This PDF contains :
About the Author,
Vedic Mathematics: An Introduction,
Maths is interesting!,
1. Miscellaneous Simple Method,
2. Criss – Cross System of Multiplication,
3. Squaring Numbers,
4. Cube Roots of Perfect Cubes,
5. Square Roots of Perfect Squares,
6. Base Method for Multiplication,
7. Base Method for Squaring,
8. Digit-Sum Method,
9. Magic Squares,
10. Dates & Calendars,
11. General Equations,
12. Simultaneous Linear Equations,
13. Square Roots of Imperfect Squares,
14. Cubing Numbers,
15. Base Method of Division,
16. Division (Part Two),
Appendix A : Multiplication of five digit numbers, Appendix B : Multiplication of Algebraic Identities,
Appendix C : Zeller’s Rule (To find the day on any date),
Appendix D : Pythagorean Values,
Appendix E : Divisibility Tests,
Appendix F : Raising to fourth and higher powers, Appendix G : Co-ordinate Geometry,
Frequently Asked Questions (FAQ’s),
WHAT THE MEDIA IS SAYING ABOUT
‘For a 19 year old Dhaval Bathia packs in quite a cerebral punch. A unique blend of talent, insight,
hard work and sheer determination. A ROCKER’
— Education Times, The Times of India
A WHIZ KID. Dhaval Bathia needs just a split second to recall details of 100 year old calendars,
hundreds of phone numbers and to solve complex arithmetical problems.
— Bombay Times, The Times of India
A unique achiever. We wish him all the best for his future projects.
— Lohana Shakti
Dhaval Bathia is simply JHAKAAS. His workshops are commendable.
— Maharashtra Times
By using simple word formulae; he manages to solve complex arithmetical calculations within
A young achiever. His seminars receive a tremendous response even from the CORPORATE WORLD.
— Education World
20-year-old Dhaval Bathia is the king of Vedic mathematics
White-lightning success in Vedic Mathematics
‘A college student who is a Mathematics faculty!’…..he has coached many MBA students too….
‘He can recollect all facts of any subject by reading them just once!’
‘The manner in which the student community has purchased his book ‘How To Top Exams and
Enjoy Studies’ might well assure him a career in counseling.’
— Navbharat Times
After attending his seminars, students have found a sharp increase in their confidence level and their
performance in exams has also improved
Dhaval Bathia’s seminar was definitely a high point. He tantalized the crowd with his mathematical
— JAM Magazines
Our country not only produces great talent but also nurtures it so that it reaches its fullest potential
which enables it to reach the sky. Since times immemorial India has not only given great intellectuals,
sportspersons, revolutionaries, socialists but also great artists, painters, scientists to the countrymen
and society in general. Adding another name to this great stardom is such an amazing talent born in
the Bathia family of Maharashtra. Shri Dhaval Bathia has researched on some unique and exciting
techniques of Vedic Mathematics.
— Jazbaat Magazine
This youngster has mastered Mathematics the Vedic way.
— The Indian Express
He solves problems on addition, subtraction, multiplication, division, roots and any other such
— Gujarat Samachar
The book ‘How To Top Exams and Enjoy Studies’ is written in easy language and any student can
follow it effortlessly. It tells you how to make studies enjoyable and at the same time focus on your
goals and objectives.
A Young Achiever. His clientele includes leading companies like BSES, Ramratan Group of
companies, Welspun, Challenge Finance etc.
The scientific systems in the book (How to Top Exams and Enjoy Studies) create a paradigm shift
from hard work to smart work. The whole emphasis is laid on the fact that education is a process to be
enjoyed and cherished.
‘He is an inspiration to the youth’
— FM Gold
‘Dhaval Bathia is a revolution...’
— S TV
Vedic Mathematics Made Easy
Published by Dhaval Bathia
Copyright by Dhaval Bathia
About the Author
Vedic Mathematics: An Introduction
Maths is interesting!
1. Miscellaneous Simple Method
2. Criss – Cross System of Multiplication
3. Squaring Numbers
4. Cube Roots of Perfect Cubes
5. Square Roots of Perfect Squares
6. Base Method for Multiplication
7. Base Method for Squaring
8. Digit-Sum Method
9. Magic Squares
10. Dates & Calendars
11. General Equations
12. Simultaneous Linear Equations
13. Square Roots of Imperfect Squares
14. Cubing Numbers
15. Base Method of Division
16. Division (Part Two)
Appendix A : Multiplication of five digit numbers
Appendix B : Multiplication of Algebraic Identities
Appendix C : Zeller’s Rule (To find the day on any date)
Appendix D : Pythagorean Values
Appendix E : Divisibility Tests
Appendix F : Raising to fourth and higher powers
Appendix G : Co-ordinate Geometry
Frequently Asked Questions (FAQ’s)
Since a very young age, I had a passion for learning mindpower sciences. In my high school years I
used to read books on NLP, SILVA, Psycho-cybernetics, Mind-mapping, quick calculations, etc. Most
of my time was spent reading books and the remaining time was spent impressing friends with my
mental abilities. I used to do (so-called) strange things like multiplying three-digit numbers in my
mind, memorizing a pack of well shuffled cards, predicting the days on which dates in the century
would fall and so on. These activities were more as a source of entertainment to me than anything
else. In my wildest dreams I had never imagined that one day I would become a faculty on these
People often ask me how I as a college student landed up in this world of Vedic mathematics and
training people in memory improvement and study skills. Well, honestly speaking, I am not sure
myself exactly when I started these activities — because there was no specific starting point but rather
a process that evolved.
When I finished my tenth standard exams, I had a long vacation before college would begin. With
the desire that I do something productive in my days of leisure, my father brought some books for me
on quick calculations. He knew that I was very good at mental mathematics and so probably he
thought that I needed to polish my abilities and some good books would help. I read the books with
keen interest and understood the systems. But as time passed by, I completely forgot about it.
About a year later, one of our family friends who is the Principal of a school came to my place.
She asked me whether I would do a seminar for her standard ten students on quick calculation and
especially Vedic Mathematics. Obviously, my first reaction was a straight ‘No’ because I was anxious
about addressing an audience. Not that I suffered from stage-fear or anything like that, but the thought
of formally teaching a whole group of similarly aged students was not instantly acceptable. The lady
did not try to convince me any further but only said to me, “Think again, you are losing an
The next day I called her up with an affirmative answer. The seminar was arranged after three
days. When I entered the auditorium, I was filled with some skepticism and pessimism as among my
audience were students who played cricket with me on weekends and now I had to teach them a
subject which they hated with all their heart!
But, as it is often said, you never know what destiny has in store for you….
I vividly remember the two hours of that introductory lecture. As I unfolded the techniques to the
audience, I could actually see that they were bedazzled with its workings. As the lecture progressed,
the excitement and enthusiasm increased amongst the crowd and even their mathematics professors
joined them. They became deeply engrossed in the lecture and never realized when the two hours had
passed by. I received some wonderful feedback from the students. They seemed to be all charged up.
And believe it or not, one girl actually came and told me that she lost all her fear for mathematics
after attending my lecture. The professors asked me to come once again. All in all, it was indeed the
most memorable experience of my life.
That same night, when I came home I became absolutely thoughtless for a moment. Then I
pondered over the day’s events and realized that it was not me who deserved the credit for the
wonderful feedback that the students gave but it was the system, it was Vedic Mathematics!
From the very next day, I became sort-of famous in my community with the word spreading
around that a very young boy is teaching “magic” in mathematics. Since then, phone calls started
coming from parents, teachers and other schools. I was so overwhelmed with this wonderful response
that I was suddenly filled with a tremendous sense of gratitude towards Vedic mathematics. I thought,
since this science has given me such a lot of happiness and a feeling of self-fulfillment, it is only fair
that I engage in the cause of making it popular and easily accessible to the general public.
And since then, there has been no looking back….
Four years have passed since I gave my first seminar and till date I have conducted hundreds of
sessions interacting with thousands of people from all parts of the country. Many leading schools and
colleges have implemented the systems that I teach. The management of these institutions has
reported that students have shown praiseworthy improvement in their exam performance after
learning these systems.
As time passed by, I started giving special training to people on topics like ‘how to improve your
memory’ where I taught how to remember dates in history, geographical maps, chemical equations,
long numbers, faces of people, etc.; then there were seminars on study skills, where topics like using
the subconscious mind for study, frequent revision techniques, and the art of writing answers in exams
Since most of the content of my seminars was oral, I often received complaints from people that
they failed to retain the techniques in the long run. I was urged to publish my systems in a written
format so that it could be used as a reference work whenever required. They urged me to write a book
on the techniques that I teach. Thus, out of popular demand I published my first book at the age of 19
entitled How To Top Exams & Enjoy Studies.
Many readers of my first book wrote to me that they liked the introductory techniques of Vedic
Mathematics in my book but they wanted to learn the systems in detail. They insisted that I deal with
the subject in a comprehensive manner. Initially, I was a bit skeptical about writing another book
when I had just published my first. Being a college student, there was a lot of commitment involved in
my mainstream studies, extra-curricular activities and pursuing my professional degrees. Thus, I had
to wait for a while before I commenced writing the typescript of this book.
My very first concern when I started writing this book on Vedic mathematics was how should I
deal with the subject. There were tons of books already present in the market and the Internet was also
filled with innumerable research articles on Vedic mathematics. When such a lot of information was
already available, I had to start searching for a strong reason to write another book. Not that I had to
prove myself to someone or something like that, but a strong purpose would help me give a focus to
the contents of the book. I would be able to write the book in such a way that it had a certain appeal
and meaning to the reader. I was not interested in writing another book on the same subject, but was
keen on giving a strong identity to my book.
Upon discussion with other colleagues who teach Vedic Mathematics and on the basis of feedback
received from the participants of my seminars, I made the following realizations:
• Most of the students who wanted to learn Vedic Mathematics were students appearing for
competitive exams like MBA-CAT, CET, GMAT, UPSC, etc. and they wanted to learn Vedic
Mathematics so that it would help them crack these exams. They wanted to learn techniques which
would help them solve mathematical problems in the least time possible.
• Secondly, many schools had started implementing the methods of Vedic Mathematics in
classroom lectures. They needed a book that would make the entire subject so simple that even an
average student would be able to understand it. The schools wanted a book which would guide
beginners and help them understand the subject right from the basics.
