of 687 is 3.
● Keep findig the digit sum of the result + unitl its less then 10
● 0 and 9 are requivalent
Look and undevstand some more example :
To find the digit sum of 18 for the example we just 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 how to get the digit sum of the following members
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 a digit sum.
So for the digit sum of 29 we add 2 + 9 11 but since 11 is a 2digit number we add again 1+11
So for the digit sum of 29 we can write
29 2 + 9 11 1 + 1 1
slide 27: 26
Similarity for 49 4 + 9 13 1 + 3 4
So the digit sum of 49 is 4.
Number 14 Digit sum 1 + 4 5 Single digit 5
19 1 + 9 10 1
39 3+ 9 12 3
58 5 + 8 13 4
407 4 + 0 + 7 11 2
CASTING OUT NINE
Adding 9 to a number does not affect its digit sum
So 5 59 95 959 all have digit sum of 5.
For example to find out the digit sum of 4939 we can cast out nines and just add up the 3 and 4 so
digit sum is 7 or using the longer method we add all digit 4 + 9 + 3+ 9 25 2 + 57
There is another way of casting out the nines from number when you are finding its digit sum.
Casting out of 9’s and digit totalling 9 comes under the Sutra when the samuccaya is the same it is
zero.
So in 465 as 4 and 5 total nine they are cast out and the digit sum is 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 Nine Point Circle
1326 3
25271 8
9643 4
23674 4
128541 3
1275 6
6317892 9 or 0
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.
1
2
3
4
5
6
7
9
8
17
26
16 25
15
24
14
23
13
22
12
11
10
18
27
19
20
21
slide 28: 27
9. Check Method
Digit sum can be used to check that the answers are correct.
Example: Find 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 answer should also be 8
Digit sum of 448 so according to this check the answer is probably correct.
There are four steps to use digit sum to check the answers:
1. Do the sum.
2. Write down the digit sums of the numbers being added.
3. Add the digit sums.
4. Check whether the two answers are same in digit sums.
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 and 2 respectively
3. Adding 8 and 2 gives 10 digits sum of 101+01
4. Digit sum of 397 is
3 + 9 + 7 19 1 + 9 10 1 + 0 1
Which confirm the answer
CAUTION
Check the following sum:
279 9
121 4
490 4
Here an estimation can help you to find the result more accurate if by mistage you write 400 in place
of 490 then it will show 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 incorrect This shows
that the digit sum does not always find errors. It usually works but not always. We will be looking at
another checking device i.e. 11  check method.
Note : The difference of 9 and its multiples in the answer make errors. So keep in mind a rough
estimation.
slide 29: 28
Practice Problems
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 sum of a two digit number is 9 and the first figure is twice the second what is it
3. Give three two 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 9’s to find the digit sums of the numbers below.
Number
465
274
3456
7819
86753
4017
59
6. Add the following and check your answer using digit sum check
1 66 + 77 2 57 + 34
3 94 + 89 4 304 + 233
5 787 + 132 6 389 + 414
7 5131 + 5432 8 456 + 654
slide 30: 29
Chapter 3 Eleven Check Method
We have already used the digit sum check that helps to show if a calculation is correct. This method works
because adding the digit in a number gives the remainder of the number after division by 9.
A similar method works by using remainders of numbers after division by 11 rather than 9
Alternate digit sum or Elevencheck Method
Suppose we want another check for 2434 × 23 55982 it can be done in the following steps
Step1: Alternately add and subtract starting from right moving towards left the digits of each
numbers as described below
Number Alternating signs Digit sum
2434 –2 + 4 – 3 + 4 3
23 –2 + 3 1
55982 +5 – 5 + 9 – 8 + 2 3
Step 2: Now multiply the Digit Sum to get the product 3 × 1 3 Since the Digit Sum of the product
and the two numbers is the same the answer is correct as per 11 check method.
Two digit and Negative number in the digit sum checking the sum of addition
4364 + 1616
Left to right
4364
1916
6280
Number Alternating signs Digit sum Single digit
4364 –4 + 3 – 6 + 4 –3 113 8
1916 –1 + 9 – 1 +6 1311+2 2
6280 –6 + 2 – 8 + 0 – 12 10
11 –12 –1
11 – 1 10
Step2: Apply the following rules to get a single positive digit for the number
• Subtract the negative numbers below 11 from 11 to get its positive counterpart so – 3 11 –3 8
And –12 –12 +11 –1 11 – 1 10
• For the two digit number above 11 divide the number by 11 and get the remainder as the
positive digit sum so 13 ÷ 13 gives remainder 2. Alternately adding and subtracting digit of 13
starting from right can obtain this same result.
slide 31: 30
Step 3 : now add the Digit sums to get the sum 8 + 2 10 the answer is correct as per 11 check
method.
Two digits in the digit sum
Check subtraction problem
2819174 – 839472
2819174
839472
1979702
Step 1: Alternatively add and subtract staring from right moving towards left the digit of each numbers
as described below
Number Alternating signs Digit sum Single digit
2819174 +2–8+1–9+1–7+4 –16–16+11 –5 11–56
839472 –8+3–9+4–7+2 –15–15+11 –4 11–47
1979702 +1–9+7–9+7–0+2 –1 11–110
Step 2: Apply the following rules to get a single positive digit for the number
• The negative numbers below –11 are to be first divided by 11 to get the remainder. Than subtract
the remainder from 11 to get its positive counterpart. So –16/11 Remainder is –5 and –5
11 – 5 6 similarly –15/11 Remainder –4 11 –4 7.
• The negative number –1 11 – 1 10
Step3: Now subtract the Digit sums to get the answer 6 – 7 –1 10 the answer is correct as per
11 checked method.
Practice Problems
Get the digit sum and single digit for the following numbers.
Numbers Alternative signs Digit sums Single digit
567
1536
93823
1978712
849391
82918
5949393
176780
slide 32: 31
Using 11 check method check the following Addition problems:
1 37 + 47 84
2 55 + 28 83
3 47 + 25 72
4 29 + 36 65
5 526 + 125 651
6 1328 + 2326 3654
7 129 + 35644 35773
8 3425 + 7491 + 8834 19750
9 1423178 + 5467 + 123 + 34 1428802
10 1314 + 5345 + 65 +781 7505
Check the following subtraction problems:
1 63 – 28 35
2 813 – 345 468
3 695 – 368 372
4 3456 – 281 3175
5 7117 – 1771 5346
6 8008 – 3839 4165
7 6363 – 3388 2795
8 51015 – 27986 23029
9 14285 – 7148 7137
10 9630369 – 3690963 5939406
slide 33: 32
Chapter 4 Special Multiplication Methods
Multiplication in considered as one of the most difficult of the four mathematical operations. Students are
scared of multiplication as well as tables. Just by knowing tables up to 5 students can multiply bigger
numbers easily by some special multiplication methods of Vedic Mathematics. We should learn and
encourage children to look at the special properties of each problem in order to understand it and decide
the best way to solve the problem. In this way we also enhance the analytical ability of a child. Various
methods of solving the questions /problems keep away the monotonous and charge up student’s mind to
try new ways and in turn sharpen their brains.
