Methods - Vedic Mathematics

Contributed by:

Harshdeep Singh Wed, Apr 06, 2022 03:11 PM UTC

I. Why Vedic Mathematics?
II. Vedic Mathematical Formulae
1. Ekadhikena Purvena
2. Nikhilam navatascaramam Dasatah
3. Urdhva - tiryagbhyam
4. Paravartya Yojayet
5. Sunyam Samya Samuccaye
6. Anurupye - Sunyamanyat
7. Sankalana - Vyavakalanabhyam
8. Puranapuranabhyam
9. Calana - Kalanabhyam
10. Ekanyunena Purvena
11. Anurupyena
12. Adyamadyenantya - mantyena
13. Yavadunam Tavadunikrtya Varganca Yojayet
14. Antyayor Dasakepi
15. Antyayoreva
16. Lopana Sthapanabhyam
17. Vilokanam
18. Gunita Samuccayah : Samuccaya Gunitah
III Vedic Mathematics - A briefing
1. Terms and Operations
2. Addition and Subtraction
3. Multiplication
4. Division
5. Miscellaneous Items
IV Conclusion

1. Vedic Mathematics - Methods
Preface ------------------------------------------------------------------------------------------------ 1
I. Why Vedic Mathematics? --------------------------------------------------------------------------------- 3
II. Vedic Mathematical Formulae -------------------------------------------------------------------------- 5
1. Ekadhikena Purvena -------------------------------------------------------------------------------------- 7
2. Nikhilam navatascaramam Dasatah ----------------------------------------------------------------- 18
3. Urdhva - tiryagbhyam ---------------------------------------------------------------------------------- 31
4. Paravartya Yojayet -------------------------------------------------------------------------------------- 41
5. Sunyam Samya Samuccaye ---------------------------------------------------------------------------- 53
6. Anurupye - Sunyamanyat ------------------------------------------------------------------------------ 64
7. Sankalana - Vyavakalanabhyam ---------------------------------------------------------------------- 65
8. Puranapuranabhyam ------------------------------------------------------------------------------------ 67
9. Calana - Kalanabhyam -------------------------------------------------------- -------------------------- 68
10. Ekanyunena Purvena ---------------------------------------------------------------------------------- 69
11. Anurupyena ---------------------------------------------------------------------------------------------- 75
12. Adyamadyenantya - mantyena ---------------------------------------------------------------------- 82
13. Yavadunam Tavadunikrtya Varganca Yojayet --------------------------------------------------- 86
14. Antyayor Dasakepi ------------------------------------------------------------------------------------- 93
15. Antyayoreva --------------------------------------------------------------------------------------------- 96
16. Lopana Sthapanabhyam ---------------------------------------------------------------------------- 101
17. Vilokanam ----------------------------------------------------------------------------------------------- 106
18. Gunita Samuccayah : Samuccaya Gunitah ------------------------------------------------------ 113
III Vedic Mathematics - A briefing ---------------------------------------------------------------------- 115
1. Terms and Operations --------------------------------------------------------------------------------- 116
2. Addition and Subtraction ------------------------------------------------------------------------------ 130

2. 3. Multiplication ---------------------------------------------------------------------------------------- 139
4. Division ----------------------------------------------------------------------------------------------- 144
5. Miscellaneous Items ------------------------------------------------------------------------------- 151
IV Conclusion --------------------------------------------------------------------------------------------- 158

3. Preface
The Sanskrit word Veda is derived from the root Vid, meaning to know
without limit. The word Veda covers all Veda-sakhas known to humanity. The
Veda is a repository of all knowledge, fathomless, ever revealing as it is
delved deeper.
Swami Bharati Krishna Tirtha (1884-1960), former Jagadguru Sankaracharya
of Puri culled a set of 16 Sutras (aphorisms) and 13 Sub - Sutras (corollaries)
from the Atharva Veda. He developed methods and techniques for amplifying
the principles contained in the aphorisms and their corollaries, and called it
Vedic Mathematics.
According to him, there has been considerable literature on Mathematics in
the Veda-sakhas. Unfortunately most of it has been lost to humanity as of
now. This is evident from the fact that while, by the time of Patanjali, about
25 centuries ago, 1131 Veda-sakhas were known to the Vedic scholars, only
about ten Veda-sakhas are presently in the knowledge of the Vedic scholars
in the country.
The Sutras apply to and cover almost every branch of Mathematics. They
apply even to complex problems involving a large number of mathematical
operations. Application of the Sutras saves a lot of time and effort in solving
the problems, compared to the formal methods presently in vogue. Though
the solutions appear like magic, the application of the Sutras is perfectly
logical and rational. The computation made on the computers follows, in a
way, the principles underlying the Sutras. The Sutras provide not only
methods of calculation, but also ways of thinking for their application.
This book on Vedic Mathematics seeks to present an integrated approach to
learning Mathematics with keenness of observation and inquisitiveness,
avoiding the monotony of accepting theories and working from them
mechanically. The explanations offered make the processes clear to the
learners. The logical proof of the Sutras is detailed in algebra, which
eliminates the misconception that the Sutras are a jugglery.
Application of the Sutras improves the computational skills of the learners in
a wide area of problems, ensuring both speed and accuracy, strictly based on
rational and logical reasoning. The knowledge of such methods enables the
teachers to be more resourceful to mould the students and improve their
talent and creativity. Application of the Sutras to specific problems involves
rational thinking, which, in the process, helps improve intuition that is the
bottom - line of the mastery of the mathematical geniuses of the past and the
present such as Aryabhatta, Bhaskaracharya, Srinivasa Ramanujan, etc.
1

4. This book makes use of the Sutras and Sub-Sutras stated above for
presentation of their application for learning Mathematics at the secondary
school level in a way different from what is taught at present, but strictly
embodying the principles of algebra for empirical accuracy. The innovation in
the presentation is the algebraic proof for every elucidation of the Sutra or
the Sub-Sutra concerned.
Sri Sathya Sai Veda Pratishtan
2

5. I. Why Vedic Mathematics?
Many Indian Secondary School students consider Mathematics a very difficult
subject. Some students encounter difficulty with basic arithmetical
operations. Some students feel it difficult to manipulate symbols and balance
equations. In other words, abstract and logical reasoning is their hurdle.
Many such difficulties in learning Mathematics enter into a long list if prepared
by an experienced teacher of Mathematics. Volumes have been written on the
diagnosis of 'learning difficulties' related to Mathematics and remedial
techniques. Learning Mathematics is an unpleasant experience to some
students mainly because it involves mental exercise.
Of late, a few teachers and scholars have revived interest in Vedic
Mathematics which was developed, as a system derived from Vedic principles,
by Swami Bharati Krishna Tirthaji in the early decades of the 20th century.
Dr. Narinder Puri of the Roorke University prepared teaching materials based
on Vedic Mathematics during 1986 - 89. A few of his opinions are stated
i) Mathematics, derived from the Veda, provides one line, mental and super-
fast methods along with quick cross checking systems.
ii) Vedic Mathematics converts a tedious subject into a playful and blissful one
which students learn with smiles.
iii) Vedic Mathematics offers a new and entirely different approach to the
study of Mathematics based on pattern recognition. It allows for constant
expression of a student's creativity, and is found to be easier to learn.
iv) In this system, for any problem, there is always one general technique
applicable to all cases and also a number of special pattern problems. The
element of choice and flexibility at each stage keeps the mind lively and alert
to develop clarity of thought and intuition, and thereby a holistic development
of the human brain automatically takes place.
v) Vedic Mathematics with its special features has the inbuilt potential to
solve the psychological problem of Mathematics - anxiety.
J.T.Glover (London, 1995) says that the experience of teaching Vedic
Mathematics' methods to children has shown that a high degree of
mathematical ability can be attained from an early stage while the subject is
enjoyed for its own merits.
3

6. A.P. Nicholas (1984) puts the Vedic Mathematics system as 'one of the most
delightful chapters of the 20th century mathematical history'.
Prof. R.C. Gupta (1994) says 'the system has great educational value because
the Sutras contain techniques for performing some elementary mathematical
operations in simple ways, and results are obtained quickly'.
Prof. J.N. Kapur says 'Vedic Mathematics can be used to remove math-
phobia, and can be taught to (school) children as enrichment material along
with other high speed methods'.
Dr. Michael Weinless, Chairman of the Department of Mathematics at the
M.I.U, Iowa says thus: 'Vedic Mathematics is easier to learn, faster to use and
less prone to error than conventional methods. Furthermore, the techniques
of Vedic Mathematics not only enable the students to solve specific
mathematical problems; they also develop creativity, logical thinking and
Keeping the above observations in view, let us enter Vedic Mathematics as
given by Sri Bharati Krishna Tirthaji (1884 - 1960), Sankaracharya of
Govardhana Math, Puri. Entering into the methods and procedures, one can
realize the importance and applicability of the different formulae (Sutras) and
4

7. II. Vedic Mathematical Formulae
What we call VEDIC MATHEMATICS is a mathematical elaboration of 'Sixteen
Simple Mathematical formulae from theVedas' as brought out by Sri
Bharati Krishna Tirthaji. In the text authored by the Swamiji, nowhere has
the list of the Mathematical formulae (Sutras) been given. But the Editor of the
text has compiled the list of the formulae from stray references in the text. The
list so compiled contains Sixteen Sutras and Thirteen Sub - Sutras as stated
SIXTEEN SUTRAS
5

8. THIRTEEN SUB – SUTRAS
In the text, the words Sutra, aphorism, formula are used synonymously. So are
also the words Upa-sutra, Sub-sutra, Sub-formula, corollary used.
Now we shall have the literal meaning, contextual meaning, process, different
methods of application along with examples for the Sutras. Explanation,
methods, further short-cuts, algebraic proof, etc follow. What follows relates to
a single formula or a group of formulae related to the methods of Vedic
6