With these concerns in mind, I included specific techniques mentioned in the book which are of
immense help to students appearing for competitive exams. For example, the study of cube-roots is
divided into two parts — perfect cubes and imperfect cubes. Similarly, the study on square roots has
also been divided into two parts. Students while preparing for competitive exams will encounter
perfect cubes and perfect squares in different questions and they can find their roots instantly with the
techniques described independently in these chapters.
Although not a part of traditional Vedic Mathematics, I have also included techniques on magic
squares, remembering calendars, calculating dates, etc. as problems on these are asked in such exams.
Thus, this book not only has techniques of Vedic Mathematics but also includes many other
methods which can be used by students appearing for competitive exams. My concern was to provide
utility to such students and thus I have included some of the world’s best techniques on quick
calculation. Not all these techniques are a part of Vedic Mathematics but I have not refrained from
The second concern was that of younger students who generally came up with the complaint that
they found it very difficult to understand Vedic Mathematics. This concern of theirs prompted me to
deal with the subject from the very basic level. I have taken nothing for granted and even the concept
of squaring/cubing a number has been explained with illustrative examples to ensure that there is no
doubt whatsoever in the mind of the reader.
I have used the simplest possible language throughout the book. The book is divided in three parts
— Basic, Intermediate and Advance. The chapters are arranged in the ascending order of the difficulty
involved in the calculation procedure.
Thirdly, every chapter is concluded with a small summary and practice exercise to enable the
student to test himself. The problems in the exercise are arranged in the increasing order of difficulty.
Questions in PART-A of any exercise will be simple; questions in PART-B will be slightly advanced;
questions in PART-C will be highly advanced. Thus, it will facilitate the students to practice problems
with different levels of difficulty and check the answers with the chapter-wise solutions given at the
end of the book.
Fourthly, I have used a lot of visual effects and diagrams along with explanations which will aid
the student to understand better.
And lastly, I have written the book in a conversational style so that you can understand it entirely
by yourself and will not feel the need for any external support to explain the techniques. Throughout
the book, you will feel that I am talking to you and will feel the presence of a teacher while reading
A section dealing with Frequently Asked Questions (FAQ’s) on Vedic Mathematics appears
towards the end of the book. You will find an answer to most of your queries in that section.
Do send me any of your comments and suggestions for the improvement of the book.
And there are those who give and know not pain in giving, nor do they seek joy, nor give with
mindfulness of virtue;They give as in yonder valley the myrtle breathes its fragrance into space
Through the hands of such as these God speaks and from behind their eyes
HE smiles upon the earth
— KAHLIL GIBRAN
I take this opportunity to express my sincere gratitude:
To my mother and father for everything and much more.
To my uncles Shaileshbhai, Pankajbhai and Shirishbhai for their guidance.
To my aunties Sheetal, Harsha and Jyotiben for their caring.
To my cousins Anand, Janak, Haseet, Gunja and Brinda for their support.
To my grandmother for allowing me to dip in the ocean of her wisdom.
To all relatives for nourishing me with apt compliments and constructive criticism.
To Pratik Sonawala for idolizing hardwork, patience and dedication.
To my sister for always being by my side like a shadow.
To my teachers at St. Anne’s and Narsee Monjee for chiseling me like master architects.
To Reuben, Kanwardeep, Abhishek, Rishi, Hemang, Nainil and all my friends for every ‘pat on my
back’ which acted like fuel for going ahead.
To my Reiki teacher, Usha aunty for attuning me to divine energy.
To my family at ‘Landmark Education Corporation’ for their amazing insights into life.
To all people from the media for acting like the breeze that carried the fragrance of my messages
across vale and brook.
To all readers of my first book, HOW TO TOP EXAMS & ENJOY STUDIES, for altering their sails
in the direction of my wind.
To all people who have created obstacles in my work: had you not given me autumn, how would I
have realized the value of spring?
To my Gurus. They showered their rain of knowledge on me when I was a bud so that I blossomed
and become a beautiful flower. And even after I had blossomed, their teachings stayed with me in the
form of dewdrops.
To my publishers. I gave them a grain of sand and they made it a pearl.
And to you, dear reader, for the opportunity to create a symphony of our musical notes….
ABOUT THE AUTHOR
DHAVAL BATHIA is a twenty-one year old student. He graduated from the prestigious Narsee
Monjee College of Mumbai. He has been accredited with the titles of the Best Student, Best Writer,
Outstanding Orator and Outstanding Organizer in his college.
He published his first book — ‘How To Top Exams & Enjoy Studies’ at the age of 19 and it has
become a bestseller.
Dhaval is one of the world’s youngest faculties in the field of mind power sciences. He is also one
of the youngest faculties to train students in Vedic Mathematics. He gave his first seminar on Vedic
Mathematics at the age of 16. At seventeen, he started training Vedic Mathematics and other mind-
power sciences to professors in various educational institutions and urged them to spread his systems
amongst the student community. He has been invited as a guest speaker in many different schools,
colleges and educational institutions across the country. He also conducts seminars for Corporates and
is possibly India’s youngest corporate trainer. He started training corporates at the age of 18 and his
clientele includes a number of leading companies like BSES (now Reliance Energy), WELSPUN,
RAMRATAN GROUP OF COMPANIES etc. He has been invited as a visiting faculty in many
management institutions and has trained MBA aspirants in Mathematics.
Dhaval started the world’s first television serial on Vedic Mathematics spanning millions of
viewers through his show. He also pioneered the concept of Vedic Mathematics and other study skills
on WAP Technology.
Apart from oratory, he has a passion for writing. Dhaval’s articles have appeared in a number of
newspapers and magazines. He writes regularly for different websites. He also edits an online
newsletter called ‘SMART IDEAS’ which has a huge global subscription base.
He is a practitioner of REIKI and uses his spiritual knowledge in his daily routine which includes,
self-hypnosis, meditation, sub-conscious programming and visualization.
His pleasing personality and sharp oratory combined with his sense of humour makes him a
favourite of his students.
You can write to him at [email protected]
VEDIC MATHEMATICS: AN
You must have heard of Vedic Mathematics but wondered what this meant and what it was all about.
Vedic Mathematics is the collective name given to a set of sixteen mathematical formulae
discovered by Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaj. Each formula deals with a
different branch of Mathematics. These sixteen formulae can be used to solve problems ranging from
arithmetic to algebra to geometry to conics to calculus. The formulae are complete by themselves and
applicable to virtually any kind of mathematical problems. Complex mathematical questions which
otherwise take numerous steps to solve can be solved with the help of a few steps and in some cases
without any intermediate steps at all! And these systems are so simple that even people with an
average knowledge of mathematics can easily understand them. Once the formula (which is called
‘sutra’ in Vedic Mathematics) is learnt, it can be applied to a certain category of problems, such as
multiplication, division, fractions, and so on.
The founder of Vedic Mathematics, Jagadguru Swami Bharati Krishna Tirthaji Maharaj, was born
in 1884. He was an exceptionally brilliant student right from childhood and always secured first place
in all his classes throughout his education. He had an extraordinary proficiency over Sanskrit, English,
mathematics and other subjects. After having completed his B.A. and M.A., he started writing articles
on different subjects. Since then much of Swamiji’s life was spent in studying the different spiritual
sciences and Vedantic philosophy. He used to go to the nearby forests and practice various forms of
meditation and ponder deeply on these subjects. The formulae (sutras) of Vedic Mathematics were
discovered by him in the same manner, viz. through deep intuitive meditation.
It is important to note that the word ‘Vedic’ is used as an adjective in connection with the Vedas.
We all know that there are four Vedas: Rigveda, Samaveda, Yajurveda and Atharvaveda. Each of these
deals with a specific set of subjects. Out of these Vedas, the ‘Atharvaveda’ dealt with subjects of
architecture, engineering and general mathematics. However, according to historians, what we
generally call Vedic Mathematics in parlance with the findings of Swamiji is not mentioned anywhere
in the Vedas, not even the Atharvaveda which deals with mathematical subjects. Then, the obvious
question arises: Why is the word Vedic used to describe this discovery when it has no direct relation
with the Vedas?
In fact, the use of the word Vedic as an adjective to the systems of Swamiji has aroused a certain
amount of controversy. However, the followers and disciples of Swamiji have strong arguments.
According to Swamiji, the word veda means the fountainhead and illimitable storehouse of all
knowledge. This means that the Vedas should have all the knowledge that is needed by a man for his
perfect all-round success. Thus, the word Vedic was used by Swamiji as an adjective to his discovery.
It is said that Swamiji discovered these Vedic formulae between 1911-1918. I have used the word
‘discovered’ because it has a particular significance in the context of Swamiji’s work. Unlike the other
sciences, which are discovered empirically, the discovery of Vedic Mathematics is very intuitive.
Swamiji did not discover the formulae of Vedic Mathematics by deduction but he discovered them out
of deep meditation. It is said that he had practiced meditation in deep silence around the forests of
Sringeri for a period of eight years, where some higher source of intelligence revealed to him these
‘secrets’. Swamiji was a spiritually realized personality and so it was possible for him to gain insights
through intuitive revelation. Swamiji also told his disciples that he had reconstructed the sixteen
formulae from Atharvaveda after tireless research and deep ‘tapas’. Thus, whereas one does not find
the sutras in the Atharvaveda, they were actually reconstructed on the basis of intuitive revelation
from scattered references of some content of the Atharvaveda.
Swamiji had written sixteen manuscripts and each manuscript dealt with each of the sixteen sutras.
These manuscripts were kept by him at the place of one of his disciples. However, the manuscripts
were lost from their place of deposit! There was no other written evidence of the sutras apart from the
lost manuscripts. Swamiji was not much disturbed at the loss and said that he could rewrite the whole
chunk of research again out of his memory.
In 1957, Swamiji wrote an introductory volume to the sixteen sutras. He planned to write further
volumes but he subsequently developed cataract in both eyes and his failing health did not allow him
to fulfill his plans. As time passed his health continued to deteriorate, and in 1960 he achieved
Thus, because of the loss of the manuscripts the world was bereft of great knowledge which it
would have otherwise greatly cherished. What Swamiji has left behind (in comparison to his research)
is just the tip of the iceberg. He created the manuscript of his introductory work around 1958 in the
USA and his introductory work has scattered references to the sixteen sutras and sub-sutras.