Easy way for multiplication
Sutra:Vertically and Cross wise :
For speed and accuracy tables are considered to be very important. Also students think why to do lengthy
calculations manually when we can do them faster by calculators. So friends/ teachers we have to take
up this challenge and give our students something which is more interesting and also faster than a
calculator. Of course it’s us the teachers/parents who do understand that more we use our brain more
alert and active we will be for that is the only exercise we have for our brain.
Example 1: 7 x8
Step 1: Here base is 10
7 – 3 7 is 3 below 10 also called deficiencies
× 8 – 2 8 is 2 below 10 also called deficiencies
Step 2: Cross subtract to get first figure or digit of the answer: 7 – 2 5 or 8 – 3 5 the two
difference are always same.
Step 3 : Multiply vertically i.e. –3 × –2 6 which is second part of the answer.
So 7 – 3
8 – 2 i.e. 7 × 8 56
5 / 6
Example 2: To find 6 × 7
Step 1 : Here base is 10
6 – 4 6 is 4 less than 10 i.e. deficiencies
7 – 3 7 is 3 less than 10 i.e. deficiencies
Step 2: Cross subtraction : 6 – 3 3 or 7 – 4 3 both same
Step 3: – 3 × – 4 + 12 but 12 is 2 digit number so we carry this 1 over to 3 obtained in 2 step
6– 4
7 – 3
3 / 1 2 i.e. 6 × 7 42
Try these : 1 9 × 7 ii 8 × 9 iii 6 × 9 iv 8 × 6 v 7 × 7
slide 34: 33
Second Method:
Same Base Method :
When both the numbers are more than the same base. This method is extension of the above method i.e.
we are going to use same sutra here and applying it to larger numbers.
Example 1: 12 × 14
Step 1: Here base is 10
12 + 2 12 is 2 more than 10 also called surplus
14 + 4 14 is 4 more than 10also called surplus
Step 2: Cross add: 12 + 4 16 or 14 + 2 16both same which gives first part of answer 16
Step 3: Vertical multiplication: 2 × 4 8
So 12 + 2
14 +4
16 / 8So 12 × 14 168
14 + 2 12 + 4
Example 2:105x 107
Step1: Here base is 100
105 + 05 105 is 5 more than 100 or 5 is surplus
107 + 07 107 is 7 more than 100 or 7 is surplus
Base here is 100 so we will write 05 in place of 5and 07 in place of 7
Step 2: Cross add: 105 + 7 112 or 107 + 5 112 which gives first part of the answer 112
Step 3: Vertical multiplication: 05 × 07 35 two digits are allowed
As the base in this problem is 100 so two digits are allowed in the second part.
So 105 × 107 11235
Example 3: 112 x 115
Step 1: Here base is 100
112 + 12 2 more than 100 i.e. 12 is surplus
115 + 15 15 more than 100 i.e. 15 is surplus
Step 2: Cross add: 112 + 15 127 115 + 12 to get first part of answer
i.e.127
Step 3: Vertical multiplication 12 × 15 Oh my godIt’s such a big number. How to get product
of this Again use the same method to get the product.
12 + 2
15 + 5
12 + 5 15 + 2 17/ 1 0 17 + 1 / 0 180 i.e. 12 × 15 180
But only two digits are allowed here so 1 is added to 127 and we get 127 + 1 128
So 112 × 115 128 80
slide 35: 34
Try these: i12 × 14 ii 14 × 17 iii 17 × 19 iv 19 × 11 v 11 × 16 vi 112 × 113 vii 113 × 117
viii 117 × 111 ix 105 × 109 x 109 × 102 xi 105 × 108 xii 108 × 102 xiii 102 × 112 xiv 112
× 119 xv 102 × 115
Both numbers less than the same base:
Same sutra applied to bigger numbers which are less than the same base.
Example1: 99 × 98
Step 1: Check the base: Here base is 100 so we are allowed to have two digits on the right hand side.
∴ 99 – 01 1 less than 100 i.e. 01 deficiency
98 – 02 2 less than 100 i.e. 0 2 deficiency
Step 2: Cross – subtract: 99 – 02 97 98 – 01 both same so first part of answer is 97
Step3: Multiply vertically – 01 × – 02 02 As base is 100 so two digits are allowed in second part
So 99 × 98 9702
Example 2 : 89 × 88
Step1: Here base is 100
So 89 – 11 i.e. deficiency 11
88 – 12 i.e. deficiency 12
Step2: Cross subtract: 89 – 12 77 88 – 11both same
So first part of answer can be 77
Step 3:Multiply vertically – 11 × – 12
Again to multiply 11 × 12 apply same rule
11 + 1 10 + 1
12 + 2 10 + 2
11 + 2 13 12 + 1 / 1 × 2 12 so 11 × 12 1 32 as only two digits are allowed on right hand
side so add 1to L.H.S.
So L.H.S. 77 + 1 78
Hence 89 × 88 7832
Example 3: 988 × 999
Step 1: As the numbers are near 1000 so the base here is 1000 and hence three digits allowed on the
right hand side
988 – 012 012 less than 1000 i.e. deficiency 0 12
999 – 001 001 less than 1000 i.e. deficiency 00 1
Step 2: Cross – subtraction: 988 – 001 987 999 – 012 987
So first part of answer can be 987
Step 3: Multiply vertically: –012 xs – 001 012 three digits allowed
∴ 988 × 999 987012
How to check whether the solution is correct or not by 9 – check method.
slide 36: 35
Example 1: 99 × 98 9702 Using 9 – check method.
As 99 0 Product L.H.S. 0 × 8 0 taking 9 0
98 8
R.H.S.
9702 7 2 9 0 + 9702 9
both are same
As both the sides are equal answer may be correct.
Example 2: 89 × 88 7832
89 8
88 8 + 8 16 1 + 6 7 add the digits
L.H.S. 8 × 7 56 5 + 6 11 2 1 + 1
R.H.S. 7832 8 + 3 11 1 + 1 2
As both the sides are equal so answer is correct
Example 3: 988 × 999 987012
988 8 + 8 16 1 + 6 7
999
0
As 0 × 7 0 LHS
987012
0 As 7 + 2 9 0 8 + 1 9 0 also 9 0
∴ RHS 0
As LHS RHS So answer is correct.