9. 1. Ekadhikena Purvena
The Sutra (formula) Ekādhikena Pūrvena means: “By one more than the
previous one”.
i) Squares of numbers ending in 5 :
Now we relate the sutra to the ‘squaring of numbers ending in 5’. Consider the
example 252.
Here the number is 25. We have to find out the square of the number. For the
number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more
than the previous one', that is, 2+1=3. The Sutra, in this context, gives the
procedure'to multiply the previous digit 2 by one more than itself, that is, by 3'.
It becomes the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S
(right hand side) of the result is52, that is, 25.
Thus 252 = 2 X 3 / 25 = 625.
In the same way,
352= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;
652= 6 X 7 / 25 = 4225;
1052= 10 X 11/25 = 11025;
1352= 13 X 14/25 = 18225;
Apply the formula to find the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.
Algebraic proof:
a) Consider (ax + b)2 Ξ a2. x2+ 2abx + b2.
Thisidentity for x = 10 and b = 5 becomes
(10a + 5) 2 = a2 . 102 + 2. 10a . 5 + 52
= a2 . 102+ a.102+ 52
= (a2+ a ) . 102 +52
7

10. = a (a + 1) . 102 + 25.
Clearly 10a + 5 represents two-digit numbers 15, 25, 35, -------,95 for the values a = 1, 2, 3, --
-----,9 respectively. In such a case the number (10a + 5)2 is of the form whose L.H.S is a (a + 1)
and R.H.S is 25, that is, a (a + 1) / 25.
Thus any such two digit number gives the result in the same fashion.
Example: 45 = (40 + 5)2, It is of the form (ax+b)2 for a = 4, x=10
and b = 5. giving the answer a (a+1) / 25
that is, 4 (4+1) / 25 + 4 X 5 / 25 = 2025.
b) Any three digit number is of the form ax2+bx+c for x =10, a ≠ 0, a, b, c Є
Now (ax2+bx+ c) 2 = a2 x4 + b2 x2 + c2 + 2abx3 + 2bcx + 2cax2
= a2 x4+2ab.x3+(b2+ 2ca)x2+2bc . x+ c2.
This identity for x = 10, c = 5becomes (a . 102 + b .10 + 5) 2
= a2.104+ 2.a.b.103 + (b2+ 2.5.a)102+ 2.b.5.10 + 52
= a2.104+ 2.a.b.103 + (b2 + 10 a)102 + b.102+ 52
= a2.104+ 2ab.103+ b2.102+ a . 103 + b 102+ 52
= a2.104 + (2ab + a).103 + (b2+ b)102 +52
= [ a2.102 +2ab.10 + a.10 + b2 + b] 102+ 52
= (10a + b) ( 10a+b+1).102 + 25
8

11. = P (P+1) 102+ 25, where P = 10a+b.
Hence any three digit number whose last digit is 5 gives the same result as in
(a) for P=10a + b, the ‘previous’ of 5.
Example : 1652= (1 . 102 + 6 . 10 + 5) 2.
It is of the form (ax2+bx+c)2for a = 1, b = 6, c = 5 and x = 10. It gives the
answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’. The
answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.
Apply Ekadhikena purvena to find the squares of the numbers 95, 225,
375, 635, 745, 915, 1105, 2545.
ii) Vulgar fractions whose denominators are numbers ending in NINE :
We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion
of such vulgar fractions into recurring decimals, Ekadhika process can be
effectively used both in division and multiplication.
a) Division Method : Value of 1 / 19.
The numbers of decimal places before repetition is the difference of numerator
and denominator, i.e.,, 19 -1=18 places.
For the denominator 19, the purva (previous) is 1.
Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2.
The sutra is applied in a different context. Now the method of division is as
Step. 1 : Divide numerator 1 by 20.
i.e.,, 1 / 20 = 0.1 / 2 = .10 ( 0 times, 1 remainder)
Step. 2 : Divide 10 by 2
9

12. i.e.,, 0.005( 5 times, 0 remainder )
Step. 3 : Divide 5 by 2
i.e.,, 0.0512 ( 2 times, 1 remainder )
Step. 4 : Divide 12 i.e.,, 12 by 2
i.e.,, 0.0526 ( 6 times, No remainder )
Step. 5 : Divide 6 by 2
i.e.,, 0.05263 ( 3 times, No remainder )
Step. 6 : Divide 3 by 2
i.e.,, 0.0526311(1 time, 1 remainder )
Step. 7 : Divide 11 i.e.,, 11 by 2
i.e.,, 0.05263115 (5 times, 1 remainder )
Step. 8 : Divide 15 i.e.,, 15 by 2
i.e.,, 0.052631517 ( 7 times, 1 remainder )
Step. 9 : Divide 17 i.e.,, 17 by 2
i.e.,, 0.05263157 18 (8 times, 1 remainder )
Step. 10 : Divide 18 i.e.,, 18 by 2
i.e.,, 0.0526315789 (9 times, No remainder )
Step. 11 : Divide 9 by 2
i.e.,, 0.0526315789 14 (4 times, 1 remainder )
Step. 12 : Divide 14 i.e.,, 14 by 2
i.e.,, 0.052631578947 ( 7 times, No remainder )
Step. 13 : Divide 7 by 2
10

13. i.e.,, 0.05263157894713 ( 3 times, 1 remainder )
Step. 14 : Divide 13 i.e.,, 13 by 2
i.e.,, 0.052631578947316 ( 6 times, 1 remainder )
Step. 15 : Divide 16 i.e.,, 16 by 2
i.e.,, 0.052631578947368 (8 times, No remainder )
Step. 16 : Divide 8 by 2
i.e.,, 0.0526315789473684 ( 4 times, No remainder )
Step. 17 : Divide 4 by 2
i.e.,, 0.05263157894736842 ( 2 times, No remainder )
Step. 18 : Divide 2 by 2
i.e.,, 0.052631578947368421 ( 1 time, No remainder )
Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving
0 __________________ . .
1 / 19 =0.052631578947368421 or 0.052631578947368421
Note that we have completed the process of division only by using ‘2’. Nowhere
the division by 19 occurs.
b) Multiplication Method: Value of 1 / 19
First we recognize the last digit of the denominator of the type 1 / a9. Here the
last digit is 9.
For a fraction of the form in whose denominator 9 is the last digit, we take the
case of 1 / 19 as follows:
For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2.
Therefore 2 is the multiplier for the conversion. We write the last digit in the
numerator as 1 and follow the steps leftwards.
Step. 1 : 1
11

14. Step. 2 : 21(multiply 1 by 2, put to left)
Step. 3 : 421(multiply 2 by 2, put to left)
Step. 4 : 8421(multiply 4 by 2, put to left)
Step. 5 : 168421 (multiply 8 by 2 =16, 1 carried over, 6 put to left)
Step. 6 : 1368421 ( 6 X 2 =12,+1 [carry over]
= 13, 1 carried over, 3 put to left )
Step. 7 : 7368421 ( 3 X 2, = 6 +1 [Carryover]
= 7, put to left)
Step. 8 : 147368421 (as in the same process)
Step. 9 : 947368421 ( Do – continue to step 18)
Step. 10 : 18947368421
Step. 11 : 178947368421
Step. 12 : 1578947368421
Step. 13 : 11578947368421
Step. 14 : 31578947368421
Step. 15 : 631578947368421
Step. 16 : 12631578947368421
Step. 17 : 52631578947368421
Step. 18 : 1052631578947368421
Now from step 18 onwards the same numbers and order towards left continue.
Thus 1 / 19 = 0.052631578947368421
It is interesting to note that we have
i) not at all used division process
12

15. ii) instead of dividing 1 by 19 continuously, just multiplied 1 by 2 and continued to multiply
the resultant successively by 2.
Observations :
a) For any fraction of the form 1 / a9 i.e.,, in whose denominator 9 is the digit in the units place
and a is the set of remaining digits, the value of the fraction is in recurring decimal form and the
repeating block’s right most digit is 1.
b) Whatever may be a9, and the numerator, it is enough to follow the said process with (a+1)
either in division or in multiplication.
c) Starting from right most digit and counting from the right, we see ( in the given example 1 /
Sum of 1st digit + 10th digit = 1 + 8 = 9
Sum of 2nd digit + 11th digit = 2 + 7 = 9
- - - - - - - - -- - - - - - - - - - - - - - - - - - -
Sum of 9th digit + 18th digit = 9+ 0 = 9
From the above observations, we conclude that if we find first 9 digits, further digits can
be derived as complements of 9.
i) Thus at the step 8 in division process we have 0.052631517 and next step. 9 gives
Now the complements of the numbers
0, 5, 2, 6, 3, 1, 5, 7, 8 from 9
9, 4, 7, 3, 6, 8, 4, 2, 1 follow the right order
i.e.,, 0.052631578947368421
Now taking the multiplication process we have
Step. 8 : 147368421
13

16. Step. 9 : 947368421
Now the complements of 1, 2, 4, 8, 6, 3, 7, 4, 9 from 9
i.e.,, 8, 7, 5, 1, 3, 6, 2, 5, 0 precede in successive steps, giving the answer.
0.052631578947368421.
d) When we get (Denominator – Numerator) as the product in the multiplicative
process, half the work is done. We stop the multiplication there and
mechanically write the remaining half of the answer by merely taking down
complements from 9.
e) Either division or multiplication process of giving the answer can be put in a
single line form.
Algebraic proof :
Any vulgar fraction of the form 1 / a9 can be written as
1 / a9 = 1 / ( (a + 1 ) x - 1 ) where x = 10
1
= ________________________
(a+1) x [1 - 1/(a+1)x ]
1
= ___________
[1 - 1/(a+1)x]-1
(a+1)x
1
= __________ [1 + 1/(a+1)x + 1/(a+1)x2+ ----------]
(a+1)x
= 1/(a+1)x + 1/(a+1)2x2 +1/(a+1)3x3+ ----ad infinitum
= 10-1(1/(a+1))+10-2(1/(a+1)2)+10-3(1/(a+1)3) + ---ad infinitum
14