Whatever little matter of Vedic Mathematics is available to us today is of such quality and genius
that it has taken the world by storm, and people have literally been bewildered by the workings of the
techniques. One can only wonder what would have happened to the mathematical world if his
complete works were accessible to us….
Maths is Interesting!
Every second person whom I meet tells me that he hates mathematics. Because they hate mathematics,
they get poor scores in their exams and the poor scores further aggravate their hatred. Thus it becomes
a vicious circle!
Over the years I realized that people often came to my seminars with a negative mindset towards
mathematics. Since an early age, their mind is conditioned to believe that maths is boring and
difficult. Thus, it becomes inevitable to create a ‘paradigm shift’ towards the subject. Unless they are
excited about the techniques that will follow, it becomes a Herculean task to arouse enthusiasm
towards the seminar.
With a desire to create a curiosity and yearning for the subject, we researched on a set of
techniques which we call ‘Mental Magic.’ These techniques are so exciting and powerful that they
leave the audience bewildered. They get tantalized with its workings and their entire attitude towards
mathematics changes in a few minutes. After the session on Mental Magic is over, the audience is all
eager, excited and ready to grasp the techniques of Vedic Mathematics. This really simplifies my
In this chapter, we are going to study many of these techniques which will change your attitude
towards mathematics. A few of these are trade secrets, but for the first time I am revealing them in my
In such sessions, I predict the date of birth of anybody from the audience. I also predict strange
things like how many brothers/sisters he has, how many children he has, how much money one has in
his pockets etc. without him disclosing any information! I also do a technique where I can get the
answer to a problem without even knowing the problem !
Sounds interesting ?! Read ahead……..
(A) How to predict a person’s Date Of Birth
With this technique you can predict the date of birth of any number of people simultaneously. You can
try this stunt with your family members, friends, relatives, colleagues and even in parties.
(a) Ask the people to take the number of the month in which they were born (January is 1, February
is 2 and so on…..)
(b) Next, ask them to double the number
(c) Add 5 to it
(d) Multiply it by 5
(e) Put a zero behind the answer
(f) Add their date of birth (If they are born on 5th January then add 5)
After the steps are over, ask them to tell you the final answer. And lo! Just by listening to their
final answer you can predict their date of birth!
From the answer that you get from each member of the audience,
• Mentally subtract 50 from the last two digits and you will have the date
• Subtract 2 from the remaining digits and you will have the month.
Thus, you will easily get his date of birth.
Let us suppose I was a member of the audience. My date of birth is 26th June. Then I would have
worked out the steps as follows:
Thus my final answer is 876. Now, let us see how we can deduct my date of birth from the final
answer. We will subtract 50 from the last two digits to get the date. Next, we will subtract 2 from the
remaining digits to get the month.
On similar lines if the total was 765, 1481 and 1071 the birthdates would be 15th May, 31th
December and 21st August respectively.
With the knowledge of this technique, you can predict the birth-date of hundreds of people
(Note: There are many such mathematical ways by which you can predict a person’s date of birth but
the method given above is one of the simplest methods)
(B) How to predict a person’s pocket money
This technique will help you to find how much money a person has in his pocket/wallet etc. It can be
tried on a group of people simultaneously.
(a) Ask him to take the amount he has in his pocket (just the rupees, ignore the paisa)
(b) Next, ask him to add 5 to it.
(c) Multiply the answer by 5
(d) Double the answer so obtained
(e) Finally, ask him to add his favourite one digit number (from 0-9)
After the steps are over ask them to tell you the final answer. And just by listening to the final
answer you will come to know the amount he has in his pocket!
• Ignore the digit in the unit’s place.
• From the remaining number, subtract 5 and you will come to know the amount he has in his
Let us suppose a person has 20 Rupees in his pocket. He would work out the steps as given below:
Thus, his final answer would be 257. Now, let us see how we can find out the amount he has from
the final answer. As mentioned earlier, we will ignore the digit in the unit’s place (in this case it is 7).
Now the remaining number is 25. From 25, we subtract 5 and get the amount as Rs. 20. And thus our
answer is confirmed.
Similarly if the total was 1052, 53 and 160 the amount would be 100, 0 and 11 Rupees
(From 1052, we ignore the last digit 2 and take only 105. From 105, we subtract 5 and get the
answer as 100 and so on…)
I use this technique in my seminars to give the audience some relief from complex calculations.
There are many ways by which we can predict the amount a person has in his pocket but I prefer this
technique because it neither involves any complex calculations nor is it too obvious for the audience
to guess the secret.
(C) How to find the answer without knowing the question!
Using this technique you can find the total of a set of five numbers without knowing the numbers.
In my seminars, I ask a member of the audience to give me a three-digit number. Let us suppose
someone he gives me the number as 801. Then, I write the number 801 and after leaving four lines I
write the final answer as 2799.
Thus, I have got the final answer without knowing all the numbers. Next, I ask him to give me
another three digit number. Let us suppose he gives me 354. Now, its my turn. Within one second, I
write the next number as 645.
Now, again it’s the turn of the audience. Let us suppose he gives me 800. Now, again it is my turn.
Before he blinks his eye, I write my number as 199.
Thus, I have all five numbers in place. When I ask him to check the total of the five numbers, he is
amazed to find out that it is 2799 which I had predicted by looking at the first number only !
Thus, I got the answer without knowing the question. Further, after every step I was writing my
three-digit number within one second (without any time for calculation)
Even you can do such ‘Magic Totals.’ You can ask the audience to give you any three-digit number
and based on that you can get the final total without looking at the other numbers.
• From the first number that the audience gives you, subtract 2 and always put 2 in the
beginning. This becomes your final answer. For example, if the number is 801, we subtract 2 and
get 799. Next, we put 2 in the beginning and our final answer becomes 2799. If the number is 567,
we subtract 2 and get 565. Next, we put 2 in the beginning and write the final answer as 2565.
Thus, you will have the final answer.
• Next, we will see the secret of finding the subsequent steps. Subtract each digit of the number
that the audience gives you from 9 and you will have your number. In the example mentioned
above, when the audience gives me 354, I subtract each of the digits 3, 5 and 4 from 9 and get my
answer as 6, 4 and 5. So my number is 645. When the audience gives me the number 800, I subtract
each of the digits 8, 0 and 0 from 9 and get my answer as 1, 9 and 9. So my number is 199.
In this way, you can have the intermediary steps.
Let us have a look at another example.
If the audience gives you a number as 600, you will subtract 2 from it (598) and put 2 in the
beginning. Thus, your final answer is 2598. Next, let us assume the audience gives you the number
481. We subtract each of the digits 4, 8 and 1 from 9 and get our answer as 518 instantly. Next, let us
suppose the audience gives you the number 909. We subtract each of the digits 9, 0 and 9 from 9 and
get our answer as 090. When you check the final total, you will be surprised to find that it is indeed
Thus, I have taught you three exciting techniques of mathematics which you can use to impress
your friends and colleagues. The intention of including this chapter in the book was twofold. First,
since most people have got habituated with the calculator, it is a good opportunity to tune your mind
to mental calculation. Secondly, the wonderful response that you will get by using these techniques on
people will definitely change your attitude towards mathematics.
Note: Since the techniques of this chapter fall beyond the scope of Vedic Mathematics, I have not
elaborated on them. However, if you are keen on learning many such ‘mental magic’ techniques like
predicting how much many brothers & sisters a person has, what is a person’s year of birth and such
other techniques, then fill the feedback form given at the end of the book and mail it to me with the
words ‘Mental Magic’ written on the envelop. I will send across the techniques via e-mail.
Miscellaneous Simple Method
In Vedic Mathematics, there are two types of techniques: specific techniques and general techniques.
The specific techniques are those which are fast and effective but can be applied only to a particular
combination of numbers. For example, the technique of squaring numbers ending with 5 is a specific
technique because it can be used to square only those numbers that end with 5. It cannot be used to
square any other type of number. On the other hand, the technique of multiplication as given by the
Criss-Cross System is a general technique, as it can be used to multiply numbers of any possible
combination of digits.
Thus, general techniques have a much wider scope of application than specific techniques because
they deal with a wider range of numbers. In this book, we will give more emphasis to general
techniques as they provide a much wider utility. Chapter 1 deals with the specific techniques; from
Chapter 2 onwards we will study general techniques.
We will discuss the following techniques in this chapter:
(a) Squaring of numbers ending with 5
(b) Squaring of numbers between 50-60
(c) Multiplication of numbers with a series of 9’s
(d) Multiplication of numbers with a series of 1’s
(e) Multiplication of numbers with similar digits in the multiplier
(f) Subtraction using the rule ‘All from 9 and the last from 10’
The first technique that we will discuss is how to instantly square numbers whose last digit is 5.
Remember, squaring is multiplying a number by itself. When we multiply 6 by 6 we get the answer
36. This 36 is called the square of 6.
(a) Squaring of numbers ending with ‘5’
Squaring is multiplying a number by itself. Let us have a look at how to square numbers ending in 5.
(Q) Find the square of 65
• In 65, the number apart from 5 is 6.
• After 6 comes 7. So, we multiply 6 by 7 and write down the answer 42.
• Next, we multiply the last digits, viz. (5 × 5) and write down 25 to the right of 42 and complete
• Our answer is 4225.
(Q) Find the square of 75
Apart from 5 the number is 7. The number that comes after 7 is 8. We multiply 7 with 8 and write
the answer 56. Next, we multiply the last digits (5 × 5) and put 25 beside it and get our answer as
5625. Thus, (75 × 75) is 5625.
(Q) Find the square of 95
Apart from 5 the digit is 9. After 9 comes 10. When 9 is multiplied by 10 the answer is 90. Finally,
we vertically multiply the right hand most digits (5 × 5) and write the answer 25 beside it. Thus, the
square of 95 is 9025
(Q) Find the square of 105
The previous examples that we solved were of two-digits each. But the same technique can be
extended to numbers of any length. In the current example, we will try to determine the square of a
three-digit number – 105.
Apart from 5 the digits are 1 and 0, that is, 10. After 10 comes 11. We multiply 10 with 11 and
write the answer as 110. We suffix 25 to it and write the final answer as 11025. The square of 105 is
So you can see how simple it is to square numbers ending with a five! In fact, you can mentally
calculate the square of a number ending with a 5. Just multiply the non-five numbers with the next
number and then multiply the last digits (5 × 5) and add 25 after it.