Try These:
i 97 × 99 ii 89 × 89 iii 94 × 97 iv 89 × 92 v 93 × 95 vi 987 × 998 vii 997 × 988 viii 988
× 996 ix 983 × 998 x 877 × 996 xi 993 × 994 xii 789 × 993 xiii 9999 × 998 xiv 7897 × 9997
xv 8987 × 9996.
Multiplying bigger numbers close to a base: number less than base
Example 1: 87798 x 99995
Step1: Base here is 100000 so five digits are allowed in R.H.S.
87798 – 12202 12202 less than 100000 deficiency is 12202
99995 – 00005 00005 less than100000 deficiency is 5
Step 2: Cross – subtraction: 87798 00005 87793
Also 99995 – 12202 87793 both same
So first part of answer can be 87793
Step 2 : Multiply vertically: –12202 × – 00005 + 61010
∴ 87798 × 99995 8779361010
slide 37: 36
Checking:
87798
total 8 + 7 + 7 + 8 30 3 single digit
99995
total 5
LHS 3 x 5 15 total 1 + 5 6
RHS product
8779361010
total 15 1 + 5 6
L.H.S R.H.S. So correct answer
Example 2 : 88777 × 99997
Step 1: Base have is 100000 so five digits are allowed in R.H.S.
88777 – 11223 i.e. deficiency is 11223
99997 – 00003 i.e. deficiency is 3
Step 2: Cross subtraction: 88777 – 00003 88774 99997 – 11223
So first part of answer is 88774
Step 3: Multiply vertically: – 11223 × – 00003 + 33669
∴ 88777 × 99997 8877433669
Checking:
88777 total 8 + 8 + 7 + 7 + 7 37 + 10 1
99997
total 7
∴ LHS 1 × 7 7
RHS 8877433669 8 + 8 + 7 + 7 + 4 34 3 + 4 7
i.e. LHS RHS So correct answer
Try These:
i 999995 × 739984 ii 99837 × 99995 iii 99998 × 77338 iv 98456 × 99993 v 99994 × 84321
Multiply bigger number close to base numbers more than base
Example 1: 10021 × 10003
Step 1: Here base is 10000 so four digits are allowed
10021 + 0021 Surplus
10003 + 0003 Surplus
Step 2: Cross – addition 10021 + 0003 10024 10003 + 0021 both same
∴ First part of the answer may be 10024
Step 3: Multiply vertically: 10021 × 0003 0063 which form second part of the answer
∴ 10021 × 10002 100240063
slide 38: 37
Checking:
10021 1+ 2 + 1 + 1 4
10003 1 + 3 4
∴ LHS 4 × 4 16 1 + 6 7
RHS
100240063
1 + 2 + 4 7
As LHS RHS So answer is correct
Example 2: 11123 × 10003
Step 1: Here base is 10000 so four digits are allowed in RHS
11123 + 1123 surplus
10003 + 0003 surplus
Step 2: Cross – addition: 11123 + 0003 11126 10003 + 1123 both equal
∴ First part of answer is 11126
Step 3: Multiply vertically: 1123 × 0003 3369 which form second part of answer
∴ 11123 × 10003 111263369
Checking:
11123 1 + 1 + 1 + 2 + 3 8
10003 1 + 3 4 and 4 × 8 32 3 + 2 5
∴ LHS 5
R.H.S 111263369 1 + 1 + 1 + 2 5
As L.H.S R.H.S So answer is correct
Try These:
i 10004 × 11113 ii 12345 × 111523 iii 11237 × 10002 iv 100002 × 111523 v 10233 × 10005
Numbers near different base: Both numbers below base
Example 1: 98 × 9
Step 1: 98 Here base is 100 deficiency 02
9 Base is 10 deficiency 1
∴ 98 – 02 Numbers of digits permitted on R.H.S is 1 digits in lower base
Step 2: Cross subtraction: 98
1
88
It is important to line the numbers as shown because 1 is not subtracted from 8 as usual but from
9 so as to get 88 as first part of answer.
Step 3: Vertical multiplication: 02 x 1 2 one digits allowed
∴ Second part 2
∴ 98 × 9 882
slide 39: 38
Checking:
Through 9 – check method
98 8 9 0 LHS 98 × 9 8 × 0 0
RHS 882 8 + 8 + 2 18 1 + 8 9 0
As LHS RHS So correct answer
Example 2: 993 × 97
Step 1: 993 base is 1000 and deficiency is 007
97 base is 100 and deficiency is 03
∴ 993 – 007 digits in lower base 2 So 2 digits are permitted on
× 97 – 03 RHS or second part of answer
Step 2: Cross subtraction:
993
– 03
963
Again line the number as shown because 03 is subtracted from 99 and not from 93 so as to get 963
which from first part of the answer.
Step 3: Vertical multiplication: –007 – –03 21 only two digits are allowed in the second part
of answer So second part 21
∴ 993 × 97 96321
Checking: through 9 – check method
993 3 97 7
∴ L.H.S. 3 × 7 21 2 + 1 3
R.H.S. 96321 2 + 1 3
As LHS RHS so answer is correct
Example 3 : 9996 base is 10000 and deficiency is 0004
988 base is 1000 and deficiency is 012
∴ 9996 – 0004 digits in the lower base are 3 so3digits
× 988 – 012 permitted on RHS or second part of answer
Step 2 : Cross – subtraction:
9996
– 012
9876
Well again take care to line the numbers while subtraction so as to get 9876 as the first part of the
answer.