17. This series explains the process of ekadhik.
Now consider the problem of 1 / 19. From above we get
1 / 19 = 10-1 (1/(1+1)) + 10-2 (1/(1+1)2) + 10-3 (1/(1+1)3) + ----
( since a=1)
= 10-1 (1/2) + 10-2 (1/2)2 + 10-3 (1/3)3 + ----------
= 10-1 (0.5) + 10-2 (0.25) + 10-3 (0.125)+ ----------
= 0.05 + 0.0025 + 0.000125 + 0.00000625+ - - - - - - -
= 0.052631 - - - - - - -
Example1 :
1. Find 1 / 49 by ekadhikena process.
Now ‘previous’ is 4. ‘One more than the previous’ is 4 + 1 = 5.
Now by division right ward from the left by ‘5’.
1 / 49 = .10 - - - - - - - - - - - -(divide 1 by 50)
= .02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder )
= .0220 - - - - - - --(divide 20 by 5, 4 times)
= .0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder )
= .020440 -- - -- - ( divide 40 by 5, 8 times )
= .020408 - - - - - (divide 8 by 5, 1 time, 3 remainder )
= .02040831 - - - -(divide 31 by 5, 6 times, 1 remainder )
= .02040811 6 - - - - - - - continue
15

18. = .0204081613322615306111222244448 - -- - - - -
On completing 21 digits, we get 48
i.e.,,Denominator - Numerator = 49 – 1 = 48 stands.
i.e, half of the process stops here. The remaining half can be obtained as
complements from 9.
.
Thus 1 / 49 = 0.020408163265306122448
.
979591836734693877551
Now finding 1 / 49 by process of multiplication left ward from right by 5, we
1 / 49 = ----------------------------------------------1
= ---------------------------------------------51
= -------------------------------------------2551
= ------------------------------------------27551
= ---- 483947294594118333617233446943383727551
i.e.,,Denominator – Numerator = 49 – 1 = 48 is obtained as 5X9+3
( Carry over ) = 45 + 3 = 48. Hence half of the process is over. The remaining
half is automatically obtained as complements of 9.
Thus 1 / 49 = ---------------979591836734693877551
.
= 0.020408163265306122448
.
979591836734693877551
16

19. Example 2: Find 1 / 39 by Ekadhika process.
Now by multiplication method, Ekadhikena purva is 3 + 1 = 4
1 / 39 = -------------------------------------1
= -------------------------------------41
= ----------------------------------1641
= ---------------------------------25641
= --------------------------------225641
= -------------------------------1025641
Here the repeating block happens to be block of 6 digits. Now the rule
predicting the completion of half of the computation does not hold. The
complete block has to be computed by ekadhika process.
Now continue and obtain the result. Find reasons for the non–applicability of
the said ‘rule’.
Find the recurring decimal form of the fractions 1 / 29, 1 / 59,
1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible. Judge whether
the rule of completion of half the computation holds good in such cases.
Note : The Ekadhikena Purvena sutra can also be used for conversion of vulgar
fractions ending in 1, 3, 7 such as 1 / 11, 1 / 21, 1 / 31 - - -- ,1 / 13, 1 / 23, - -
- -, 1 / 7, 1 / 17, - - - - - by writing them in the following way and solving them.
17

20. 2. Nikhilam navatascaramam Dasatah
The formula simply means : “all from 9 and the last from 10”
The formula can be very effectively applied in multiplication of numbers, which
are nearer to bases like 10, 100, 1000i.e., to the powers of 10 . The procedure
of multiplication using the Nikhilam involves minimum number of steps, space,
time saving and only mental calculation. The numbers taken can be either less
or more than the base considered.
The difference between the number and the base is termed as deviation.
Deviation may be positive or negative. Positive deviation is written without the
positive sign and the negative deviation, is written using Rekhank (a bar on the
number). Now observe the following table.
Number Base Number – Base Deviation
14 10 14 - 10 4
_
8 10 8 - 10 -2 or 2
__
97 100 97 - 100 -03 or 03
112 100 112 - 100 12
___
993 1000 993 - 1000 -007 or 007
1011 1000 1011 - 1000 011
Some rules of the method (near to the base) in Multiplication
a) Since deviation is obtained by Nikhilam sutra we call the method as Nikhilam
Eg : 94. Now deviation can be obtained by ‘all from 9 and the last from 10’
sutrai.e., the last digit 4 is from 10 and remaining digit 9 from 9 gives 06.
b) The two numbers under consideration are written one below the other. The
deviations are written on the right hand side.
Eg : Multiply 7 by 8.
Now the base is 10. Since it is near to both the numbers,
7
we write the numbers one below the other. 8
18

21. -----
Take the deviations of both the numbers from
the base and represent _
7 3
_
Rekhank or the minus sign before the deviations 8 2
------
------
or 7 -3
8 -2
-------
-------
or remainders 3 and 2 implies that the numbers to be multiplied are both less
than 10
c) The product or answer will have two parts, one on the left side and the other
on the right. A vertical or a slant linei.e., a slash may be drawn for the
demarcation of the two parts i.e.,
(or)
d) The R.H.S. of the answer is the product of the deviations of the numbers. It
shall contain the number of digits equal to number of zeroes in the base.
_
i.e., 7 3
_
8 2
_____________
/ (3x2) = 6
Since base is 10, 6 can be taken as it is.
e) L.H.S of the answer is the sum of one number with the deviation of the
other. It can be arrived at in any one of the four ways.
i) Cross-subtract deviation 2 on the second row from the original number7 in
the first row i.e., 7-2 = 5.
ii) Cross–subtract deviation 3 on the first row from the original number8 in the
19

22. second row (converse way of(i))
i.e., 8 - 3 = 5
iii) Subtract the base 10 from the sum of the given numbers.
i.e., (7 + 8) – 10 = 5
iv) Subtract the sum of the two deviations from the base.
i.e., 10 – ( 3 + 2) = 5
Hence 5 is left hand side of the answer.
_
Thus 7 3
_
8 2
‾‾‾‾‾‾‾‾‾‾‾‾
5/
Now (d) and (e) together give the solution
_
7 3 7
_
8 2 i.e., X 8
‾‾‾‾‾‾‾ ‾‾‾‾‾‾
5/ 6 56
f) If R.H.S. contains less number of digits than the number of zeros in the base,
the remaining digits are filled up by giving zero or zeroes on the left side of the
R.H.S. If the number of digits are more than the number of zeroes in the base,
the excess digit or digits are to be added to L.H.S of the answer.
The general form of the multiplication under Nikhilam can be shown as follows :
Let N1 and N2 be two numbers near to a given base in powers of 10, andD1 and
D2 are their respective deviations from the base. ThenN1 X N2 can be
represented as
Case (i) : Both the numbers are lower than the base.We have already
considered the example 7 x 8 , with base 10.
20

23. Now let us solve some more examples by taking bases 100 and 1000
Ex. 1: Find 97 X 94. Here base is 100. Now following the rules, the working is
as follows:
Ex. 2: 98 X 97 Base is 100.
Ex. 3: 75X95. Base is 100.
Ex. 4: 986 X 989. Base is 1000.
Ex. 5: 994X988. Base is 1000.
21

24. Ex. 6: 750X995.
Case ( ii) : Both the numbers are higher than the base.
The method and rules follow as they are. The only difference is the positive
deviation. Instead of cross – subtract, we follow cross – add.
Ex. 7: 13X12. Base is 10
Ex. 8: 18X14. Base is 10.
Ex. 9: 104X102. Base is 100.
104 04
102 02
‾‾‾‾‾‾‾‾‾‾‾‾
106 / 4x2 = 10608 ( rule -f )
‾‾‾‾‾‾‾‾‾‾‾‾
Ex. 10: 1275X1004. Base is 1000.
22

25. 1275 275
1004 004
‾‾‾‾‾‾‾‾‾‾‾‾
1279/ 275x4 = 1279 / 1100 ( rule -f )
____________ = 1280100
Case ( iii ): One number is more and the other is less than the base.
In this situation one deviation is positive and the other is negative. So the
product of deviations becomes negative. So the right hand side of the answer
obtained will therefore have to be subtracted. To have a clear representation
and understanding a vinculum is used. It proceeds into normalization.
Ex.11: 13X7. Base is 10
Note : Conversion of common number into vinculum number and vice versa.
Eg :
__
9 = 10 –1 = 11
_
98 = 100 – 2 = 102
_
196 = 200 – 4 = 204
_
32 = 30 – 2 = 28
_
145 = 140 – 5 = 135
_
322 = 300 – 22 = 278. etc
The procedure can be explained in detail using Nikhilam Navatascaram Dasatah,
Ekadhikenapurvena, Ekanyunena purvena in the foregoing pages of this book.]
Ex. 12: 108 X 94. Base is 100.
23

26. Ex. 13: 998 X 1025. Base is 1000.
Algebraic Proof:
Case ( i ):
Let the two numbers N1 and N2 be less than the selected base say x.
N1 = (x-a), N2 = (x-b). Here a and b are the corresponding deviations of the
numbersN1 and N2 from the base x. Observe that x is a multiple of 10.
Now N1 X N2 = (x-a) (x-b) = x.x – x.b – a.x + ab
= x (x – a – b ) + ab. [rule – e(iv), d ]
= x [(x – a) – b] + ab = x (N1–b) + ab[rule–e(i),d]
or = x [(x – b) – a] = x (N2 – a) + ab. [rule –e (ii),d]
x (x – a – b) + ab can also be written as
x[(x – a) + (x – b) – x] + ab = x[N1+N2 – x] + ab [rule –e(iii),d].
A difficult can be faced, if the vertical multiplication of the deficit digits or
deviationsi.e., a.b yields a product consisting of more than the required digits.
Then rule-f will enable us to surmount the difficulty.
Case ( ii ) :
When both the numbers exceed the selected base, we have N1 = x + a,N2 = x +
b, x being the base. Now the identity (x+a) (x+b) = x(x+a+b) + a.b holds
good, of course with relevant details mentioned in case(i).
24

27. Case ( iii ) :
When one number is less and another is more than the base, we can use (x-
a)(x+b) = x(x–a+ b)–ab. and the procedure is evident from the examples
Find the following products by Nikhilam formula.
1) 7 X 4 2) 93 X 85 3) 875 X 994
4) 1234 X 1002 5) 1003 X 997 6) 11112 X 9998
7) 1234 X 1002 8) 118 X 105
Nikhilam in Division
Consider some two digit numbers (dividends) and same divisor 9. Observe the
following example.
i) 13 ÷ 9 The quotient (Q) is 1, Remainder (R) is 4.
since 9 ) 13 ( 1
9
____
4
ii) 34 ÷ 9, Q is 3, R is 7.
iii) 60 ÷ 9, Q is 6, R is 6.
iv) 80 ÷ 9, Q is 8, R is 8.
Now we have another type of representation for the above examples as given
i) Split each dividend into a left hand part for the Quotient and right - hand part
for the remainder by a slant line or slash.
Eg. 13 as 1 / 3, 34 as 3 / 4 , 80 as 8 / 0.
ii) Leave some space below such representation, draw a horizontal line.
25