A few more examples are given below:
Thus we see that the technique holds true in all the examples.
The technique of squaring numbers ending with 5 is a very popular technique. Some educational
boards have included it in their curriculum. In Vedic Mathematics, there is an extension to this
principle which is not known to many people. This formula of Vedic Mathematics tells us that the
above rule is applicable not only to the squaring of numbers ending in 5 but also to the multiplication
of numbers whose last digits add to 10 and the remaining digits are the same.
Thus, there are two conditions necessary for this multiplication. The first condition is that the last
digits should add to 10 and the second condition is that the remaining digits should be the same.
Let us have a look at a few examples:
In the above examples, it can be observed that the last digits in each case add up to 10 and the
remaining digits are the same. Let us take the first example…
Here the last digits are 6 and 4 which add up to 10. Secondly, the remaining digits are the same,
viz. ‘6’ and ‘6’. Thus, we can find the square of this number by the same principle which we used in
squaring numbers ending with a 5.
• First, multiply the number 6 by the number that follows it. After 6 comes 7. Thus, (6 × 7) is 42.
• Next, we multiply the right-hand most digits (6 × 4) and write the answer as 24. The complete
answer is 4224.
In the second example, we have to multiply 107 by 103. In this case the last digits 7 and 3 add to
10 and the remaining digits are the same. We will obtain the product using the same procedure.
• First, we will multiply the number 10 by the number that follows it, 11, and write the answer as
• Next, we multiply the right-hand most digits, viz. 7 and 3, and write the answer as 21. The
complete answer is 11021
In the third example, we have to multiply 91 by 99
• We multiply 9 by the number that follows it, 10, and write the answer as 90.
• We multiply the numbers (1 × 9) and write the answer as 09. The final answer is 9009.
(Note: The right hand part should always be filled-in with a two-digit number. Thus, we have to
convert the number 9 to 09.)
In the last example, we have to multiply 51 by 59
• We multiply 5 with the next number 6 and write the answer as 30.
• Next, we multiply (1 × 9) and write the answer as 09. The final answer is 3009
This formula of Vedic Mathematics works for any such numbers whose last digits add up to ten
and the remaining digits are the same. The same formula works while squaring numbers ending with 5
because when you square two numbers ending with 5, then the right hand most digits add to 10 (5 plus
5) and the remaining digits are the same (since we are squaring them).
Let us look at a few other examples where the right-hand most digits add to 10 and the remaining
digits are the same.
This was the first specific technique that we studied. The next technique that we will discuss is
also related to squaring. It is used to square numbers that lie between 50 and 60
(b) Squaring of numbers between 50 and 60
We have taken four different examples above. We will be squaring the numbers 57, 56, 52 and 53
respectively. We can find the answer to the questions by taking two simple steps as given below:
(1) Add 25 to the digit in the units place and put it as the left-hand part of the answer.
(2) Square the digits in the units place and put it as the right-hand part of the answer. (If it is a
single digit then convert it to two digits)
(Q) Find the square of 57
• In the first example we have to square 57. In this case we add 25 to the digit in the units place,
viz., 7. The answer is 32 which is the LHS (left-hand side) of our answer. (Answer at this stage is
• Next, we square the digit in the units place ‘7’ and get the answer as 49. This 49 we put as the
right hand part of our answer. The complete answer is 3249.
(Q) Find the square of 56
In the second example, we add 25 to 6 and get the LHS as 31. Next, we square 6 and get the answer
36 which we put on the RHS. The complete answer is 3136.
(Q) Find the square of 52
In the third example, we add 2 to 25 and get the LHS as 27. Next, we square 2 and get the answer 4
which we will put on the RHS. However, the RHS should be a two-digit number. Hence, we convert 4
to a two-digit number and represent it as 04. The complete answer is 2704
(Q) Find the square of 53
In the last example, we add 3 to 25 and get the answer as 28. Next, we square 3 and get the answer
as 9. As mentioned in rule B, the answer on the RHS should be converted to two digits. Thus, we
represent the digit 9 as 09. The complete answer is 2809.
On similar lines we have:
(c) Multiplication of numbers with a series of 9’s
In my seminars, I often have an audience challenge round. In this round, the audience members ask
me to perform various mental calculations and give them the correct answer. They generally ask me to
multiply numbers which involve a lot of 9’s in them. The general perception is that the higher the
number of 9’s the tougher it will be for me to calculate. However, the truth is exactly the opposite —
the higher the number of 9’s in the question, the easier it is for me to calculate the correct answer. I
use two methods for this. The first method is given below and the second method is explained in the
chapter ‘Base Method of Multiplication’.
Using the method given below, we can multiply any given number with a series of nines. In other
words, we can instantly multiply any number with 99, 999, 9999, 99999, etc.
The technique is divided into three cases. In the first case, we will be multiplying a given number
with an equal number of nines. In the second case we will be multiplying a number with a higher
number of nines. In the third case, we will be multiplying a number with a lower number of nines.
(Multiplying a number with an equal number of nines)
(Q) Multiply 654 by 999
• We subtract 1 from 654 and write half the answer as 653.
Answer at this stage is 653____
• Now we will be dealing with 653. Subtract each of the digits six, five and three from nine and
write them in the answer one by one.
• Nine minus six is 3. Nine minus five is 4. Nine minus three is 6.
• The answer already obtained was 653 and now we suffix to it the digits 3, 4 and 6. The complete
answer is 653346
(Q) Multiply 9994 by 9999
We subtract one from 9994 and write it as 9993. This becomes our left half of the answer. Next, we
subtract each of the digits of 9993 from 9 and write the answer as 0006. This becomes the right half of
the answer. The complete answer is 99930006
(Q) Multiply 456789 by 999999
We subtract 1 from 456789 and get the answer 456788. We write this down on the left hand side.
Next, we subtract each of the digits of 456788 (left hand side) from 9 and get 543211 which becomes
the right hand part of our answer. The complete answer is 456788543211
The simplicity of this method can be vouched from the examples given above. Now we move to
Case 2. In this case, we will multiply a given number with a higher number of nines.
(Multiplying a number with a higher number of nines)
(Q) Multiply 45 with 999
There are three nines in the multiplier. However, the multiplicand 45 has only two digits. So we
add a zero and convert 45 to 045 and make it a three digit number. After having done so, we can carry
on with the procedure explained in Case 1.
First we subtract 1 from 045 and write it down as 044. Next, we subtract each of the digits of 044
from 9 and write the answer as 955. The complete answer is 044955.
(Q) Multiply 888 with 9999
We convert 888 to 0888 and make the digits equal to the number of nines in the multiplier. Next,
we subtract 1 from 0888 and write the answer as 0887. Finally, we subtract each digit of 0887 from 9
and write the answer as 9112. The final answer is 08879112 which is 8879112
(Q) Multiply 123 by 99999
The multiplicand is a three-digit number and the multiplier is a five-digit number. Therefore, we
add two zeros in the multiplicand so that the digits are equal in the multiplicand and the multiplier.
We now subtract 1 from 00123 and write the left hand part of the answer as 00122. Next, we
subtract each of the digits of the left hand part of the answer from 9 and write it down as 99877 as the
right hand part of the answer. The complete answer is 12299877
We can see that this technique is not only simple and easy to follow, but it also enables one to
calculate the answer in the mind itself. This is the uniqueness of these systems. As you read the
chapters of this book, you will realize how simple and easy it is to find the answer to virtually any
problem of mathematics that one encounters in daily life and especially in the exams. And the
approach is so different from the traditional methods of calculation that it makes the whole process
Case 3 of this technique deals with multiplying a number with a lower number of nines. There is a
separate technique for this in Vedic Mathematics and requires the knowledge of the Nikhilam Sutra
(explained later in this book). However, at this point of time, we can solve such problems using our
normal practices of instant multiplication.
(Q) Multiply 654 by 99
In this case the number of digits are more than the number of nines in the multiplier. Instead of
multiplying the number 654 with 99 we will multiply it with (100-1). First we will multiply 654 with
100 and then we will subtract from it 654 multiplied by
(Q) Multiply 80020 by 999
We will multiply 80020 with (1000 - 1).
This method is so obvious that it needs no further elaboration.
(d) Multiplication of numbers with a series of 1’s
In the previous technique we saw how to multiply numbers with a series of 9’s. In this technique we
will see how to multiply numbers with a series of 1’s. Thus, the multiplier will have numbers like 1,
11, 111, etc.
Let us begin with the multiplier 11.
(Q) Multiply 32 by 11
• First we write the right-hand most digit 2 as it is.
(Answer = _______2)
• Next, we add 2 to the number in left 3 and write 5.
(Answer = ______52)
• Last, we write the left hand most digit 3 as it is.
Thus, the answer is 352.
(Q) Multiply 43 by 11
Write the last digit 3 as it is. Next we add 3 to 4 and get 7. Finally we write 4 as it is. The complete
answer is 473.
(Q) Multiply 64 by 11
In this example we write down the last digit 4 as it is. Next, we add 4 to 6 and get the answer 10.
Since, 10 is a two-digit answer, we write down the 0 and carry over 1. Finally, we add 1 to 6 and make
it 7. The complete answer is 704.
(Q) Multiply 652 by 11
The logic of two digit numbers can be expanded to higher numbers. In the given example we have
to multiply 652 by 11.
• We write down the last digit of the answer as 2. (Answer = ____2)
• Next, we add 2 to 5 and make it 7. (Answer = ____72)
• Next, we add 5 to 6 and make it 11. We write down 1 and carry over 1. (Answer is ____172)
• Last, we take 6 and add the one carried over to make it 7. (Final answer is 7172)
(Q) Multiply 3102 by 11
• We write down 2 as it is. (Answer is ____2)
• We add 2 to 0 and make it 2 (Answer is ____22)
• We add 0 to 1 and make it 1. (Answer is ____122)
• We add 1 to 3 and make it 4 (Answer is ____4122)
• We write the first digit 3 as it is (Final answer is 34122)
When we multiply a number by 11 we write the last digit as it is. Then we move towards the left
and continue to add two digits at a time till we reach the last digit which is written as it is.
Since the multiplier 11 has two 1’s we add maximum two digits at a time. When the multiplier is
111 we will add maximum three digits at a time because the multiplier 111 has three digits. When the
multiplier is 1111 we will add maximum four digits at a time since the multiplier 1111 has four digits.