Step3 : Vertical multiplication: –0004 × –012 048
slide 40: 39
Remember three digits are permitted in the second part i.e. second part of answer 048
∴ 9996 × 988 9876048
Checking:9 – check method
9996 6 988 8 + 8 + 16 1 + 6 7
∴ LHS 6 × 7 42 4 + 2 6
RHS 9876045 8 + 7 15 1 + 5 6
As LHS RHS so answer is correct
When both the numbers are above base
Example 1: 105 × 12
Step 1: 105 base is 100 and surplus is 5
12 base is 10 and surplus is 2
∴ 105 + 05 digits in the lower base is 1 so 1 digit is permitted in the second part of answer
12 + 2
Step 2: Cross – addition:
105
+ 2
125 again take care to line the numbers properly so as to get 125
∴ First part of answer may be 125
Step 3: Vertical multiplication : 05 × 2 10 but only 1 digit is permitted in the second part so 1
is shifted to first part and added to 125 so as to get 126
∴ 105 × 12 1260
Checking:
105 1 + 5 6 12 1 + 2 3
∴ LHS 6 × 3 18 1 + 8 9 0
∴ RHS 1260 1 + 2 + 6 90
Example 2: 1122 × 104
Step1: 1122 – base is 1000 and surplus is 122
104 – base is 100 and surplus is 4
∴ 1122 + 122
104 + 04 digits in lower base are 2 so 2digits are permitted in the second part of answer
Step 2: Cross – addition
1122
+ 04 again take care to line the nos. properly so as to get 1162
1162
slide 41: 40
∴ First part of answer may be 1162
Step 3: Vertical multiplication: 122 × 04 4 88
But only 2 – digits are permitted in the second part so 4 is shifted to first part and added to 1162
to get 1166 1162 + 4 1166
∴ 1122 × 104 116688
Can be visualised as: 1122 + 122
104 + 04
1162 / ← 4 88 116688
+ 4 /
Checking:
1122 1 + 1 + 2 + 2 + 6 104 1 + 4 5
∴ LHS 6 × 5 30 3
RHS 116688 6 + 6 12 1 + 2 3
As LHS RHS So answer is correct
Example 3: 10007 × 1003
Now doing the question directly
10007 + 0007 base 10000
× 1003 + 003 base 1000
10037 / 021 three digits per method in this part
∴ 10007 × 10003 10037021
Checking : 10007 1 + 7 8 1003 1 + 3 4
∴ LHS 8 × 4 32 3 + 2 5
RHS 10037 021 1 + 3 + 1 5
As LHS RHS so answer is correct
Try These:
i 1015 × 103 ii 99888 × 91 iii 100034 × 102 iv 993 × 97 v 9988 × 98 vi 9995 × 96 vii 1005
× 103 viii 10025 × 1004 ix 102 × 10013 x 99994 × 95
VINCULUM: “Vinculum” is the minus sign put on top of a number e.g. 5 41 63 etc. which means
–5 40 – 1 60 – 3 respectively
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 avoided.
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.
slide 42: 41
Converting from positive to negative form or from normal to vinculum form:
Sutras: All from 9 the last from 10 and one more than the previous one
9 1 1 i.e. 10 – 1 8 12 7 13 6 14 19 21 29 3 1
28 32 36 44 40 – 4 38 42
Steps to convert from positive to vinculum form:
1 Find out the digits that are to be converted i.e. 5 and above.
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.
Numbers with several conversions:
159 241 i.e. 200 – 41
168 232 i.e. 200 – 32
237 243 i.e. 240 – 7
1286 1314 i.e. 1300 – 14
2387129 24 1313 1
here only the large digits are be changed
From vinculum back to normal form:
Sutras: “All from 9 and last from ten” and “one less than then one before”.
1 1 09 10 – 1 13 07 10 – 3 24 16 20 – 4 241 200 – 41 159 162 160 – 2 158
222 200 – 22 178 1314 1300 – 14 1286 2413131 2387129
can be done in part as
13 1 130 – 1 129 and 2413 2400 – 13 2387
∴
2413131 2387129.
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
Try These: Convert the following to their vinculum form:
i 91 ii 4427 iii 183 iv 19326 v 2745 vi 7648 vii 81513 viii 763468 ix 73655167 x
83252327
Try These: From vinculum back to normal form.
i 14 i 21 iii 23 iv 231 v 172 vi 1413 vii 2312132 viii 241231
ix 632233 1 x 14142323
slide 43: 42
When one number is above and the other below the base
Example1: 102 × 97
Step 1: Here base is 100
102 + 02 02 above base i.e. 2 surplus
97 – 03 03 below base i.e. 3 deficiency
Step 2: Divide the answer in two parts as 102 / + 02
97 / – 03
Step 3: Right hand side of the answer is + 02 × – 03 – 06 06
Step 4: Left hand side of the answer is 102 – 3 99 97 + 02 same both ways
∴ 102 × 97 9906 9894 i.e. 9900 – 6 9894
Checking: 102 1 + 2 3 9 7 7
∴ L.H.S. 3 × 7 21 1 + 2 3
∴ R.H.S 9894 8 + 4 12 1 + 2 3
As L.H.S. R.H.S. So answer is correct
Example 2 : 1002 × 997
1002 + 002 006 1000 – 6 994 and 1 carried from 999 to 999 reduces to 998
997 – 003
999 006
∴ 1002 × 997 998 994
When base is not same:
Example1: 988 × 12
988 – 012 base is 1000 deficiency 12
12 + 2 base is 10 surplus is 2 1 digit allowed in R.H.S.
1188 – 2 024
1186 24
∴ 988 × 12 1186 4 11856 because 4 10 – 4 6
Checking: 9 88 8 + 8 16 1 + 6 7 12 1 + 2 3
∴ LHS 7 × 3 21 2 + 1 3
R.H.S 11856 1 + 5 + 6 12 1 + 2 3
As LHS RHS So answer is correct
Example 2: 1012 × 98
1012 1012 + 012 base is 1000 12 surplus +ve sign
– 02 98 – 02 base is 100 2 deficiency –ve sign
992 992 24 As 012 × – 02 – 24 2 digits allowed in RHS of
slide 44: 43
Answer
∴ 1012 × 98 99224 99176 As 992200 – 24 99176
Checking:1012 1 + 1 + 2 4 98 8
LHS 4 × 8 32 3 + 2 5
RHS 99176 1 + 7 + 6 14 1 + 4 5
As RHS LHS so answer is correct
Try These:
i 1015 × 89 ii 103 × 97 iii 1005 × 96 iv 1234 × 92 v 1223 × 92 vi 1051 × 9 vii 9899 × 87
viii 9998 × 103 ix 998 × 96 x 1005 × 107
Sub – base method:
Till now we have all the numbers which are either less than or more than base numbers. i.e.10 100 1000
10000 etc. now we will consider the numbers which are nearer to the multiple of 10 100 10000 etc.
i.e. 50 600 7000 etc. these are called subbase.
Example: 213 × 202
Step1: Here the sub base is 200 obtained by multiplying base 100 by 2
Step 2: R. H. S. and L.H.S. of answer is obtained using base method.
213 + 13
202 + 02
215 13 × 02 26
Step 3: Multiply L.H.S. of answer by 2 to get 215 × 2 430
∴ 213 × 202 43026
∴
Example 2: 497 × 493
Step1: The Subbase here is 500 obtained by multiplying base 100 by 5.
Step2: The right hand and left hand sides of the answer are obtained by using base method.
Step3: Multiplying the left hand side of the answer by 5.