28. Eg. 1/3 3/4 8/0
______ , ______ , ______
iii) Put the first digit of the dividend as it is under the horizontal line. Put the
same digit under the right hand part for the remainder, add the two and place
the sumi.e., sum of the digits of the numbers as the remainder.
1/3 3/4 8/0
1 3 8
______ , ______ , ______
1/4 3/7 8/8
Now the problem is over. i.e.,
13 ÷ 9 gives Q = 1, R = 4
34 ÷ 9 gives Q = 3, R = 7
80 ÷ 9 gives Q = 8, R = 8
Proceeding for some more of the two digit number division by 9, we get
a) 21 ÷ 9 as
9) 2 / 1 i.e Q=2, R=3
2
‾‾‾‾‾‾
2 / 3
b) 43 ÷ 9 as
9) 4 / 3 i.e Q = 4, R = 7.
4
‾‾‾‾‾‾
4 / 7
The examples given so far convey that in the division of two digit numbers by 9,
we canmechanically take the first digit down for the quotient – column and that,
by adding the quotient to the second digit, we can get the remainder.
Now in the case of 3 digit numbers, let us proceed as follows.
9 ) 104 ( 11 9 ) 10 / 4
99 1 / 1
26

29. ‾‾‾‾‾‾ as ‾‾‾‾‾‾‾
5 11 / 5
9 ) 212 ( 23 9 ) 21 / 2
207 2 / 3
‾‾‾‾‾ as ‾‾‾‾‾‾‾
5 23 / 5
9 ) 401 (44 9 ) 40 / 1
396 4 / 4
‾‾‾‾‾ as ‾‾‾‾‾‾‾‾
5 44 / 5
Note that the remainder is the sum of the digits of the dividend. The first digit
of the dividend from left is added mechanically to the second digit of the
dividend to obtain the second digit of the quotient. This digit added to the third
digit sets the remainder. The first digit of the dividend remains as the first digit
of the quotient.
Consider 511 ÷ 9
Add the first digit 5 to second digit 1 getting 5 + 1 = 6. Hence Quotient is 56.
Now second digit of 56 i.e., 6 is added to third digit 1 of dividend to get the
remainder i.e., 1 + 6 = 7
9) 51 / 1
5/ 6
‾‾‾‾‾‾‾
56 / 7
Q is 56, R is 7.
Extending the same principle even to bigger numbers of still more digits, we can
get the results.
Eg : 1204 ÷ 9
i) Add first digit 1 to the second digit 2. 1 + 2 = 3
ii) Add the second digit of quotient 13. i.e., 3 to third digit ‘0’ and obtain the
Quotient. 3 + 0 = 3, 133
27

30. iii) Add the third digit of Quotient 133 i.e.,3 to last digit ‘4’ of the dividend and
write the final Quotient and Remainder. R = 3 + 4 = 7, Q = 133
In symbolic form 9 ) 120 / 4
13 / 3
‾‾‾‾‾‾‾‾
133 / 7
Another example.
9 ) 13210 /1 132101 ÷ 9
gives
1467 / 7 Q = 14677, R = 8
‾‾‾‾‾‾‾‾‾‾
14677 / 8
In all the cases mentioned above, the remainder is less than the divisor. What
about the case when the remainder is equal or greater than the divisor?
9) 3 / 6 9) 24 / 6
3 2 / 6
‾‾‾‾‾‾ or ‾‾‾‾‾‾‾‾
3 / 9 (equal) 26 / 12 (greater).
We proceed by re-dividing the remainder by 9, carrying over this Quotient to
the quotient side and retaining the final remainder in the remainder side.
9) 3 / 6 9 ) 24 / 6
/ 3 2 / 6
‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾
3 / 9 26 / 12
‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾
4 / 0 27 / 3
Q = 4, R=0 Q = 27, R = 3.
When the remainder is greater than divisor, it can also be represented as
9 ) 24 / 6
2 / 6
‾‾‾‾‾‾‾‾
26 /1 / 2
/1
‾‾‾‾‾‾‾‾
1 /3
28

31. ‾‾‾‾‾‾‾‾
27 / 3
Now consider the divisors of two or more digits whose last digit is 9,when
divisor is 89.
We Know 113 = 1 X 89 + 24, Q =1, R = 24
10015 = 112 X 89 + 47, Q = 112, R = 47.
Representing in the previous form of procedure, we have
89 ) 1 / 13 89 ) 100 / 15
/ 11 12 / 32
‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾
1 / 24 112 / 47
But how to get these? What is the procedure?
Now Nikhilam rule comes to rescue us. The nikhilam states “all from 9 and the
last from 10”. Now if you want to find 113 ÷ 89, 10015 ÷ 89, you have to apply
nikhilam formula on 89 and get the complement 11.Further while carrying the
added numbers to the place below the next digit, we have to multiply by this
89 ) 1 / 13 89 ) 100 /15
‾‾
/ 11 11 11 / first digit 1 x 11
‾‾‾‾‾‾‾‾
1 / 24 1/ 1 total second is 1x11
22 total of 3rd digit is 2 x 11
‾‾‾‾‾‾‾‾‾‾
112 / 47
What is 10015 ÷ 98 ? Apply Nikhilam and get 100 – 98 = 02. Set off the 2
digits from the right as the remainder consists of 2 digits. While carrying the
added numbers to the place below the next digit, multiply by 02.
98 ) 100 / 15
‾‾
02 02 / i.e., 10015 ÷ 98 gives
0 / 0 Q = 102, R = 19
/ 04
‾‾‾‾‾‾‾‾‾‾
102 / 19
29

32. In the same way
897 ) 11 / 422
‾‾‾
103 1 / 03
/ 206
‾‾‾‾‾‾‾‾‾
12 / 658
gives 11,422 ÷ 897, Q = 12, R=658.
In this way we have to multiply the quotient by 2 in the case of 8, by 3 in the
case of 7, by 4 in the case of 6 and so on. i.e., multiply the Quotient digit by the
divisors complement from 10. In case of more digited numbers we apply
Nikhilam and proceed. Any how, this method is highly useful and effective for
division when the numbers are near to bases of 10.
* Guess the logic in the process of division by 9.
* Obtain the Quotient and Remainder for the following problems.
1) 311 ÷ 9 2) 120012 ÷ 9 3) 1135 ÷ 97
4) 2342 ÷ 98 5) 113401 ÷ 997
6) 11199171 ÷ 99979
Observe that by nikhilam process of division, even lengthier divisions involve no
division or no subtraction but only a few multiplications of single digits with
small numbers and a simple addition. But we know fairly well that only a special
type of cases are being dealt and hence many questions about various other
types of problems arise. The answer lies in Vedic Methods.
30

33. 3. Urdhva - tiryagbhyam
Urdhva – tiryagbhyam is the general formula applicable to all cases of
multiplication and also in the division of a large number by another large
number. It means
(a) Multiplication of two 2 digit numbers.
Ex.1: Find the product 14 X 12
i) The right hand most digit of the multiplicand, the first number (14) i.e.,4 is
multiplied by the right hand most digit of the multiplier, the second number
(12)i.e., 2. The product 4 X 2 = 8 forms the right hand most part of the answer.
ii) Now, diagonally multiply the first digit of the multiplicand (14) i.e., 4 and
second digit of the multiplier (12)i.e., 1 (answer 4 X 1=4); then multiply the
second digit of the multiplicand i.e.,1 and first digit of the multiplier i.e., 2
(answer 1 X 2 = 2); add these two i.e.,4 + 2 = 6. It gives the next, i.e., second
digit of the answer. Hence second digit of the answer is 6.
iii) Now, multiply the second digit of the multiplicand i.e., 1 and second digit of
the multiplieri.e., 1 vertically, i.e., 1 X 1 = 1. It gives the left hand most part of
the answer.
Thus the answer is 16 8.
Symbolically we can represent the process as follows :
The symbols are operated from right to left .
Step i) :
31

34. Step ii) :
Step iii) :
Now in the same process, answer can be written as
23
13
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
2 : 6 + 3 : 9 = 299 (Recall the 3 steps)
41
X 41
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
16 : 4 + 4 : 1 = 1681.
32

35. What happens when one of the results i.e., either in the last digit or in the
middle digit of the result, contains more than 1 digit ? Answer is simple. The
right – hand – most digit there of is to be put down there and the
preceding,i.e., left –hand –side digit or digits should be carried over to the left
and placed under the previous digit or digits of the upper row. The digits carried
over may be written as in Ex. 4.
Ex.4: 32 X 24
Step (i) : 2X4=8
Step (ii) : 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16.
Here 6 is to be retained. 1 is to be carried out to left side.
Step (iii) : 3 X 2 = 6. Now the carried over digit 1 of 16 is to be added.
i.e., 6 + 1 = 7.
Thus 32 X 24 = 768
We can write it as follows
32
24
‾‾‾‾
668
1
‾‾‾‾
768.
33

36. Note that the carried over digit from the result (3X4) + (2X2) = 12+4 = 16
i.e.,1 is placed under the previous digit 3 X 2 = 6 and added.
After sufficient practice, you feel no necessity of writing in this way and simply
operate or perform mentally.
Ex.5 28 X 35.
Step (i) : 8 X 5 = 40. 0 is retained as the first digit of the answer and 4 is
carried over.
Step (ii) : 2 X 5 = 10; 8 X 3 = 24; 10 + 24 = 34; add the carried over 4 to
34. Now the result is 34 + 4 = 38. Now 8 is retained as the second digit of the
answer and3 is carried over.
Step (iii) : 2 X 3 = 6; add the carried over 3 to 6. The result 6 + 3 = 9 is the
third or final digit from right to left of the answer.
Thus 28 X 35 = 980.
48
47
‾‾‾‾‾‾
1606
65
‾‾‾‾‾‾‾
2256
Step (i): 8 X 7 = 56; 5, the carried over digit is placed below the second
Step (ii): ( 4 X 7) + (8 X 4) = 28 + 32 = 60; 6, the carried over digit is
placed below the third digit.
Step (iii): Respective digits are added.
Algebraic proof :
a) Let the two 2 digit numbers be (ax+b) and (cx+d). Note that x = 10. Now
consider the product
34