We have already seen how to multiply numbers by 11. Let us have a look at how to multiply
numbers by 111.
(Q) Multiply 203 by 111
• We write down the digit in the units place 3 as it is in the answer
• We move to the left and add (3 + 0) = 3
• We move to the left and add (2 + 0 + 3) = 5 (maximum three digits)
• We move to the left and add (0 + 2) = 2
• We take the last digit 2 and write down as it is
(Q) Multiply 201432 by 111
• The (2) in the units place of the multiplicand is written as the units digit of the answer
• We move to the left and add (2 + 3) and write 5
• We move to the left and add (2 + 3 + 4) and write 9
• We move to the left and add (3 + 4 + 1) and write 8
• We move to the left and add (4 + 1 + 0) and write 5
• We move to the left and add (1 + 0 + 2) and write 3
• We move to the left and add (0 + 2) and write 2
• We move to the left and write the single digit (2) in the answer.
Thus, the complete answer obtained by each of the steps above is 22358952
The simplicity of this method is evident from the examples. In most cases you will get the answer
within a minute. In fact, the beauty of this technique is that it converts a process of multiplication to
Using the same method, we can multiply any number by a series of 1’s.
If you want to multiply a number by 1111 you can use the method given above. The only
difference will be that we will add maximum four numbers at a time (because there are four ones in
1111) and when the multiplier is 11111 we will be multiplying maximum five digits at a time. An
example of the former type is given below:
(Q) Multiply 210432 by 1111
Thus, the product of (210432 × 1111) = 233789952
(e) Multiplication of numbers with a series of similar digits in multiplier
This technique is basically an extension of the previous technique. In technique ‘c’ we saw how to
multiply a number with a series of nines and in technique ‘d’ we saw how to multiply a number with a
series of ones.
A question may arise regarding how to multiply numbers with a series of 2’s, like 2222, or with a
series of 3’s, like 333, and such other numbers.
Let us have a look at a few examples.
(Q) Multiply 333 by 222
The question asks us to multiply 333 by 222. Now, carefully observe the logic that we apply in this
Therefore, multiplication of 333 by 222 is the same as multiplication of 666 by 111. But, we have
already studied the procedure of multiplying a number by 111 in the previous subtopic. Our answer
will be as follows:
Therefore, 333 × 222 is 73936
(Q) Multiply 3021 by 333
The multiplicand is a normal number and the multiplier is a series of 3’s. We do not know how to
multiply a number with a series of 3’s but we know how to multiply a number with a series of 1’s.
Thus, we will represent the expression in such a manner that the multiplier is 111.
We have already learnt how to multiply 9063 by 111. On the basis of our knowledge we can easily
complete the multiplication
On the basis of the above examples it can be seen that there is no need of explaining the procedure
of multiplication. The procedure is the same as observed in the previous sub-topic. Basically, we have
to convert a series of 2’s, 3’s, 4’s, etc. in the multiplier to a series of 1’s by dividing it by a certain
number. Next, we have to multiply the multiplicand by the same number
(a) Multiply 1202 by 44
In this case, we have divided the multiplier 44 by 4 to obtain a series of 1’s (11). Since we have
divided the multiplier by 4 we will multiply the multiplicand 1202 by 4. Thus, we have the new
multiplicand as 4808. When 4808 is multiplied by 11 the answer is 52888. This is also the answer to
the original question of 1202 by 44.
(b) Multiply 2008 by 5555
Ans: The product of 2008 multiplied by 5555 is 11154440
(c) Multiply 10503 by 888
Ans: The product of 10503 multiplied by 888 is 9326664.
(f) Subtraction using the rule ‘All from 9 and the last from 10’
Subtraction using the rule ‘All from 9 and the last from 10’ is one of the elementary techniques of
Vedic Mathematics. Basically, it is used to subtract any number from a power of ten. The powers of
ten include numbers like 10, 100, 1000, 10000, etc.
So if you want to learn a method by which you can quickly subtract a number from a power of ten,
then this technique can come to your aid.
When we go to the market to buy something, we generally give a hundred rupee note to the
shopkeeper and calculate the change that we should get after deducting the total amount of groceries.
In such a situation, this technique can come to our aid.
(Q) Subtract 54.36 from 100
We are asked to subtract 54.36 from 100. In this case, we generally start from the right and
subtract 6 from 0. But, we realize that it is not possible to subtract 6 from 0 and so we move to the
number in the left and then borrow one and give it to zero and make it ten and so on.
This whole procedure is slightly cumbersome and there is a possibility of making a mistake too.
Vedic Mathematics provides a very simple alternative. The approach of Vedic Mathematics is
explained by the rule ‘All from 9 and the last from 10.’ It means that we have to subtract each digit
from nine and subtract the last digit from 10. This will give us the answer.
The number to be subtracted is 54.36. We have to subtract all the digits from nine except for the
last digit which will be subtracted from ten. Thus,
The final answer is 45.64.
(Q) Subtract 3478.2281 from 10000
In this case, we will subtract the digits 3, 4, 7, 8, 2, 2 and 8 from 9 and the last digit 1 from 10. The
respective answers will be 6, 5, 2, 1, 7, 7, 1 and 9. Thus, the final answer is 6521.7719
The examples prove the simplicity and efficiency of this system. In examples (d) and (e) we added
zeros in the number below so that we get the accurate answer.
In this chapter we have seen 6 simple yet quick techniques of Vedic Mathematics. I wanted to
begin the book with these easy techniques so that we can prepare ourselves for the comprehensive
techniques that will follow in the forthcoming chapters.
You must have observed that the techniques that we have employed in this chapter work with a
totally different approach. We have found answers to our questions using a completely different
approach. In all the chapters of this book you will discover that the method used in solving the
problems is far more efficient than the normal systems that we have been using, thus enabling us to
produce outstanding results.
Q. (1) Find the product in the following numbers whose last digits add to ten
(a) 45 × 45
(b) 95 × 95
(c) 111 × 119
(d) 107 × 103
Q. (2) Find the squares of the following numbers between fifty and sixty
Q. (3) Find the product of the following numbers which are multiplied by a series of nines
(a) 567 × 999
(b) 23249 × 99999
(c) 66 × 9999
(d) 302 × 99999
Q. (4) Find the product of the following numbers which are multiplied by a series of ones
(a) 32221 × 11
(b) 64609 × 11
(c) 12021 × 111
(d) 80041 × 111
Q. (5) Find the product of the following numbers which are multiplied by a series of same numbers
(a) 7005 × 77
(b) 1234 × 22
(c) 2222 × 222
(d) 1203 × 333
Q. (6) Subtract the following numbers from a given power of ten
(a) 1000 minus 675.43
(b) 10000 minus 7609.98
(c) 10000 minus 666
(d) 1000 minus 2.653
The traditional system of multiplication taught to students in schools and colleges is a universal
system, i.e., it is applicable to all types of numbers. The traditional system can also be used for
numbers of any length.
Let us have a look at an example:
This is the traditional way of multiplication which is taught to students in schools. This system of
multiplication is perfect and works for any combination of numbers.
In Vedic Mathematics too, we have a similar system but it helps us to get the answer much faster.
This system is also a universal system and can be used for any combination of numbers of any length.
This system of multiplication is given by the ‘Urdhva-Tiryak Sutra.’ It means ‘vertically and
cross-wise’. The applications of this system are manifold, but in this chapter we shall confine our
study only to its utility in multiplying numbers. We shall call it the Criss-Cross system of
Let us suppose we want to multiply 23 by 12. According to the traditional system, our answer would
With the Criss-Cross system, we can get the answer in just one step as given below:
Let us have a look at the modus operandi of this system:
We multiply the digits in the ones place, that is, 3 × 2 = 6. We write 6 in the ones place of the
Now, we cross multiply and add the products, that is, (2 × 2) + (3 × 1) = 7. We write the 7 in the
tens place of the answer.
Now we multiply the ones digits, that is, 2 × 1 = 2.
The completed multiplication is:
Let us notate the three steps involved in multiplying a 2-digit number by a 2-digit number.
We shall have a look at one more example.
Let us multiply 31 by 25
• First we multiply 1 by 5 vertically and get the answer as 5
• Then, we cross-multiply (3 × 5) + (2 × 1) and get the answer as 17. We write down 7 in the tens
place of the answer and carry over 1.
• Lastly, we multiply (3 × 2) and get the answer as 6. But, we have carried over 1. So, the final
answer is 7.
Given below are a few examples of 2-digit multiplication where there is no carrying-over involved
An example of 2-digit numbers where there is a carry-over involved:
• First, we multiply 3 by 4. The answer is 12. We write down 2 and carry over 1. The answer at this
• Next, we cross multiply (2 × 4) and add it to (3 × 1). The total is 11. Now, we add the 1 which we
carried over. The total is 12. So, we write 2 and carry over 1. (The answer at this stage is _____22)
• Last, we multiply (2 × 1) and get the answer as 2. To it, we add the 1 that is carried over and get
the final answer as 3.
(The completed answer is 322)
Thus, we see how the Criss-Cross system of multiplication helps us get our answer in just one line!
And, the astonishing fact is that this same system can be expanded to multiplication of numbers of
higher digits too.
And in every case, we will be able to get the answer in a single line.
Let us have a look at the multiplication process involved in multiplying a three-digit number by
another three-digit number.
Let us multiply two three-digit numbers where there is no carry over involved:
As suggested by step (a), we multiply 1 into 2 and get the answer as 2.
Next, we cross-multiply (2 × 2) and add it to (1 × 0). Thus, the final answer is 4.
In step (c), we multiply (1 × 2) and (2 × 0) and (3 × 1). We add the three answers thus obtained to
get the final answer 5.
In step (d), we multiply (1 × 0) and (3 × 2). The final answer is 6.
In step (e), we multiply the left-hand most digits (1 × 3) and get the answer as 3.
Thus, it can be seen that the product obtained by multiplying two 3-digit numbers can be obtained
in just one line with the help of the Criss-Cross system.
We shall quickly have a look at how to multiply two 3-digit numbers where there is a carry over
involved. Obviously, the process of carrying over is the same as we use in normal multiplication.
• We multiply 4 by 5 and get the answer as 20. We write down 0 and carry over 2.
(The answer at this stage is ________0)
• (2 × 5) is 10 plus (4 x 5) is 20. The total is 30 and we add the 2 carried over to get 32. We write
down 2 and carry over 3.