497 –03
493 –07
Same 497–07 490 21
493 – 03 490
490 × 5
2450
∴ 497 × 493 245021
slide 45: 44
Example 3: 206 × 197
Subbase here is 200 so multiply L.H.S. by 2
206 + 06
197 – 03
206 – 3 203 –18
197 + 06 203 × 2 18
406
∴ 206 × 197
40618
40582
Example 4: 212 × 188
Sub – base here is 200
212 + 12
188 12
200 – 12 200 144
188 + 12 200
× 2
400 –1 399
∴ 212 × 188 399 44 39856
Checking:11 – check method
+ – +
2 1 2 2 + 2 – 1 3
+ – +
1 8 8 1 – 8 + 8 1
L.H.S. 3 × 1 3
+ – + – +
R.H.S. 3 9 8 5 6 3
As L.H.S R.H.S. So answer is correct.
Try these
1 42 × 43 2 61 × 63 3 8004 × 8012 4 397 × 398 5 583 × 593
6 7005 × 6998 7 499 × 502 8 3012 × 3001 9 3122 × 2997 10 2999 × 2998
Doubling and Making halves
Sometimes while doing calculations we observe that we can calculate easily by multiplying the number
by 2 than the larger number which is again a multiple of 2. This procedure in called doubling:
slide 46: 45
35 × 4 35 × 2 + 2 × 35 70 + 70 140
26 × 8 26 × 2 + 26 × 2 + 26 × 2 + 26 × 2 52 + 52 + 52 + 52
52 × 2 + 52 × 2 104 × 2 208
53 × 4 53 × 2 + 53 × 2 106 × 2 212
Sometimes situation is reverse and we observe that it is easier to find half of the number than calculating
5 times or multiples of 5. This process is called
Making halves:
4. 1 87 × 5 87 × 5 × 2/2 870/2 435
2 27 × 50 27 × 50 × 2/2 2700/2 1350
3 82 × 25 82 × 25 × 4/4 8200/4 2050
Try These:
1 18 × 4
2 14 × 18
3 16 × 7
4 16 × 12
5 52 × 8
6 68 × 5
7 36 × 5
8 46 × 50
9 85 × 25
10 223 × 50
11 1235 × 20
12 256 × 125
13 85 × 4
14 102 × 8
15 521 × 25
Multiplication of Complimentary numbers :
Sutra: By one more than the previous one.
This special type of multiplication is for multiplying numbers whose first digitsfigure are same and
whose last digitsfiguresadd up to 10100 etc.
Example 1: 45 × 45
Step I: 5 × 5 25 which form R.H.S. part of answer
Step II: 4 × next consecutive number
slide 47: 46
i.e. 4 ×5 20 which form L.H.S. part of answer
∴ 45 × 45 2025
Example 2: 95 × 95 9 × 10 90/25 → 5
2
i.e. 95 × 95 9025
Example 3: 42 × 48 4 × 5 20/16 → 8 × 2
∴ 42 × 48 2016
Example 4: 304 × 306 30 × 31 930/24 → 4 × 6
∴ 304 × 306 93024
Try These:
1 63 × 67
2 52 × 58
3 237 × 233
4 65 × 65
5 124 × 126
6 51 × 59
7 762 × 768
8 633 × 637
9 334 × 336
10 95 × 95
Multiplication by numbers consisting of all 9’s :
Sutras: ‘By one less than the previous one’ and ‘All from 9 and the last from 10’
When number of 9’s in the multiplier is same as the number of digits in the multiplicand.
Example 1 : 765 × 999
Step I : The number being multiplied by 9’s is first reduced by 1
i.e. 765 – 1 764 This is first part of the answer
Step II : “All from 9 and the last from 10” is applied to 765 to
get 235 which is the second part of the answer.
∴ 765 × 999 764235
When 9’s in the multiplier are more than multiplicand
Example II : 1863 × 99999
Step I : Here 1863 has 4 digits and 99999 have 5digits we suppose 1863 to be as 01863. Reduce this
by one to get 1862 which form the first part of answer.
slide 48: 47
Step II: Apply ‘All from 9 and last from 10’ to 01863 gives 98137which form the last part of answer
∴ 1863 x 99999 186298137
When 9’s in the multiplier are less than multiplicand
Example 3 : 537 x 99
Step I: Mark off two figures on the right of 537 as 5/37 one
more than the L.H.S. of it i.e. 5+1 is to be subtracted from the whole
number 537 – 6 531this forms first part of the answer
Step II: Now applying “all from 9 last from 10” to R.H.S.
part of 5/37 to get 63 100 – 37 63
∴ 537 x 99 53163
Try these
1 254 × 999 2 7654 × 9999 3 879 × 99 4 898 × 9999
5 423 × 9999 6 876 × 99 7 1768 × 999 8 4263 × 9999
9 30421 × 999 10 123 × 99999
Multiplication by 11
Example 1: 23 × 11
Step 1 : Write the digit on L.H.S. of the number first. Here the number is 23 so 2 is written first.
Step 2 : Add the two digits of the given number and write it in between. Here 2 + 3 5
Step 3 : Now write the second digit on extreme right. Here the digit is 3. So 23 × 11 253
OR
23 × 11 2 / 2+3 / 3 253
Here base is 10 so only 2 digits can be added at a time
Example 2: 243 × 11
Step 1: Mark the first second and last digit of given number
First digit 2 second digit 4 last digit 3
Now first and last digits of the number 243 form the first and last digits of the answer.
Step 2: For second digit from left add first two digits of the number i.e. 2 + 4 6
Step 3: For third digit add second and last digits of the number i.e. 3 + 4 7
So 243 × 11 2673
OR
243 × 11 2 / 2 + 4 / 4 + 3 / 3 2673
Similarly we can multiply any bigger number by 11 easily.
Example 3: 42431 × 11
slide 49: 48
42431 × 11 4 / 4 + 2 / 2 + 4 / 4 + 3 / 3 + 1 / 1 466741
If we have to multiply the given number by 111
Example 1: 189 × 111
Step 1: Mark the first second and last digit of given number
First digit 1 second digit 8 last digit 9
Now first and last digits of the number 189 may form the first and last digits of the answer
Step 2: For second digit from left add first two digits of the number i.e. 1 + 8 9
Step 3: For third digit add first second and last digits of the number to get 1 + 8 + 9 18 multiplying
by 111 so three digits are added at a time
Step 4: For fourth digit from left add second and last digit to get 8 + 9 17
As we cannot have two digits at one place so 1 is shifted and added to the next digit so as to get 189
× 111 20979
OR
1 1 + 8 9 1 + 8 + 9 8 + 9 9
9 + 1 18 1 7
1 + 1 2 1 0 18 + 1
1 9
∴189 × 111 20979
Example 2 : 2891 × 111
2 2 + 8 2 + 8 + 9 8 + 9 + 1 9 + 1
10 + 2 19 + 1 18 + 1 1 0 1
1 2 2 0 1 9
2891 × 111 320901
Try These:
1 107 × 11 2 15 × 11 3 16 × 111 4 112 × 111
5 72 × 11 6 69 × 111 7 12345 × 11 8 2345 × 111
9 272 × 11 10 6231 × 111.