37. (ax + b) (cx + d) = ac.x2 + adx + bcx + b.d
= ac.x2 + (ad + bc)x + b.d
Observe that
i) The first term i.e., the coefficient of x2 (i.e., 100, hence the digit in the100th
place) is obtained by vertical multiplication of a and c i.e.,the digits in10th place
(coefficient of x) of both the numbers;
ii) The middle term, i.e., the coefficient of x (i.e., digit in the 10th place) is
obtained by cross wise multiplication of a and d; and of b and c; and the
addition of the two products;
iii) The last (independent of x) term is obtained by vertical multiplication of the
independent terms b and d.
b) Consider the multiplication of two 3 digit numbers.
Let the two numbers be(ax2 + bx + c) and (dx2 + ex + f). Note that x=10
Now the product is
ax2 + bx + c
x dx2 + ex + f
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
ad.x4+bd.x3+cd.x2+ae.x3+be.x2+ce.x+af.x2+bf.x+cf
= ad.x4 + (bd + ae). x3 + (cd + be + af).x2 + (ce + bf)x + cf
35

38. Note the following points :
i) The coefficient of x4 , i.e., ad is obtained by the vertical multiplication of the
firstcoefficient from the left side :
ii)The coefficient of x3 , i.e., (ae + bd) is obtained by the cross –wise
multiplication of the first two coefficients and by the addition of the two
iii) The coefficient of x2 is obtained by the multiplication of the first coefficient
of the multiplicand(ax2+bx +c) i.e., a; by the last coefficient of the multiplier
(dx2 +ex +f)i.e.,f ; of the middle one i.e., b of the multiplicand by the middle
one i.e., e of the multiplier and of the last onei.e., c of the multiplicand by the
first one i.e., d of the multiplier and by the addition of all the three productsi.e.,
af + be +cd :
iv) The coefficient of x is obtained by the cross wise multiplication of the second
coefficienti.e., b of the multiplicand by the third one i.e., f of the multiplier, and
conversely the third coefficienti.e., c of the multiplicand by the second
coefficient i.e., e of the multiplier and by addition of the two products,i.e., bf +
ce ;
36

39. v) And finally the last (independent of x) term is obtained by the vertical
multiplication of the last coefficients c and f i.e., cf
Thus the process can be put symbolically as (from left to right)
Consider the following example
124 X 132.
Proceeding from right to left
i) 4 X 2 = 8. First digit = 8
ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6 is retained and 1 is carried over
to left side. Second digit = 6.
iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried over 1 of above
step is added i.e., 12 + 1 = 13. Now 3 is retained and 1 is carried over to left
37

40. side. Thus third digit = 3.
iv) ( 1X 3 ) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step is added
i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6
v) ( 1 X 1 ) = 1. As there is no carried over number from the previous step it is
retained. Thus fifth digit = 1
124 X 132 = 16368.
Let us work another problem by placing the carried over digits under the first
row and proceed.
234
x 316
‾‾‾‾‾‾‾
61724
1222
‾‾‾‾‾‾‾
73944
i) 4 X 6 = 24 : 2, the carried over digit is placed below the second digit.
ii) (3 X 6) + (4 x 1) = 18 + 4 = 22 ; 2, the carried over digit is placed below
third digit.
iii) (2 X 6) + (3 X 1) + (4 X 3) = 12 + 3 + 12 = 27 ; 2, the carried over digit is
placed below fourth digit.
iv) (2 X 1) + ( 3 X 3) = 2 + 9 = 11; 1, the carried over digit is placed below
fifth digit.
v) ( 2 X 3 ) = 6.
vi) Respective digits are added.
Note :
1. We can carry out the multiplication in urdhva - tiryak process from left to
right or right to left.
2. The same process can be applied even for numbers having more digits.
3. urdhva –tiryak process of multiplication can be effectively used in
multiplication regarding algebraic expressions.
38

41. Example 1 : Find the product of (a+2b) and (3a+b).
Example 2 :
3a2 + 2a + 4
x 2a2 + 5a + 3
‾‾‾‾‾‾‾‾‾‾‾‾‾‾
i) 4X3 = 12
ii) (2 X 3) + ( 4 X 5 ) = 6 + 20 =26 i.e., 26a
iii) (3 X 3) + ( 2 X 5 ) + ( 4 X 2 ) = 9 + 10 + 8 = 27 i.e., 27a2
iv) (3 X 5) + ( 2 X 2 ) = 15 + 4 = 19 i.e., 19 a3
v) 3X2 = 6 i.e.,6a4
Hence the product is 6a4 +19a3 + 27a2 + 26a + 12
Example 3 : Find (3x2 + 4x + 7) (5x +6)
Now 3.x2 + 4x + 7
0.x2 + 5x + 6
‾‾‾‾‾‾‾‾‾‾‾‾
i) 7 X 6 = 42
ii) (4 X 6) + (7 X 5) = 24 + 35 = 59 i.e., 59x
iii) (3 X 6) + (4 X 5) + (7 X 0) = 18 + 20 + 0 = 38 i.e., 38x2
iv) (3 X 5) + (0 X 4) = 15 + 0 = 15 i.e., 15x3
v) 3 X 0 = 0
39

42. Hence the product is 15x3 +38x2 + 59x + 42
Find the products using urdhva tiryagbhyam process.
1) 25 X 16 2) 32 X 48 3) 56 X 56
4) 137 X 214 5) 321 X 213 6) 452 X 348
7) (2x + 3y) (4x + 5y) 8) (5a2 + 1) (3a2 + 4)
9) (6x2 + 5x + 2 ) (3x2 + 4x +7) 10) (4x2 + 3) (5x + 6)
Urdhva – tiryak in converse for division process:
As per the statement it an used as a simple argumentation for division process
particularly in algebra.
Consider the division of (x3 +5x2 + 3x + 7) by (x – 2)process by converse of
urdhva – tiryak :
i) x3 divided by x gives x2 . x3 + 5x2 + 3x + 7
It is the first term of the Quotient. ___________________
x–2
2
Q=x +-----------
ii) x2 X – 2 = - 2x2 . But 5x2 in the dividend hints7x2 more since 7x2 – 2x2 =
5x2 . This ‘more’ can be obtained from the multiplication of x by 7x. Hence
second term of Q is 7x.
x3 + 5x2 + 3x + 7
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ gives Q = x2 + 7x + - - - - - - - -
x–2
iii)We now have – 2 X 7x = -14x. But the 3rd term in the dividend is 3x for
which ‘17x more’ is required since 17x – 14x =3x.
Now multiplication of x by 17 gives 17x. Hence third term of quotient is 17
40

43. Thus
x3 + 5x2 + 3x + 7
_________________ gives Q= x2 + 7x +17
x–2
iv) Now last term of Q, i.e., 17 multiplied by –2 gives 17X–2 = -34 but the
relevant term in dividend is 7. So 7 + 34 = 41 ‘more’ is required. As there no
more terms left in dividend, 41 remains as the remainder.
x3 + 5x2 + 3x + 7
________________ gives Q=x2 + 7x +17 and R = 41.
x–2
Find the Q and R in the following divisions by using the converse
process of urdhva – tiryagbhyam method :
1) 3x2 – x – 6 2) 16x2 + 24x +9
‾‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾‾‾
3x – 7 4x+3
3) x3+ 2x2 +3x + 5 4) 12x4 – 3x2 – 3x + 12
‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
x-3 x2 + 1
4. Paravartya Yojayet
'Paravartya – Yojayet' means 'transpose and apply'
(i) Consider the division by divisors of more than one digit, and when the
divisors are slightly greater than powers of 10.
Example 1 : Divide 1225 by 12.
Step 1 : (From left to right ) write the Divisor leaving the first digit, write the
other digit or digits using negative (-) sign and place them below the divisor
as shown.
12
-2
‾‾‾‾
Step 2 : Write down the dividend to the right. Set apart the last digit for the
remainder.
41

44. i.e.,, 12 122 5
-2
Step 3 : Write the 1st digit below the horizontal line drawn under
thedividend. Multiply the digit by –2, write the product below the 2nd digit
and add.
i.e.,, 12 122 5
-2 -2
‾‾‾‾‾ ‾‾‾‾
10
Since 1 x –2 = -2and 2 + (-2) = 0
Step 4 : We get second digits’ sum as ‘0’. Multiply the second digits’ sum
thus obtained by –2 and writes the product under 3rd digit and add.
12 122 5
-2 -20
‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾
102 5
Step 5 : Continue the process to the last digit.
i.e., 12 122 5
-2 -20 -4
‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾
102 1
Step 6: The sum of the last digit is the Remainder and the result to its left is
Thus Q = 102 andR = 1
Example 2 : Divide 1697 by 14.
14 169 7
-4 -4–8–4
‾‾‾‾ ‾‾‾‾‾‾‾
121 3
Q = 121, R = 3.
Example 3 : Divide 2598 by 123.
Note that the divisor has 3 digits. So we have to set up the last two
42