(The answer at this stage is _______20)
• (1 × 5) is 5 plus (2 × 5) is 10 plus (4 × 3) is 12. The total is 27 and we add the 3 carried over to
get the answer as 30. We write down 0 and carry over 3.
(The answer at this stage is ____020)
• (3 × 2) is 6 plus (1 × 5) is 5. The total is 11 and we add the 3 carried over. The final answer is 14.
We write down 4 and carry over 1.
(The answer at this stage is ______4020)
• Finally, (1 × 3) is 3. Three plus 1 carried over is 4. The final answer is 44020.
A few examples with their completed answers are given below:
MULTIPLICATION OF HIGHER-ORDER NUMBERS
We have seen how to multiply 2-digit and 3-digit numbers. We can expand the same logic and
multiply bigger numbers. Let us have a look at how to multiply 4-digit numbers.
The following are the steps involved in multiplying 4-digit numbers:
Suppose we want to multiply 1111 by 1111. Then, there will be 7 steps involved in the complete
multiplication as suggested above from steps ‘a’ to ‘g’
Let us have a look at one more example:
Example: 2104 multiplied by 3072
(a) (4 × 2) = 8
(b) (7 × 4) + (0 × 2) = 28 (2 carry-over)
(c) (0 × 4) + (0 × 7) + (1 × 2) + (2 carried) = 4
(d) (3 × 4) + (0 × 0) + (7 × 1) + (2 × 2) = 23 (2 carry over)
(e) (3 × 0) + (0 × 1) + (7 × 2) + (2 carried) = 16 (1 carry over)
(f) (2 × 0) + (3 × 1) + (1 carried) = 4
(g) (2 × 3) = 6
We have seen the multiplication of 2-digit, 3-digit and 4-digit numbers. A question may arise
regarding multiplication of numbers with an unequal number of digits.
Let us suppose you want to multiply 342 by 2009. Here, we have one number that has three digits
and another number that has four digits. Now, for such a multiplication which technique will you use?
Will you use the technique used for multiplying numbers of 3- digits or the technique used for
multiplying numbers of 4 digits?
To multiply 342 by 2009, write the number 342 as 0342 and then multiply it by 2009. Use the
technique used for multiplying four-digit numbers.
Thus, if we want to multiply 312 by 64, we will write 64 as 064 and then multiply it by 312 using
the technique of 3-digit multiplication.
THE CHARACTERISTICS OF CRISS-CROSS MULTIPLICATION
Now, we shall learn a few characteristics of the Criss-Cross (urdhva-tiryak) system of
multiplication. The knowledge of these characteristics will help us to easily calculate any answer.
To understand the characteristics, we shall carefully observe the steps used in multiplying 2-digit,
3-digit and 4-digit numbers. Let us have a look at them:
The steps used for multiplying 2-digit, 3-digit and 4-digit numbers is 3, 5, and 7 respectively. The
number of steps used for any multiplication can be found using the formula:
‘2 multiplied by (number of digits) minus 1’
Thus, when we multiply 2-digit numbers, the steps used are 2 × 2 – 1 = 3 and therefore we use 3
steps. When we multiply 4-digit numbers, the steps used are 2 × 4 - 1 = 7. When we multiply 5-digit
numbers, the steps used will be 2 × 5 - 1 = 9.
(If there are an unequal number of digits in the multiplicand and the multiplier, they should first be
made equal by inserting 0’s at the appropriate places and then the formula must be used.)
Most students will be able to guess the trend of the steps used in this system of multiplication by
the mere observation of the examples that we have solved. This is because they follow a systematic
The second characteristic of the Criss-Cross system of multiplication is that the number of steps
used will always be an odd number. Amongst these odd number of steps, the first and last steps are
mirror images of one another, the second and second-to-last steps are mirror images of one another
and so on.
In every case there will be a central step, which will be independent (that is, without any
corresponding mirror image.)
For example, in 2-digit multiplication, steps ‘a’ and ‘c’ are mirror images and step ‘b’ is a central
figure. In 3-digit multiplication, steps ‘a’ and ‘e’ are mirror images, ‘b’ and ‘d’ are mirror images, and
‘c’ is a central figure.
In our daily practice, we mainly deal with small numbers of 2, 3 and 4-digits. Thus, I have given
special emphasis to them. However, to test whether you have understood the logic of this system, I
suggest that in the empty space provided next, you notate the steps that we will require if we were to
multiply 5-digit numbers. Once you are done, you can check your work with the answer mentioned in
APPENDIX A at the end of the book.
Don’t proceed until you have checked the answer in APPENDIX A.
Most of you might have made a mistake in step ‘e’ and solved the other steps correctly. If you got
all the steps correct then it is indeed praiseworthy. Those who made a mistake in other steps need not
get disheartened: with enough practice you shall be able to solve sums effortlessly.
I would again like to draw your attention to the trend of the steps used in the multiplication
procedure. Since we are multiplying two five-digit numbers, the number of steps according to the
formulae — 2 multiplied by the number of digits minus 1 — will be 9, and thus we have used steps
from (a) to (i).
Further, you will observe that steps (a) and (i) are mirror-images, steps (b) and (h) are mirror-
images, steps (c) and (g) are mirror images, steps (d) and (f) are mirror images and step (e) is a central
The Criss-Cross system not only helps us to get answers quickly but also helps us to eliminate all
the intermediate steps used in the multiplication process. This quality of the system can be of
immense aid to students giving various competitive and professional exams.
Let us take a hypothetical situation:
We know that in competitive exams, we are given a question with four alternatives out of which
we have to select the correct one.
(Q) What is the product of 121 by 302?
(a) 36522 (b) 36592 (c) 36542 (d) 36544
In this question we are asked to calculate the product of 121 by 302. Now, I read the question and
start the multiplication process using the Criss-Cross system.
First, I multiply the extreme digits 1 by 2 and get the final answer as 2. My answer at this stage is
Next, I cross multiply (2 × 2) and add it to (1 × 0). My final answer is 4. The answer at this stage is
The moment I get the digits 4 and 2 as the last two digits of my answer, I discontinue the
multiplication process and instantly tick option (c) as the correct answer to the problem.
The reason is that the last two digits of the other alternatives are 22, 92 and 44 but I know the last
two digits of my answer are 42 and so the correct answer can only be option (c).
This was just an example. The idea which I am trying to convey is that while solving any
multiplication problem, the moment you get a part of the answer which is unique to any one of the
given alternatives, you can instantly mark that alternative as the correct answer.
This advantage is not available with the traditional system as it compels you to do the whole
multiplication procedure with the intermediary steps.
This comparison proves the fact that the urdhva-tiryak sutra (which I call the Criss-Cross system)
helps us to solve any multiplication problem instantly. There are numerous methods in mathematics
which help you to multiply numbers quickly but most of them can be used only for a particular
category of numbers, like numbers adding up to 10, numbers between 11 and 20, etc. However, this
system is a universal system and can be used for any combination of numbers.
In the next page, I have given the steps for multiplying six-digit and seven-digit numbers. In this
manner we can go on and on and on, with eight-digit, nine-digit, ten-digit numbers, etc., but I guess
the techniques that we have studied up to this point will suffice us to get a thorough understanding of
The concept that we have studied here can be applied in multiplying algebraic expressions too. For
example, if you want to find the product of a + 6b multiplied by 2a + 3b you can use the Criss-Cross
system. For solved examples refer to Appendix B.
TECHNIQUE FOR MULTIPLYING SIX-DIGIT NUMBERS
TECHNIQUE FOR MULTIPLYING SEVEN-DIGIT NUMBERS
Questions in PART-A are very easy questions and you will be able to solve them just by looking at the
numbers without any rough work
Questions in PART-B are of average difficulty and you will be able to solve them with some thought
Questions in PART-C will compel you to do some writing work to get the answer.
Squaring can be defined as ‘multiplying a number by itself.’
There are many different ways of squaring numbers. Many of these techniques have their roots in
multiplication as squaring is simply a process of multiplication.
Examples: 32 is 3 multiplied by 3 which equals 9
42 is four multiplied by 4 which equals 16
The techniques that we will study are:
(A) Squaring of numbers using the Criss-Cross system
(B) Squaring of number using formulae
(A) SQUARING OF NUMBERS USING CRISSCROSS SYSTEM
The Urdhva-Tiryak Sutra (the Criss-Cross system) is by far the most popular system of squaring
numbers amongst practitioners of Vedic Mathematics. The reason for its popularity is that it can be
used for any type of numbers.
Ex: Find the square of 23
(a) First, we multiply 3 by 3 and get the answer as 9.
(Answer at this stage is _____9)
(b) Next, we cross multiply (2 × 3) and add it with (2 × 3).
The final answer is 12. We write down 2 and carry over 1.
(Answer at this stage is _____29)
(c) Thirdly, we multiply (2 × 2) and add the 1 to it. The answer is 5.
The final answer is 529
Similarly, numbers of higher orders can be squared. Refer to the chapter on Criss-Cross system for
(B) FORMULA METHOD
There are various formulae used in general mathematics to square numbers instantly. Let us discuss
them one by one.
This method is generally to square numbers which are near multiples of 10. In this method, a given
number is expanded in such a manner that the value of ‘a’ is a number which can be easily squared
and the value of ‘b’ is a small number which too can be easily squared.
(Q) Find the square of 1009
We represent the number 1009 as 1000 + 9. Thus, we have converted it into a form of (a + b) where
the value of a is 1000 and the value of b is 9.
(Q) Find the square of 511
The number 511 will be written as 500 + 11
The second formula that we will discuss is also very well known to the students. It is a part of the
regular school curriculum. This formula is used to square numbers which can be easily expressed as a
difference of two numbers ‘a’ and ‘b’ in such a way that the number ‘a’ is one which can be easily
squared and the number ‘b’ is a small number which too can be easily squared. The formula is
This formula is very much like the first one. The only difference is that the middle term carries a
negative sign in this formula.
(Q) Find the square of 995
We will express the number 995 as (1000 – 5)
(Q) Find the square of 698
We will express the number 698 as (700 - 2)
Thus, we see that the two formulae can help us find the squares of any number above and below a
round figure respectively. There is another formulae which is used to find the square of numbers, but
it is not so popular. I discuss it below.