Note: This method can be extended to number of any size and to multiplying by 1111 11111 etc. This
multiplication is useful in percentage also. If we want to increase a member by 10 we multiply it
by 1.1
slide 50: 49
General Method of Multiplication.
Sutra: Vertically and crosswise.
Till now we have learned various methods of multiplication but these are all special cases wherenumbers
should satisfy certain conditions like near base or sub base complimentary to each other etc. Now we
are going to learn about a general method of multiplication by which we can multiply any two numbers
in a line. Vertically and crosswise sutra can be used for multiplying any number.
For different figure numbers the sutra works as follows:
Two digit – multiplication
Example: Multiply 21 and 23
Step1: Vertical one at a time 2 1 1 × 3 3 3
2 3
Step2: Cross –wise two at a time 2 1 2 × 3 + 2 × 1 8
2 3 8 3
Step3: Vertical one at a time 2 1 2 × 2 4
2 3 4 8 3
∴ 21 × 23 483
Multiplication with carry:
Example: Multiply 42 and 26
Step1: Vertical 42 2 × 6 12 12
26
Step2: Crosswise 4 2 4 × 6 + 2 × 2 2
8 1
2
2 6 24 + 4 28
Step3: Vertical 42 4 × 2 8 8 8 2
26 + 2 1
10 2 9
∴ 42 × 26 1092
Three digit multiplication:
Example: 212 × 112


↓


↓


↓


↓
slide 51: 50
Step1: Vertical one at a time 212 2 × 2
112 4 4
Step2: Crosswise 2 1 2 2 × 1 + 2 × 1 4 4
two at a time 1 1 2 2 + 2 4
Step3: Vertical and crosswise 2 1 2
three at a time 1 1 2
2 × 2 + 2 × 1 + 1 × 1 4 + 2 + 1 7 7 4 4
Step4: cross wise 2 1 2 2 × 1 + 1 × 1 3 7 4 4
Two at a time 1 1 2 2 + 1 3
Step 5: vertical one at a time 2 1 2 2 × 1 2 2 3 7 4 4
1 1 2
∴ 212 × 112 23744
Three digits Multiplication with carry:
Example: 816 × 223
8 1 6 8 × 2 8 × 2 + 2 × 1 8 × 3 + 6 × 2 + 1 × 2 3 × 1 + 2 × 6 6 × 3 18
16 + 2 18 24 + 12 + 2 3 + 12 15
38
2 2 3 16
16 + 2 18 + 3 38 + 1 15 + 1 1 8
2 1 3 9 1 6
∴ 816 × 223 181968
Checking by 11 – check method
+ – + – +
8 1 6 14 – 1 1 3 3 – 1 2
+  +
2 2 3 3
∴ L.H.S. 3 × 2 6


↓


↓
↑


↓
↑





↓
slide 52: 51
 +  +  +  +
1 8 1 9 6 8 1 7 7 – 1 6
As L.H.S. R.H.S.
∴ Answer is correct
Try These:
1 342 × 514 2 1412 × 4235 3 321 × 53 4 2121 × 2112 5 302 × 415
6 1312 × 3112 7 5123 × 5012 8 20354 × 131 9 7232 × 125 10 3434 × 4321
Number Split Method
As you have earlier used this method for addition and subtraction the same may be done for multiplication
also.
For example :
263 26 3 Note : The split allows us to add 36 + 24
× 2 × 2 ×2 and 42 + 39 both of which can be done
52 6 mentally
Multiplication of algebraic expressions:
Sutra: Vertically and crosswise
Example1 : x + 3 x + 4
x + 3 x × x 4x + 3x 4 × 3
x
2
7x 12
x + 4 x
2
7 × 12
x
2
+ 7x + 12
Example2: 2x + 5 3x + 2
2x + 5
2x × 3x4x + 15x 10
6x
2
19x
3x + 2
6x
2
+19x + 10




↓




↓




↓




↓
slide 53: 52
Example3: x
2
+ 2x + 5 x
2
– 3x + 1
x
2
+ 2x – 5 x
4
–3x
3
+2x
3
x
2
+ 5x
2
–6x
2
2x –15x 5 × 1
x
2
– 3x + 1 –x
3
0 – 13x 5
x
4
– x
3
– 13x + 5 x
4
–x
3
0 –13x 5
Try These:
1 2x – 1 3x + 2
2 2x + 1 x
2
+ 3x – 5
3 5x+ 5 7x – 6
4 x + 5 x
2
– 2x + 3
5 x – 4 x
2
+ 2x + 3
6 x
2
+ 4x – 5 x + 5
7 x
3
– 5 x
2
+ 3
8 x
2
– 2x + 8 x
4
– 2
9 x
2
– 7x + 4 x
3
– 1
10 x
3
– 5x
2
+ 2 x
2
+ 1


↓


↓
slide 54: 53
Chapter 5 Squaring and square Roots
Square of numbers ending in 5 :
Sutra: ‘By one more than previous one”
Example: 75 × 75 or 75
2
As explained earlier in the chapter of multiplication we simply multiply 7 by the next number i.e. 8 to
get 56 which forms first part of answer and the last part is simply 25 5
2
. So 75 × 75 5625
This method is applicable to numbers of any size.
Example: 605
2
60 × 61 3660 and 5
2
25
∴ 605
2
366025
Square of numbers with decimals ending in 5
Example : 7.5
2
7 × 8 56 0.5
2
0.25
7.5
2
56.25 Similar to above example but with decimal
Squaring numbers above 50:
Example: 52
2
Step1: First part is calculated as 5
2
+ 2 25 + 2 27
Step2: Last part is calculated as 2
2
04 two digits
∴ 52
2
2704
Squaring numbers below 50
Example : 48
2
Step1: First part of answer calculated as: 5
2
– 2 25 – 2 23
Step2: second part is calculated as : 2
2
04
∴ 48
2
2304
Squaring numbers near base :
Example : 1004
2
Step1: For first part add 1004and 04 to get 1008
Step2: For second part4
2
16 016 asbase is 1000 a three digit no.