45. digits of the dividend for the remainder.
123 25 98 Step ( 1 ) & Step ( 2 )
-2-3
‾‾‾‾‾ ‾‾‾‾‾‾‾‾
Now proceed the sequence of steps write –2 and –3 as follows :
123 25 98
-2-3 -4 -6
‾‾‾‾‾ -2–3
‾‾‾‾‾‾‾‾‾‾
21 15
Since 2 X (-2, -3)= -4 , -6;5 – 4 = 1
and (1 X (-2,-3); 9 – 6 – 2 = 1; 8 – 3 = 5.
Hence Q = 21 and R = 15.
Example 4 : Divide 239479 by 11213. The divisor has 5 digits. So the last
4digits of the dividend are to be set up for Remainder.
1 1213 23 9 479
-1-2-1-3 -2 -4-2-6 with 2
‾‾‾‾‾‾‾‾ -1-2-1-3 with 1
‾‾‾‾‾‾‾‾‾‾‾‾‾
21 40 0 6
Hence Q = 21, R = 4006.
Example 5 : Divide 13456 by 1123
1 12 3 134 5 6
-1–2–3 -1-2-3
‾‾‾‾‾‾‾ -2-4 –6
‾‾‾‾‾‾‾‾‾‾‾‾‾
1 2 0–2 0
Note that the remainder portion contains –20, i.e.,, a negative quantity. To
over come this situation, take 1 over from the quotient column, i.e.,, 1123
over to the right side, subtract the remainder portion 20 to get the actual
Thus Q = 12 – 1 = 11, and R = 1123 - 20 = 1103.
43

46. Find the Quotient and Remainder for the problems using
paravartya – yojayet method.
1) 1234 ÷ 112 2) 11329 ÷ 1132
3) 12349÷ 133 4) 239479÷1203
Now let us consider the application of paravartya – yojayet in algebra.
Example 1 : Divide 6x2 + 5x + 4 by x – 1
X- 1 6x2 + 5x + 4
‾‾‾‾‾‾
1 6+ 11
‾‾‾‾‾‾‾‾‾‾‾‾
6x + 11 + 15 ThusQ = 6x+11,R=15.
Example 2 : Divide x3 –3x2 + 10x – 4 by x - 5
X- 5 x3 – 3x2 + 10x – 4
‾‾‾‾‾
5 5 + 10 100
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
x2 + 2x + 20, + 96
Thus Q= x2 + 2x + 20, R = 96.
The procedure as a mental exercise comes as follows :
i) x3 / xgives x2 i.e.,, 1 the first coefficient in the Quotient.
ii) Multiply 1 by + 5,(obtained after reversing the sign of second term in the
Quotient) and add to the next coefficient in the dividend. It gives 1 X( +5) =
+5, adding to the next coefficient, i.e.,, –3 + 5 = 2. This is next coefficient in
iii) Continue the process : multiply 2 by +5, i.e.,, 2 X +5 =10, add to the
next coefficient 10 + 10 = 20. This is next coefficient in Quotient. Thus
Quotient is x2 + 2x + 20
iv) Nowmultiply 20 by + 5i.e.,, 20 x 5 = 100. Add to the next (last) term,
100 + (-4) = 96, which becomesR,i.e.,, R =9.
44

47. Example 3:
x4 –3x3 + 7x2 + 5x + 7
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
x+4
Now thinking the method as in example ( 1 ), we proceed as follows.
x+4 x4 - 3x3 +7x2 + 5x + 7
‾‾‾‾‾
-4 - 4 + 28 - 140 + 540
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
x3 - 7x2 + 35x - 135 547
Thus Q = x3 – 7x2 + 35x – 135 and R = 547.
or we proceed orally as follows:
x4 / x gives 1 as first coefficient.
i) -4 X 1 = - 4 : add to next coefficient – 4 + (-3) = - 7 which gives next
coefficient in Q.
ii) – 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q.
iii) 35 X - 4 = - 140 : then – 140 + 5 = - 135, the next coefficient in Q.
iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R.
Thus Q = x3 – 7x2 + 35x – 135 , R = 547.
Note :
1. We can follow the same procedure even the number of terms is more.
2. If any term is missing, we have to take the coefficient of the term as zero
and proceed.
Now consider the divisors of second degree or more as in the following
Example :4 2x4 – 3x3 – 3x + 2 by x2 + 1.
Here x2 term is missing in the dividend. Hence treat it as 0 .x2 or 0 .
And the x term in divisor is also absent we treat it as 0 . x. Now
45

48. x2 +1 2x4 -3x3 + 0 . x2 - 3x + 2
x2+ 0 . x + 1 0 -2
‾‾‾‾‾‾‾‾‾‾‾‾
0 -1 0 +3
0 +2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
2 -3 -2 0 4
Thus Q = 2x2 - 3x - 2andR = 0 . x + 4 = 4.
Example 5 : 2x5 – 5x4 + 3x2 – 4x + 7by x3 – 2x2 + 3.
We treat the dividend as 2x5 –5x4 + 0. x3 + 3x2 – 4x + 7 and divisor as
x3 –2x2 + 0 . x + 3 and proceed as follows :
x3 – 2x2 + 0 . x + 3 2x5 – 5x4 + 0.x3 + 3x2 – 4x + 7
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
2 0 -3 4 -60
-2
0 + 3
-4 0 + 6
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
2 - 1 -2 - 7 - 1 +13
Thus Q = 2x2 – x – 2, R = - 7x2 – x + 13.
You may observe a very close relation of the method paravartya in this
aspect with regard to REMAINDER THEOREM and HORNER PROCESS of
Synthetic division. And yet paravartya goes much farther and is capable of
numerous applications in other directions also.
Apply paravartya – yojayet to find out the Quotient and Remainder in
each of the following problems.
1) (4x2 + 3x + 5)÷(x+1)
2) (x3 – 4x2 + 7x + 6) ÷ (x – 2)
3) (x4 – x3 + x2 + 2x + 4) ÷(x2 - x – 1)
4) (2x5 +x3 – 3x + 7) ÷ (x3 + 2x – 3)
5) (7x6 + 6x5 – 5x4 + 4x3 –3x2 + 2x – 1)÷ (x-1)
46

49. Paravartya in solving simple equations :
Recall that 'paravartya yojayet' means 'transpose and apply'. The rule
relating to transposition enjoins invariable change of sign with every change
of side. i.e., + becomes - and conversely ; and X becomes ÷ and conversely.
Further it can be extended to the transposition of terms from left to right and
conversely and from numerator to denominator and conversely in the
concerned problems.
Type ( i ) :
Consider the problem 7x – 5 = 5x + 1
7x – 5x = 1 + 5
i.e.,, 2x = 6 x = 6 ÷ 2 = 3.
Observe that the problem is of the type ax + b = cx + d from which we get
by ‘transpose’ (d – b), (a – c) and
d - b.
x= ‾‾‾‾‾‾‾‾
a-c
In this example a = 7, b = - 5, c = 5, d = 1
Hence 1 – (- 5) 1+5 6
x = _______ = ____ = __ = 3
7–5 7-5 2
Example 2: Solve for x,3x + 4 = 2x + 6
d-b 6-4 2
x = _____ = _____ = __ = 2
a-c 3-2 1
Type ( ii ) :Consider problems of the type (x + a) (x+b) = (x+c) (x+d). By
paravartya, we get
cd - ab
x = ______________
(a + b) – (c + d)
47

50. It is trivial form the following steps
(x + a) (x + b) = (x + c) (x + d)
x2 + bx + ax + ab = x2 + dx + cx + cd
bx + ax – dx – cx = cd – ab
x( a + b – c – d) = cd – ab
cd – ab cd - ab
x = ____________ x = _________________
a+b–c–d ( a + b ) – (c + d.)
Example 1 : (x – 3) (x – 2 ) = (x + 1 ) (x + 2 ).
By paravartya
cd – ab 1 (2) – (-3) (-2)
x= __________ = ______________
a + b – c –d -3–2–1–2
2-6 -4 1
= _______ = ___ = __
-8 -8 2
Example 2 : (x + 7) (x – 6) = (x +3) (x – 4).
Now cd - ab (3) (-4) – (7) (-6)
x= ___________ = ________________
a+b–c–d 7 + (-6) – 3 - (-4)
- 12 + 42 30
= ____________ = ___ = 15
7–6–3+4 2
Note that if cd - ab = 0 i.e.,, cd = ab, i.e.,, if the product of the absolute
terms be the same on both sides, the numerator becomes zero giving x = 0.
For the problem (x + 4) (x + 3) = (x – 2 ) ( x – 6 )
Solution is x = 0 since 4 X 3 = - 2 X - 6. = 12
Type ( iii) :
Consider the problems of the type ax + b m
______ = __
cx + d n
48

51. By cross – multiplication,
n ( ax + b) = m (cx + d)
nax + nb = mcx + md
nax - mcx = md – nb
x( na – mc ) = md – nb
md - nb
x = ________
na - mc.
Now look at the problem once again
ax + b m
_____ = __
cx+ d n
paravartya gives md - nb, na - mc and
md - nb
x = _______
na - mc
Example 1: 3x + 1 13
_______ = ___
4x + 3 19
md - nb 13 (3) - 19(1) 39 - 19 20
x = ______ = ____________ = _______ = __
na- mc 19 (3) - 13(4) 57 - 52 5
= 4
Example 2: 4x + 5 7
________ = __
3x + 13/2 8
(7) (13/2) -(8)(5)
x = _______________
(8)(4) - (7)(3)
(91/2) - 40 (91 - 80)/2 11 1
= __________ = _________ = ______ = __
32 – 21 32 – 21 2 X 11 2
49

52. Type (iv) :Consider the problems of the type m n
_____ + ____ = 0
x+a x+b
Take L.C.M and proceed.
m(x+b) + n (x+a)
______________ = 0
(x + a) (x +b)
mx + mb + nx + na
________________ = 0
x + a)(x + b)
(m + n)x + mb + na = 0 (m + n)x = - mb - na
-mb - na
x = ________
(m + n)
Thus the problem m n
____ + ____ = 0, by paravartya process
x+a x+b
gives directly
-mb - na
x = ________
(m + n)
Example 1 : 3 4
____ + ____ = 0
x+4 x–6
gives -mb - na
x = ________ Note that m = 3, n = 4, a = 4, b = - 6
(m + n)
-(3)(-6) – (4) (4) 18 - 16 2
= _______________ = ______ = __
( 3 + 4) 7 7
50