We know that:
This is the formula that we will be using: a2 = (a + b) (a –b) + b2
Suppose we are asked to find the square of a number. Let us call this number ‘a’. Now in this case we
will use another number ‘b’ in such a way that either (a + b) or (a – b) can be easily squared.
(Q) Find the square of 72
Ans: In this case, the value of ‘a’ is 72. Now, we know that
a2 = (a + b) (a – b) + b2
Substituting the value of ‘a’ as 72, we can write the above formula as:
(72)2 = (72 + b) (72 - b) + b2
We have substituted the value of ‘a’ as 72. However, we cannot solve this equation because a
variable ‘b’ is still present. Now, we have to substitute the value of ‘b’ with such a number that the
whole equation becomes easy to solve.
Let us suppose I take the value of ‘b’ as 2.
Then the equation becomes,
(72)2 = (72 + 2) (72 - 2) + (2)2
= (74) (70) + 4
In this case we can find the answer by multiplying 74 by 70 and adding 4 to it. However, if one
finds multiplying 74 by 70 difficult, we can simplify it still further. First, multiply 70 by 70 and then
multiply 4 by 70 and add both for the answer.
Let us continue the example given above:
= (70 × 70) + (4 × 70) + 4
= 4900 + 280 + 4
Thus, the square of 72 is 5184.
In this example we have taken the value of ‘b’ as 2. Because of this, the value of (a - b) became 70
and the multiplication procedure became easy (as the number 70 ends with a zero).
(Q) Find the square of 53
Ans: Using the formula a2 = (a + b) (a - b) + b2 and taking the value of ‘a’ as 53, we have:
(53)2 = (53 + b) (53 - b) + (b)2
Now we have to find a suitable value for ‘b’. If we take the value of ‘b’ as 3, the expression (53 -
3) will be 50 and hence it will simplify the multiplication procedure. So we will take the value of ‘b’
as 3 and the equation will become:
(53)2 = (53 + 3) (53 - 3) + (3)2
= (56) (50) + 9
= (50 × 50) + (6 × 50) + 9
= 2500 + 300 + 9
(Q) Find the square of 67
Ans: In this case the value of ‘a’ is 67. Next, we will substitute ‘b’ with a suitable value. In this
case, let us take the value of ‘b’ as 3 so that the value of (a + b) will become (67 + 3) which equals 70.
Thus: (67)2 = (67 + 3) (67 - 3) + (3)2
= (70) (64) + 9
= (70 × 60) + (70 × 4) + 9
= 4200 + 280 + 9
(Q) Find the square of 107
Ans: In this case, we will take the value of ‘a’ as 107 and take the value of ‘b’ as 7. The equation
(107)2 = (107 + 7) (107 - 7) + (7)2
= (114) (100) + 49
= 11400 + 49
(Q) Find the square of 94
Ans: In this example we will take the value of ‘a’ as 94. Next, we will take the value of ‘b’ as 6 so
that the value of (a + b) becomes 100.
(94)2 = (94 + 6) (94 - 6) + (6)2
= (100) (88) + 36
Q. (1) Find the squares of the following numbers using the Criss-Cross System
Q. (2) Find the squares of the numbers using the formula for (a + b)2
Q. (3) Find the squares of the numbers using the formula for (a - b)2
Q. (4) Find the squares of the following numbers using the formula: a2 = (a + b) (a - b) + b2
Cube Roots of
In problems of algebra dealing with factorization, equations of third power and in many problems of
geometry related to three dimensional figures, one will often find the need to calculate the cube root
of numbers. Calculating the cube root of a number by the traditional method is a slightly cumbersome
procedure, but the technique used by Vedic Mathematicians is so fast that one can get the answer in
two to three seconds!
The technique for solving cube roots is simply so amazing that the student will be able to correctly
predict the cube root of a number just by looking at it and without the need for any intermediate steps.
You might find it difficult to believe, but at the end of this study, you will be calculating cube
roots of complicated numbers like 262144, 12167 and 117649 in 2-3 seconds. Even primary school
students who have learnt these techniques from me are able to calculate cube-roots in a matter of
Before we delve deeper in this study, let us clear our concepts relating to cube roots.
WHAT IS CUBE ROOTING?
Let us take the number 3. When we multiply 3 by itself we are said to have squared the number 3.
Thus 3 × 3 is 9. When we multiply 4 by itself we are said to have squared 4 and thus 16 is the square
Similarly the square of 5 is 25 (represented as 52)
The square of 6 is 36 (represented as 62)
In squaring, we multiply a number by itself, but in cubing we multiply a number by itself and then
multiply the answer by the original number once again.
Thus, the cube of 2 is 2 × 2 × 2 and the answer is 8. (represented as 23)
The cube of 3 is 3 × 3 × 3 and the answer is 27. (represented as 33)
Basically, in squaring we multiply a number by itself and in cubing we multiply a number twice by
Now, since you have understood what cubing is it will be easy to understand what cube rooting is.
Cube-rooting is the procedure of determining the number which has been twice multiplied by itself to
obtain the cube. Calculating the cube-root is the reciprocal procedure of calculating a cube.
Thus, if 8 is the cube of 2, then 2 is the cube-root of 8.
If 27 is the cube of 3 then 3 is the cube root of 27
And so on.
In this chapter, we will learn how to calculate cube-roots. Thus, if you are given the number 8 you
will have to arrive at the number 2. If you are given the number 27 you will have to arrive at 3.
However, these are very basic examples. We shall be cracking higher order numbers like 704969,
At this point, I would like to make a note that the technique provided in this chapter can be used to
find the cube-roots of perfect cubes only. It cannot be used to find the cube root of imperfect cubes.
I have given below a list containing the numbers from 1 to 10 and their cubes. This list will be used
for calculating the cube roots of higher order numbers. With the knowledge of these numbers, we shall
be able to solve the cube-roots instantly. Hence, I urge the reader to memorize the list given below
before proceeding ahead with the chapter.
The cube of 1 is 1, the cube of 2 is 8, the cube of 3 is 27 and so on….
Once you have memorized the list I would like to draw your attention to the underlined numbers in
the key. You will notice that I have underlined certain numbers in the key. These underlined numbers
have a unique relationship amongst themselves.
In the first row, the underlined numbers are 1 and 1. It establishes a certain relationship that if the
last digit of the cube is 1 then the last digit of the cube root is also 1.
In the second row, the underlined numbers are 2 and 8. It establishes a relationship that if the last
digit of the cube is 8 then the last digit of the cube root is 2. Thus, in any given cube if the last digit of
the number is 8 the last digit of its cube-root will always be 2.
In the third row, the underlined numbers are 3 and 7 (out of 27 we are interested in the last digit
only and hence we have underlined only 7). We can thus conclude that if the last digit of a cube is 7
the last digit of the cube root is 3.
And like this if we observe the last row where the last digit of 10 is 0 and the last digit of 1000 is
also 0. Thus, when a cube ends in 0 the cube-root also ends in 0.
On the basis of the above observations, we can form a table as given below:
From the above table, we can conclude that all cube-roots end with the same number as their
corresponding cubes except for pairs of 3 & 7 and 8 & 2 which end with each other.
There is one more thing to be kept in mind before solving cube-roots:
Whenever a cube is given to you to calculate its cube-root, you must put a slash before the last
If the cube given to you is 103823 you will represent it as 103|823
If the cube given to you is 39304, you will represent it as 39|304
Immaterial of the number of digits in the cube, you will always put a slash before the last three
SOLVING CUBE ROOTS
We will be solving the cube root in 2 parts. First, we shall solve the right hand part of the answer
and then we shall solve the left hand part of the answer. If you wish you can solve the left hand Part-
Before the right hand part. There is no restriction on either method but generally people prefer to
solve the right hand part first.
As illustrative examples, we shall take four different cubes.
(Q) Find the cube root of 287496
• We shall represent the number 287496 as 287 496
• Next, we observe that the cube 287496 ends with a 6 and we know that when the cube ends with a
6, the cube root also ends with a 6. Thus our answer at this stage is ___6. We have thus got the
right hand part of our answer.
• To find the left hand part of the answer we take the number which lies to the left of the slash. In
this case, the number lying to the left of the slash is 287. Now, we need to find two perfect cubes
between which the number 287 lies in the number line. From the key, we find that 287 lies between
the perfect cubes 216 (the cube of 6) and 343 (the cube of 7).
• Now, out of the numbers obtained above, we take the smaller number and put it on the left hand
part of the answer. Thus, out of 6 and 7, we take the smaller number 6 and put it beside the answer
of __6 already obtained. Our final answer is 66. Thus, 66 is the cube root of 287496.
(Q) Find the cube root of 205379
• We represent 205379 as 205 379
• The cube ends with a 9, so the cube root also ends with a 9. (The answer at this stage is ______9.
• The part to the left of the slash is 205. It lies between the perfect cubes 125 (the cube of 5) and
216 (the cube of 6)
• Out of 5 and 6, the smaller number is 5 and so we take it as the left part of the answer. The final
answer is 59.
(Q) Find the cube root of 681472
• We represent 681472 as 681 472
• The cube ends with a 2, so the root ends with an 8. The answer at this stage is ______8.
• 681 lies between 512 (the cube of 8) and 729 (the cube of 9)
• The smaller number is 8 and hence our final answer is 88.
(Q) Find the cube root of 830584
• The cube ends with a 4 and the root will also end with a 4.
• 830 lies between 729 (the cube of 9) and 1000 (the cube of 10)
• Since, the smaller number is 9, the final answer is 94.
You will observe that as we proceeded with the examples, we took much less time to solve the
cube-roots. After some practice you will be able to solve the cube-roots by a mere observation of the
cube and without the necessity of doing any intermediary steps.
It must be noted that immaterial of the number of digits in the cube, the procedure for solving
them is the same.
(Q) Find the cube root of 2197.
• The number 2197 will be represented as 2 197
• The cube ends in 7 and so the cube root will end with a 3.
We will put 3 as the right hand part of the answer.
• The number 2 lies between 1 (the cube of 1) and 8 (the cube of 2)
• The smaller number is 1 which we will put as the left hand part of the answer. The final answer is
We may thus conclude that there exists only one common procedure for solving all types of perfect
In my seminars, the participants often ask what is the procedure of solving cube roots of numbers
having more than 6 digits. (All the examples that we have solved before had 6 or fewer digits.)
Well the answer to this question is that the procedure for solving the problem is the same. The only
difference in this case is that you will be expanding the number line.