∴ 1004
2
1008016
slide 55: 54
Squaring numbers near sub  base:
Example 302
2
Step1: For first part 3 302 + 02 3 × 304 912 Here sub – base is 300 so multiply by 3
Step2: For second part 2
2
04
∴ 302
2
91204
General method of squaring:
The Duplex
Sutra: “Single digit square pair multiply and double” we will use the term duplex` D’ as follows:
For 1 figureor digit Duplex is its squaree.g. D4 4
2
16
For2 digitsDuplex is twice of the product e.g. D34 2 3 x 4 24
For 3 digit number: e.g. 341
2
D3 3
2
9
D 34 2 3 × 4 24
D 341 2 3 × 1 + 4
2
6 + 16 22
9 4 2 8 1
D 41 2 4 × 1 8 2 2
D 1 1
2
1 116281
∴ 341
2
116281
Algebraic Squaring :
Above method is applicable for squaring algebraic expressions:
Example: x + 5
2
D x x
2
Dx + 5 2 x × 5 10x
D 5 5
2
25
∴ x + 5
2
x
2
+ 10x + 25
Example: x – 3y
2
D x x
2
Dx – 3y 2 x × – 3y – 6xy
D–3y –3y
2
9y
2
∴ x – 3y
2
x
2
– 6xy + 9y
2
Try these:
I 85
2
II 8
2
1
2
III 10.5
2
IV 8050
2
V 58
2
VI 52
2
VII 42
2
VIII 46
2
IX 98
2
X 106
2
XI 118
2
XII x + 2
2
XIII y – 3
2
XIV 2x – 3
2
XV 3y – 5
2
slide 56: 55
SQUARE ROOTS:
General method:
As 1
2
1 2
2
4 3
2
9 4
2
1 6 5
2
2 5 6
2
3 6
7
2
4 9 8
2
6 4 9
2
81 i.e. square numbers only have digits 145690 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
13 1 + 3 4 64 6 + 4 10 1 + 0 1 81 8 + 1 9 i.e. square number only have digit sums
of 1 4 7 and 9.
This means that square numbers cannot have certain digit sums and they cannot end with certain
figures or digits using above information 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 in spite of the fact that its digit sum is 4 and last figure is 9.
Square Root of Perfect Squares:
Example1: √5184
Step 1: Pair the numbers from right to left 5184 two pairs
Therefore answer is 2 digit numbers
7
2
49 and 8
2
64
49 is less than 51
Therefore first digit of square root is 7.
Look at last digit which is 4
As 2
2
4 and 8
2
64 both end with 4
Therefore the answer could be 72 or 78
As we know 75
2
5625 greater than 5184
Therefore √5184 is below 75
Therefore √5184 72
Example 2: √9216
Step 1: Pair the numbers from right to left 9216two pairs
Therefore answer is 2 digit numbers
9
2
81 and 10
2
100
81 is less than 92
Therefore first digit of square root is 9.
Look at last digit which is 6
As 4
2
16 and 6
2
36 both end with 6
Therefore the answer could be 94 or 96
slide 57: 56
As we know 95
2
9025 less than 9216
Therefore √9216 is above 95
Therefore √9216 96
General method
Example 1 : √2809
Step1: Form the pairs from right to left which decide the number of digits in the square root. Here
2 pairs therefore 2  digits in thesquare root
Step 2: Now √28 nearest squares is 25
So first digit is 5 from left
Step3: As 28 – 25 3 is reminder which forms 30 with the next digit 0.
Step 4: Multiply 2 with 5 to get 10 which is divisor 10 √2809
30
Now 3 × 10 30 30 Q R
10 3 0
Step 5: As 3
2
9 and 9 – 9 last digit of the number 0
∴ 2809 is a perfect square and √2809 53
Example 2:3249
Step1: Form the pairs form right to left which decided the number of digits in the square root. Here
2 pairs therefore 2 digits in the square root.
Step2: Now 32 25 5
2
so the first digit in 5 from left
Step 3: 32 – 25 7 is remainder which form 74 with the next digit 4
5 7
Step 4: Multiply 2 with 5 to get 10 which is divisor 10√3249
Now 74 Q R 7 4
107 4
Step5: 7
2
49 and 49 – 49 0 remainder is 4 which together with9 form49
∴ 3249 is a perfect square and √3249 57
Example 3: 54 75 6
Step1: Form the pairs from right to left therefore the square root of 54756 has 3digits.
Step2: 5 4 2
2
i.e. nearest square is 2
2
4
So first digit is 2 from left
Step3: As 5 – 4 1 is remainder which form 14 with the next digit 4.
slide 58: 57
Step4: Multiply 2 with 2 to get 4 which is divisor
2
4 5
1
4
2
75 6 Now 14 Q R
4 3 2
Step 5: Start with remainder and next digit we get 27.
Find 27 – 3
2
27 – 9 18 square of quotient
234
Step 6: 18 Q R 4 5
1
4
2
7
2
5
1
6
4 4 2
Now 25 – 3 × 4 × 2 25 – 24 1
1 Q R
4 0 1
16 – 4
2
16 – 16 0
∴ 54756 is a perfect square and so √5 4 7 5 6 234
Try These:
1. 2116 2. 784
3. 6724 4. 4489
5. 9604 6. 3249
7. 34856 8. 1444
9. 103041 10. 97344
slide 59: 58
CHAPTER6 DIVISION
Defining the Division terms
There are 16 balls to be distributed among 4 people How much each one will get is a problems of division.
Let us use this example to understand the terms used in
division.
Divisor: —Represent number of people we want to
distribute them or the number that we want to divide by. Here
the divisor is 4.
Dividend: Represents number of balls to be divided 16
in this case.
Quotient:Represents the number of balls in each part 4
is this case.
Remainder:What remains after dividing in equal parts
0 in this case
The remainder theorem follows from the division example
above and is expressed mathematically as follows.
Divided Divisor × Quotient + Remainder
The remainder theorem can be used to check the Division
sums in Vedic Mathematics as described in the following
sections.
Different methods are used for dividing numbers based
on whether the divisor is single digit numbers below a base
above a base or no special case.
Special methods of Division.
Number splitting
Simple Division of Divisor with single digits can be done using this method.
Example:The number 682 can be split into
6/82 and we get 3/41 because
6 and 82 are both easy to halve
Therefore 682/2 341
Example : 3648/2 becomes
36/48/2 18/24 1824
Example:1599/3 we notice that 15 and 99 can be separately by 3 so
15/99/3 5/33 533
slide 60: 59
Example: 618/6 can also be mentally done
6/18/6 103 note the 0 here
Because the 18 takes up two places
Example: 1435/7
14/35/7 2/05 205
Example: 27483/3 becomes
27/48/3/3 9/16/1 9161
Practice Problem
Divided mentally Numbers Splitting
1 2656
2 2726
3 31899
4 61266
5 32139
6 22636
7 4812
8 64818
9 840168
10 5103545
Division by 9
As we have seen before that the number 9 is special and there is very easy way to divide by 9.