53. Example 2 :
5 6
____ + _____ = 0
x+1 x – 21
gives -(5) (-21) - (6) (1) 105 - 6 99
x = ________________ = ______ = __ = 9
5+6 11 11
I . Solve the following problems using the sutra Paravartya – yojayet.
1) 3x + 5 = 5x – 3 6) (x + 1) ( x + 2) = ( x – 3) (x – 4)
2) (2x/3) + 1=x - 1 7) (x – 7) (x – 9)= (x – 3) (x – 22)
3) 7x + 2 5 8) (x + 7) (x + 9)= (x + 3 ) (x + 21)
______ = __
3x- 5 8
4) x + 1 / 3
_______ = 1
3x - 1
5) 5 2
____ + ____ = 0
x+3 x–4
1.Show that for the type of equations
m n p
____ + ____ + ____ = 0, the solution is
x+a x+b x+c
- mbc – nca – pab
x = ________________________ , if m + n + p =0.
m(b + c) + n(c+a) + p(a + b)
51

54. 2. Apply the above formula to set the solution for the problem
Problem 3 2 5
____ + ____ - ____ = 0
x+4 x+6 x+5
some more simple solutions :
m n m+n
____ + ____ = _____
x+a x+b x+c
Now this can be written as,
m n m n
____ + ____ = _____ + _____
x+a x+b x+c x+c
m m n n
____ - ____ = _____ - _____
x+a x+c x+c x+b
m(x +c) – m(x + a) n(x + b) – n(x + c)
________________ = ________________
(x + a) (x + c) (x + c) (x + b)
mx + mc – mx – ma nx + nb – nx – nc
________________ = _______________
(x + a) (x + c) (x +c ) (x + b)
m (c – a) n (b –c)
____________ = ___________
x +a x+b
m (c - a).x + m (c - a).b = n (b - c). x + n(b - c).a
x [ m(c - a)- n(b - c) ] = na(b - c) – mb (c - a)
or x [ m(c - a) + n(c - b) ] = na(b - c) + mb (a - c)
52

55. Thus mb(a - c) + na (b - c)
x = ___________________
m(c-a) + n(c-b).
By paravartya rule we can easily remember the formula.
Example 1 : solve 3
5. Sunyam Samya Samuccaye
The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that
Samuccaya is Zero.' i.e., it should be equated to zero. The term 'Samuccaya'
has several meanings under different contexts.
i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all
the terms concerned and proceed as follows.
Example 1: The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all
its terms. Hence by the sutra it is zero,i.e., x = 0.
Otherwise we have to work like this:
7x + 3x = 4x + 5x
10x = 9x
10x – 9x = 0
x=0
This is applicable not only for ‘x’ but also any such unknown quantity as
Example 2: 5(x+1) = 3(x+1)
No need to proceed in the usual procedure like
5x + 5 = 3x + 3
5x – 3x = 3 – 5
2x = -2 or x = -2 ÷ 2 = -1
Simply think of the contextual meaning of 'Samuccaya'
53

56. Now Samuccaya is ( x + 1)
x+1=0 gives x = -1
ii) Now we interpret 'Samuccaya' as product of independent terms in
expressions like (x+a) (x+b)
Example 3: ( x + 3 ) ( x + 4) = ( x – 2) ( x – 6 )
Here Samuccaya is 3 x 4 = 12 = -2 x -6
Since it is same , we derive x = 0
This example, we have already dealt in type ( ii ) of Paravartya in solving
simple equations.
iii) We interpret ' Samuccaya 'as the sum of the denominators of two fractions
having the same numerical numerator.
Consider the example.
1 1
____ + ____ = 0
3x-2 2x-1
for this we proceed by takingL.C.M.
(2x-1)+(3x–2)
____________ = 0
(3x–2)(2x–1)
5x–3
__________ = 0
(3x–2)(2x–1)
5x – 3 = 0 5x = 3
3
x = __
5
Instead of this, we can directly put the Samuccaya i.e., sum of the
i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0
54

57. giving 5x = 3 x=3/5
It is true and applicable for all problems of the type
m m
____ + _____ = 0
ax+b cx+d
Samuccaya is ax+b+cx+d and solution is ( m ≠ 0 )
-(b+d)
x = _________
(a+c)
iii) We now interpret 'Samuccaya' as combination or total.
If the sum of the numerators and the sum of the denominators be the
same, then that sum = 0.
Consider examples of type
ax+ b ax + c
_____ = ______
ax+ c ax + b
In this case, (ax+b) (ax+b) =(ax+c) (ax+c)
a2x2 + 2abx + b2 = a2x2 + 2acx +c2
2abx – 2acx = c2 – b2
x ( 2ab – 2ac ) = c2 – b2
c2–b2 (c+b)(c-b) -(c+b)
x= ______ = _________ = _____
2a(b-c) 2a(b-c) 2a
As per Samuccaya (ax+b) + (ax+c) = 0
2ax+b+c = 0
2ax = -b-c
-(c+b)
x = ______
2a Hence the statement.
55

58. Example 4:
3x+ 4 3x + 5
______ = ______
3x+ 5 3x + 4
Since N1 + N2 = 3x + 4 + 3x + 5 = 6x + 9 ,
And D1 + D2 = 3x + 4 + 3x + 5 = 6x + 9
We haveN1 + N2 = D1 + D2 = 6x + 9
Hence from Sunya Samuccaya we get 6x + 9 = 0
6x = -9
-9 -3
x = __ = __
6 2
Example 5:
5x +7 5x + 12
_____ = _______
5x+12 5x + 7
Hence N1 + N2 = 5x + 7 + 5x + 12 = 10x + 19
And D1 + D2 = 5x + 12 + 5x + 7 = 10x + 19
N1 + N2 = D1 + D2 gives 10x + 19 = 0
10x = -19
-19
x = ____
10
Consider the examples of the type, where N1 + N2 = K (D1 + D2 ), where K is
a numerical constant, then also by removing the numerical constant K, we
can proceed as above.
Example 6:
2x + 3 x +1
_____ = ______
4x + 5 2x + 3
Here N1 + N2 = 2x + 3 + x + 1 = 3x + 4
56

59. D1 + D2 = 4x + 5 + 2x + 3 = 6x + 8
= 2 ( 3x + 4 )
Removing the numerical factor 2, we get 3x + 4 on both sides.
3x + 4 = 0 3x = -4 x = - 4 / 3.
v) 'Samuccaya' with the same meaning as above, i.e., case (iv), we solve the
problems leading to quadratic equations. In this context, we take the
problems as follows;
If N1 + N2 = D1 + D2 and also the differences
N1 ~ D1 = N2 ~ D2 then both the things are equated to zero, the
solution gives the two values for x.
Example 7:
3x + 2 2x + 5
_____ = ______
2x + 5 3x + 2
In the conventional text book method, we work as follows :
3x + 2 2x + 5
_____ = ______
2x + 5 3x + 2
( 3x + 2 ) ( 3x + 2 ) = ( 2x + 5 ) ( 2x + 5 )
9x2 + 12x + 4 = 4x2 + 20x + 25
9x2 + 12x + 4 - 4x2 - 20x – 25 = 0
5x2 – 8x – 21 = 0
5x2 – 15x + 7x – 21 = 0
5x ( x – 3 ) + 7 ( x – 3 ) = 0
(x – 3 ) ( 5x + 7 ) = 0
x – 3 = 0 or 5x + 7 = 0
x = 3 or - 7 / 5
Now ‘Samuccaya’ sutra comes to help us in a beautiful way as follows :
Observe N1 + N2 = 3x + 2 + 2x + 5 = 5x + 7
D1 + D2 = 2x + 5 + 3x + 2 = 5x + 7
Further N1 ~ D1 = ( 3x + 2 ) – ( 2x + 5 ) = x – 3
N2 ~ D2 = ( 2x + 5) – ( 3x + 2 ) = - x + 3 = - ( x – 3 )
57

60. Hence 5x + 7 = 0 , x – 3 = 0
5x = -7 , x = 3
i.e., x = -7 / 5 , x = 3
Note that all these can be easily calculated by mere observation.
Example 8:
3x + 4 5x + 6
______ = _____
6x + 7 2x + 3
Observe that
N1 + N2 = 3x + 4 + 5x + 6 = 8x + 10
andD1 + D2 = 6x + 7 + 2x + 3 = 8x + 10
Further N1 ~D1 = (3x + 4) – (6x + 7)
= 3x + 4 – 6x – 7
= -3x – 3 = -3 ( x + 1 )
N2 ~ D2 = (5x + 6) – (2x + 3) = 3x + 3 = 3( x + 1)
By ‘Sunyam Samuccaye’ we have
8x + 10 = 0 3( x + 1 ) = 0
8x = -10 x+1=0
x = - 10 / 8 x = -1
=-5/4
vi)‘Samuccaya’ with the same sense but with a different context and
application .
Example 9:
1 1 1 1
____ + _____ = ____ + ____
x-4 x–6 x-2 x-8
Usually we proceed as follows.
x–6+x-4 x–8+x-2
___________ = ___________
(x–4) (x–6) (x–2) (x-8)
58

61. 2x-10 2x-10
_________ = _________
x2–10x+24 x2–10x+16
( 2x – 10 ) ( x2 – 10x + 16 ) = ( 2x – 10 ) ( x2 – 10x + 24)
2x3–20x2+32x–10x2+100x–160 = 2x3–20x2+48x–10x2+100x-240
2x3 – 30x2 + 132x – 160 = 2x3 – 30x2 + 148x – 240
132x – 160 = 148x – 240
132x – 148x = 160 – 240
– 16x = - 80
x = - 80 / - 16 = 5
Now ‘Samuccaya’ sutra, tell us that, if other elements being equal, the sum-
total of the denominators on the L.H.S. and their total on the R.H.S. be the
same, that total is zero.
Now D1 + D2 = x – 4 + x – 6 = 2x – 10, and
D3 + D4 = x – 2 + x – 8 = 2x – 10
By Samuccaya, 2x – 10 gives 2x = 10
10
x = __ = 5
2
Example 10:
1 1 1 1
____ + ____ = ____ + _____
x -8 x–9 x -5 x – 12
D1 +D2 = x – 8 + x – 9 = 2x – 17, and
D3 +D4 = x – 5 + x –12 = 2x – 17
Now 2x – 17 = 0 gives 2x = 17
17
x = __ = 8½
2
Example 11:
1 1 1 1
____ - _____ = ____ - _____
x +7 x + 10 x +6 x+9
59