Let us take an example.
We know that the cube of 9 is 729 and the cube of 10 is 1000. Now let us go a step ahead and
include the higher numbers. We know that the cube of 11 is 1331 and the cube of 12 is 1728.
(Q) Find the cube root of 1157625
• We put a slash before the last three digits and represent the number as 1157 625
• The number 1157625 ends with a 5 and so the root also ends with a 5. The answer at this stage is
• We take the number to the left of the slash, which is 1157. In the number line it lies between
1000 (the cube of 10) and 1331 (the cube of 11)
• Out of 10 and 11, we take the smaller number 10 and put it beside the 5 already obtained. Our
final answer is 105.
(Q) Find the cube root of 1404928.
• The number will be represented as 1404 928
• The number ends with a 8 and so the cube-root will end with a 2.
• 1404 lies between 1331 (the cube of 11) and 1728 (the cube of 12). Out of 11 and 12 the smaller
number is 11 which we will put beside the 2 already obtained. Hence, the final answer is 112.
The two examples mentioned above were just for explanation purposes. Under normal circumstances,
you will be asked to deal with cubes of 6 or less than 6 digits in your exams. Hence, knowledge of the
key which contains cubes of numbers from 1 to 10 is more than sufficient. However, since we have
dealt with advanced level problems also, you are well equipped to deal with any kind of situation.
As usual, we will be comparing the normal technique of calculation with the Vedic technique. In the
traditional method of calculating cube-roots we use prime numbers as divisors.
Prime numbers include numbers like 2, 3, 5, 7, 11, 13 and so on.
Let us say you want to find the cube root of 64. Then, the process of calculating the cube root of 64
is as explained below.
First, we divide the number 64 by 2 and get the answer 32.
• 32 divided by 2 gives 16
• 16 divided by 2 gives 8
• 8 divided by 2 gives 4
• 4 divided by 2 gives 2
• 2 divided by 2 gives 1
(We terminate the division when we obtain 1)
Thus, 64 = 2 × 2 × 2 × 2 × 2 × 2
To obtain the cube root, for every three similar numbers we take one number. So, for the first three
2’s, we take one 2 and for the next three 2’s we take one more 2.
When these two 2’s are multiplied with each other we get the answer 4 which is the cube root of
It can be represented as:
Hence, 4 is the cube root of 64.
Similarly, to find the cube root of 3375 by the traditional method, we can use the following
Hence, 15 is the cube-root of 3375.
After studying the above two examples, the reader will agree with me that the traditional method is
cumbersome and time-consuming compared to the method used by Vedic mathematicians. However,
you will be shocked to see the difference between the two methods when we try to calculate the cube
root of some complicated number.
Example: Find the cube root of 262144
It can be observed from the comparison that not only does the Vedic method helps us to find the
answer in one line but also helps us to find the answer directly without the need for any intermediate
steps. This characteristic of this system helps students in instantly cracking such problems in
With this comparison we terminate this study. Students are urged to solve the practice exercise
before proceeding to the next chapter.
Q. (1) Find the cube roots of the following numbers with the aid of writing material
Q. (2) Find the cube roots of the following numbers without the aid of writing material
Square Roots of
We have divided the study of square roots in two parts. In this chapter we will study how to find the
square roots of perfect squares; in the ‘Advance Level’ we will study the square roots of general
numbers. Most school and college exams ask the square roots of perfect squares. Therefore, this
chapter is very useful to students giving such exams. Students of higher classes and other researchers
will find the thirteenth chapter useful as they will be able to study the Vedic Mathematics approach to
calculate square roots of any given number — perfect as well as imperfect.
The need to find perfect square roots arises in solving linear equations, quadratic equations and
factorizing equations. Solving square roots is also useful in geometry while dealing with the area,
perimeter, etc. of geometric figures. The concepts of this chapter will also be useful in dealing with
the applications of the Theorem of Pythagoras.
The technique of finding square roots of perfect squares is similar to the technique of finding the
cube root of perfect cubes. However, the former has an additional step and hence it is discussed after
having dealt with cube roots.
WHAT IS SQUARE ROOT
To understand square roots it will be important to understand what are squares. Squaring of a number
can be defined as multiplying a number by itself. Thus, when we multiply 4 by 4 we are said to have
‘squared’ the number four.
The symbol of square is represented by putting a small 2 above the number.
From the above example we can say that 16 is the square of 4, and 4 is the ‘square root’ of 16.
Similarly, 25 is the square of 5, and 5 is the square-root of 25.
To find the square roots it is necessary to be well versed with the squares of the numbers from 1 to 10.
The squares are given below. Memorize them before proceeding ahead.
In the chapter dealing with perfect cube roots we observed that if the last digit of the cube is 1 the
last digit of the cube root is also 1. If the last digit of the cube is 2 then the last digit of the cube-root
is 8 and so on. Thus, for every number there was a unique corresponding number.
However in square roots we have more than one possibility for every number. Look at the first
row. Here, we have 1 in the number column and 1 in the square column. Similarly, in the ninth row we
have 1 in the number column and 1 (of 81) in the square column. Thus, if the number ends in 1, the
square root ends in 1 or 9 (because 1 × 1 is one and 9 × 9 is eighty-one). Do not worry if you do not
follow this immediately. You may glance at the table below as you read these explanations, then all
will be clear.
• Similar to the 1 and 9 relationship, if a number ends in 4 the square root ends in 2 or 8. (because
2 × 2 is four and 8 × 8 is sixty-four)
• If a number ends in 9, the square root ends in 3 or 7. (because 3 × 3 is nine and 7 × 7 is
• If a number ends in 6, the square root ends in 4 or 6. (because 4 × 4 is 16 and 6 × 6 is 36)
• If the number ends in 5, the square root ends in 5 (because 5 × 5 is twenty-five)
• If the number ends in 0, the square root also ends in 0 (because 10 × 10 is 100)
On the basis of such observations, we can form a table as given below:
Whenever we come across a square whose last digit is 9, we can conclude that the last digit of the
square root will be 3 or 7. Similarly, whenever we come across a square whose last digit is 6, we can
conclude that the last digit of the square root will be 4 or 6 and so on….
Now, I want you to look at the column in the left. It reads ‘Last digit of the square’ and the
numbers contained in the column are 1,4, 9, 6, 5, and 0. Note that the numbers 2, 3, 7 and 8 are absent
in the column. That means there is no perfect square which ends with the numbers 2, 3, 7 or 8. Thus
we can deduct a rule:
‘A perfect square will never end with the digits 2, 3, 7 or 8’
At this point we have well understood how to find the last digit of a square root. However, in many
cases we will have two possibilities out of which one is correct. Further, we do not know how to find
the remaining digits of the square root. So we will solve a few examples and observe the technique
used to find the complete square root.
Before proceeding ahead with the examples, I have given below a list of the squares of numbers
which are multiples of 10 up to 100. This table will help us to easily determine the square roots.
(Q) Find the square root of 7744
• The number 7744 ends with 4. Therefore the square root ends with 2 or 8. The answer at this
stage is __2 or__8.
• Next, we take the complete number 7744. We find that the number 7744 lies between 6400
(which is the square of 80) and 8100 (which is the square of 90).
The number 7744 lies between 6400 and 8100. Therefore, the square root of 7744 lies between the
numbers 80 and 90.
• From the first step we know that the square root ends with 2 or 8. From the second step we know
that the square root lies between 80 and 90. Of all the numbers between 80 and 90 (81, 82, 83, 84,
85, 86, 87, 88, 89) the only numbers ending with 2 or 8 are 82 or 88. Thus, out of 82 or 88, one is
the correct answer.
(Answer at this stage is 82 or 88)
• Observe the number 7744 as given below:
Is it closer to the smaller number 6400 or closer to the bigger number 8100 ?
If the number 7744 is closer to the smaller number 6400 then take the smaller number 82 as the
answer. However, if it is closer to the bigger number 8100, then take 88 as the answer.
In this case, we observe that 7744 is closer to the bigger number 8100 and hence we take 88 as the
The square root of 7744 is 88.
(Q) Find the square root of 9801
• The last digit of the number 9801 is 1 and therefore the last digit of the square root will be either
1 or 9. The answer at this stage is _____1 or _____9.
• Next, we observe that the number 9801 lies between 8100 (which is the square of 90) and 10000
(which is the square of 100). Thus, our answer lies between 90 and 100. Our possibilities at this
stage are: 91, 92, 93, 94, 95, 96, 97, 98, 99
• However, from the first step we know that the number ends with a 1 or 9. So, we can eliminate
the numbers that do not end with a 1 or 9.
• The two possibilities at this stage are 91 or 99. Lastly, we know that the number 9801 is closer to
the bigger number 10,000 and so we take the bigger number 99 as the answer.
(Q) Find the square root of 5184
• 5184 ends in 4. So the square root ends in either 2 or 8
(Answer = ____2 or ____8)
• 5184 is between 4900 and 6400. So the square root is between 70 and 80. Combining the first two
steps, the only two possibilities are 72 and 78
• Out of 4900 and 6400, our number 5184 is closer to the smaller number 4900 (70 × 70). Thus, we
take the smaller number 72 as the correct answer
(Q) Find the square root of 2304
• 2304 ends in 4 and so the root either ends in a 2 or in a 8
• 2304 lies between 1600 and 2500. So, the root lies between 40 and 50.
• Thus, the two possibilities are 42 and 48.
• Lastly, the number 2304 is closer to the bigger number 2500. Hence, out of 42 and 48 we take the
bigger number 48 as the correct answer.
(Q) Find the square root of 529
• 529 ends with a 9. The answer is ____3 or _____7.
• It lies between 20 and 30. The possibilities are 23 or 27.
• 529 is closer to the smaller number 400 and hence 23 is the answer.
We have seen five different examples and calculated their square roots. However, the final answer
in each case is always a two-digit number. In most exams and even in general life, one will come
across squares whose roots are a two-digit answer. Thus, the above examples are sufficient and there
is no necessity to stretch the concept further. However, we will study a couple of examples involving
big numbers so that you will understand the fundamentals thoroughly.
(Q) Find the square root of 12544
The number 12544 ends with a 4. So, the square root ends with 2 or 8. The answer at this stage is
__2 or __8.
• Further, we know that the square of 11 is 121 and so the square of 110 is 12100. Similarly, the
square of 12 is 144 and so the square of 120 is 14400.