Example : Find 25 ÷ 9
25/9 gives 2 remainder 7
The first figure of 25 is the answer
And adding the figures of 25 gives the remainders 2 + 5 7 so 25 ÷ 9 2 remainder 7. It is easy
to see why this works because every 10 contains 9 with 1 left over so 2 tens contains 2 times with 2 left
over. The answer is the same as the remainders 2. And that is why we add 2 to 5 to get remainder. It can
happen that there is another nine in the remainder like in the next example
Example: Find 66 ÷ 9
66/9 gives 6 + 6 12 or 7 or 3
We get 6 as quotient and remainder 12 and there is another nine in the remainder of 12 so we add
the one extra nine to the 6 which becomes 7 and remainder is reduced to 3 take 9 from 12 We can also
slide 61: 60
get the final remainder 3 by adding the digits in 12. The unique property of number nine that it is one
unit below ten leads to many of the very easy Vedic Methods.
This method can easily be extended to longer numbers.
Example: 3401 ÷ 9 377 remainder 8
Step 1: The 3 at the beginning of 3401 is brought straight into the answer.
93401
3
Step 2: This 3 is add to 4 in 3401 and 7 is put down
93401
37
Step 3: This 7 is then added to the 0 in 3401 and 7 is put down.
93401
377
Step 4: This 7 is then added to give the remainder
9 340/1
377/8
Divided the following by 9
1 951
2 934
3 917
4 944
5 960
6 926
7 946
8 964
9 988
10 996
Longer numbers in the divisor
The method can be easily extended to longer numbers. Suppose we want to divide the number 21 3423
by 99. This is very similar to division by 9 but because 99 has two 9’s we can get the answer in two digits
at a time. Think of the number split into pairs.
21/34/23 where the last pair is part of the remainder.
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Step 1: Then put down 21 as the first part of the answer
9921/34/23
21
Step 2: Then add 21 to the 34 and put down 55 as next part
9921/34/23
21/55
Step 3: Finally add the 55 to the last pair and put down 78 as the remainder
9921/34/23
21/55/78
So the answer is 2155 remainder 78
Example: 12/314 ÷ 98 1237
Step 1: This is the same as before but because 98 is 2 below 100 we double the last part of the answer
before adding it to the next part of the sum. So we begin as before by bringing 12 down into the
answer.
98 12/13/14
12
Step 2: Then we double 12 add 24 to 13 to get 37
98 12/13/14
12/37
Step 3: Finally double 37 added 37 × 2 74 to 14
9812/13/14
12/37/88 1237 remainder 88.
It is similarly easy to divide by numbers near other base numbers 100 1000 etc.
Example: Suppose we want to divide 236 by 88 which is close to 100. We need to know how many
times 88 can be taken from 235 and what the remainder is
Step 1: We separate the two figures on the right because 88 is close to
100 Which has 2 zeros
88 2/36
Step 2: Then since 88 is 12 below 100 we put 12 below 88 as shown
88 2/36
Step 3: We bring down the initial 2 into the answer
88 2/36
12
2
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Step 4: This 2 is multiplies Haggled 12 and the 22 is placed under the 36 as
Shown
88 2/36
12 2 / 24
Step 5: We then simply add up the last two columns.
88 2/36
12 2 r 60
In a similar way we can divide by numbers like 97 and 999.
Practice problems
Divide the following using base method
1 121416 by 99
2 213141 by 99
3 332211 by 99
4 282828 by 99
5 363432 by 99
6 11221122 by 98
7 3456 by 98
Sutra: Transpose and Apply
A very similar method allows us to divide numbers which are close to but above a base number.
Example: 1479 ÷ 123 12 remainder 13
Step 1: 123 is 23 more than base 100
Step 2: Divide 1479 in two columns therefore of 2digit each
Step 3: Write 14 down
Step 4: Multiply 1 by 23 and write it below next two digits. Add in the
Second column and put down 2.
Step 5: Add multiply this 2 the 2 3 and put 46 then add up last two
Columns
123 14 78
23 23
46
12/02
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Straight Division
The general division method also called Straight division allows us to divide numbers of any size by
numbers of any sine in one line Sri BharatiKrsnaTirthaji called this “the cowing gem of V edic Mathematics”
Sutra:  ‘vertically and crosswise’ and ‘on the flag’
Example: Divide 234 by 54
The division 54 is written with 4 raised up on the flag and a vertical line is drawn one figure from the
right hand end to separate the answer 4 from the remainder 28
23
3
4
5
4
20 16
428
Step 1: 5 into 20 goes 4 remained 3 as shown
Step 2: Answer 4 multiplied by the flagged 4 gives 16 and this 16 taken from 34 leaves the remainder
28 as shown
Example: Divide: 507 by 72
50
1
7
7 2
49 14
7 3
Step 1: 7 into 50 goes 7 remainder 1 as shown
Step 2: 7 times the flagged 2 gives 14 which we take from 17 to have remainder of 3
Split Method
Split method can be done for division also. For example :
6234 ÷ 2 62 34
÷ 2 ÷2
31 17
The split may require more parts.
30155 ÷ 5 30 15 5
÷5 ÷5 ÷5
6 03 1 6031
244506 ÷ 3 24 45 06
÷3 ÷3 ÷3
8 15 02 81502
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Practice Question
Divide the following using straight division
1 209 ÷ s52 2 621 ÷ 63
3 503 ÷ 72 4 103 ÷ 43
5 74 ÷ 23 6 504 ÷ 72
7 444 ÷ 63 8 543 ÷ 82
9 567 ÷ 93 10 97 ÷ 28
11 184 ÷ 47 12 210 ÷ 53
13 373 ÷ 63 14 353 ÷ 52
15 333 ÷ 44 16 267 ÷ 37
17 357 ÷ 59 18 353 ÷ 59
19 12233 ÷ 53
Books for Reference
1. Sri BharatiKrsnaTirthaji “Vedic Mathematics” published by MotilalBanarsidass 1965. ISBN 81
20801636.
2. Williams K.R. “Discover Vedic Mathematics.” Vedic Mathematics Research Group 1984. ISBN 1
869932013
3. Williams K.R. and M. Gaskell “The Cosmic Calculator”. MotilalBanarsidass 2002.ISBN 81208
18717.
4. Nicholas A.P. Williams J. Pickles. “Vertically and Crosswise”. Inspiration Books 1984. ISBN 1
902517032.