62. This is not in the expected form. But a little work regarding transposition
makes the above as follows.
1 1 1 1
____ + ____ = ____ + _____
x +7 x+9 x +6 x + 10
Now ‘Samuccaya’ sutra applies
D1 +D2 = x + 7 + x + 9 = 2x + 16, and
D3 +D4 = x + 6 + x + 10 = 2x + 16
Solution is given by 2x + 16 = 0 i.e., 2 x = - 16.
x = - 16 / 2 = - 8.
Solve the following problems using Sunyam Samya-Samuccaye
1. 7(x+2)+3(x+2)=6(x+2)+5(x+2)
2. ( x + 6 ) ( x + 3 ) = ( x – 9 ) ( x – 2)
3. ( x - 1 ) ( x + 14 ) = ( x + 2 ) ( x – 7)
1 1
4. ______ + ____ = 0
4x-3 x–2
4 4
5. _____ + _____ = 0
3x + 1 5x + 7
2x + 11 2x+5
6. ______ = _____
2x+ 5 2x+11
3x + 4 x+1
7. ______ = _____
6x + 7 2x + 3
60

63. 4x - 3 x+ 4
8. ______ = _____
2x+ 3 3x - 2
1 1 1 1
9. ____ + ____ = ____ + _____
x-2 x-5 x-3 x-4
1 1 1 1
10. ____ - ____ = _____ - _____
x-7 x-6 x - 10 x-9
Sunyam Samya Samuccaye in Certain Cubes:
Consider the problem ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5)3. For the solution
by the traditional method we follow the steps as given below:
( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3
x3 – 12x2 + 48x – 64 + x3 – 18x2 + 108x – 216
= 2 ( x3 – 15x2 + 75x – 125 )
2x3 – 30x2 + 156x – 280 = 2x3 – 30x2 + 150x – 250
156x – 280 = 150x – 250
156x – 150x = 280 – 250
6x = 30
x = 30 / 6 = 5
But once again observe the problem in the vedic sense
We have ( x – 4 ) + ( x – 6 ) = 2x – 10. Taking out the numerical factor 2,
we have ( x – 5 ) = 0, which is the factor under the cube on R.H.S. In such a
case “Sunyam samya Samuccaye” formula gives that x – 5 = 0. Hence x = 5
Think of solving the problem (x–249)3 + (x+247)3 = 2(x–1)3
The traditional method will be horrible even to think of.
But ( x – 249 ) + ( x + 247 ) = 2x – 2 = 2 ( x – 1 ). And x – 1. on R.H.S.
61

64. cube, it is enough to state that x – 1 = 0 by the ‘sutra’.
x = 1 is the solution. No cubing or any other mathematical operations.
Algebraic Proof :
Consider ( x – 2a)3 + ( x – 2b )3 = 2 ( x – a – b )3 it is clear that
x – 2a + x – 2b = 2x – 2a – 2b
=2(x–a–b)
Now the expression,
x3 -6x2a + 12xa2 – 8a3 + x3 – 6x2b +12xb2 – 8b3 =
2(x3–3x2a–3x2b+3xa2+3xb2+6axb–a3–3a2b–3ab2–b3)
= 2x3–6x2a–6x2b+6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3
cancel the common terms on both sides
12xa2+12xb2–8a3–8b3 = 6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3
6xa2 + 6xb2 – 12xab = 6a3 + 6b3 –6a2b – 6ab2
6x ( a2 + b2 – 2ab ) = 6 [ a3 + b3 – ab ( a + b )]
x ( a – b )2 = [ ( a + b ) ( a2 + b2 –ab ) – ( a + b )ab]
= ( a + b ) ( a2 + b2 – 2ab )
= ( a + b ) ( a – b )2
x=a+b
Solve the following using “Sunyam Samuccaye” process :
1. ( x – 3 )3 + ( x – 9 )3 = 2 ( x – 6 )3
2. ( x + 4 )3 + ( x – 10 )3 = 2 ( x – 3 )3
3. ( x + a + b – c )3 + ( x + b + c – a )3 = 2 ( x + b )3
62

65. Example :
(x + 2)3 x+1
______ = _____
(x + 3)3 x+4
with the text book procedures we proceed as follows
x3 + 6x2 + 12x +8 x+1
_______________ = _____
x3 + 9x2 + 27x +27 x+4
Now by cross multiplication,
( x + 4 ) ( x3 +6x2 + 12x + 8 ) = ( x + 1 ) ( x3 + 9x2 + 27x + 27 )
x4 + 6x3 + 12x2+ 8x + 4x3 + 24x2 + 48x + 32 =
x4 + 9x3 + 27x2 + 27x + x3 + 9x2 + 27x + 27
x4 +10x3 + 36x2 + 56x + 32 = x4 + 10x3 +36x2 + 54x + 27
56x + 32 = 54x + 27
56x – 54x = 27 – 32
2x = - 5
x=-5/2
Observe that ( N1 + D1 ) with in the cubes on
L.H.S. is x + 2 + x + 3 = 2x + 5 and
N2 + D2 on the right hand side
is x + 1 + x + 4 = 2x + 5.
By vedic formula we have 2x + 5 = 0 x = - 5 / 2.
Solve the following by using vedic method :
1.
(x + 3)3 x+1
______ = ____
(x + 5)3 x+7
63

66. 2.
(x - 5)3 x-3
______ = ____
(x - 7)3 x-9
6. Anurupye - Sunyamanyat
The Sutra Anurupye Sunyamanyat says : 'If one is in ratio, the other one is
We use this Sutra in solving a special type of simultaneous simple equations
in which the coefficients of 'one' variable are in the same ratio to each other
as the independent terms are to each other. In such a context the Sutra says
the 'other' variable is zero from which we get two simple equations in the first
variable (already considered) and of course give the same value for the
Example 1:
3x + 7y = 2
4x + 21y = 6
Observe that the y-coefficients are in the ratio 7 : 21 i.e., 1 : 3, which is
same as the ratio of independent terms i.e., 2 : 6 i.e., 1 : 3. Hence the other
variable x = 0 and 7y = 2 or 21y = 6 gives y = 2 / 7
Example 2:
323x + 147y = 1615
969x + 321y = 4845
The very appearance of the problem is frightening. But just an observation
and anurupye sunyamanyat give the solution x = 5, because coefficient of x
ratio is
323 : 969 = 1 : 3 and constant terms ratio is 1615 : 4845 = 1 : 3.
y = 0 and 323 x = 1615 or 969 x = 4845 gives x = 5.
Solve the following by anurupye sunyamanyat.
1. 12x + 78y = 12 2. 3x + 7y = 24
64

67. 16x + 96y =16 12x + 5y = 96
3. 4x – 6y = 24 4. ax + by = bm
7x – 9y = 36 cx + dy = dm
In solving simultaneous quadratic equations, also we can take the help of the
‘sutra’ in the following way:
Example 3 :
Solve for x and y
x + 4y = 10
2 2
x + 5xy + 4y + 4x - 2y = 20
x2 + 5xy + 4y2 + 4x - 2y = 20 can be written as
( x + y ) ( x + 4y ) + 4x – 2y = 20
10 ( x + y ) + 4x – 2y = 20 ( Since x + 4y = 10 )
10x + 10y + 4x – 2y = 20
14x + 8y = 20
Now x + 4y = 10
14x + 8y = 20 and 4 : 8 :: 10 : 20
from the Sutra, x = 0 and 4y = 10, i.e.,, 8y= 20 y = 10/4 = 2½
Thus x = 0 and y = 2½ is the solution.
7. Sankalana - Vyavakalanabhyam
This Sutra means 'by addition and by subtraction'. It can be applied in solving
a special type of simultaneous equations where the x - coefficients and the y
- coefficients are found interchanged.
Example 1:
45x – 23y = 113
23x – 45y = 91
In the conventional method we have to make equal either the coefficient of x
or coefficient of y in both the equations. For that we have to multiply equation
( 1 ) by 45 and equation ( 2 ) by 23 and subtract to get the value of x and
then substitute the value of x in one of the equations to get the value of y or
we have to multiply equation ( 1 ) by 23 and equation ( 2 ) by 45 and then
subtract to get value of y and then substitute the value of y in one of the
65

68. equations, to get the value of x. It is difficult process to think of.
From Sankalana – vyavakalanabhyam
add them,
i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91
i.e., 68x – 68y = 204 x–y=3
subtract one from other,
i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91
i.e., 22x + 22y = 22 x+y=1
and repeat the same sutra, we get x = 2 and y = - 1
Very simple addition and subtraction are enough, however big the
coefficients may be.
Example 2:
1955x – 476y = 2482
476x – 1955y = -4913
Oh ! what a problem ! And still
just add, 2431( x – y ) = - 2431 x – y = -1
subtract, 1479 ( x + y ) = 7395 x+y=5
once again add, 2x = 4 x=2
subtract - 2y = - 6 y=3
Solve the following problems usingSankalana – Vyavakalanabhyam.
1. 3x + 2y = 18
2x + 3y = 17
2. 5x – 21y = 26
21x – 5y = 26
3. 659x + 956y = 4186
956x + 659y = 3889
66

69. 8. Puranapuranabhyam
The Sutra can be taken as Purana - Apuranabhyam which means by the
completion or non - completion. Purana is well known in the present
system. We can see its application in solving the roots for general form of
quadratic equation.
We have : ax2 + bx + c = 0
x2 + (b/a)x + c/a = 0 ( dividing by a )
x2 + (b/a)x = - c/a
completing the square ( i.e.,, purana ) on the L.H.S.
x2 + (b/a)x + (b2/4a2) = -c/a + (b2/4a2)
[x + (b/2a)]2 = (b2 - 4ac) / 4a2
________
- b ± √ b2 – 4ac
Proceeding in this way we finally get x = _______________
2a
Now we apply purana to solve problems.
Example 1. x3 + 6x2 + 11 x + 6 = 0.
Since (x + 2 )3 = x3 + 6x2 + 12x + 8
Add ( x + 2 ) to both sides
We get x3 + 6x2 + 11x + 6 + x + 2 = x + 2
i.e.,, x3 + 6x2 + 12x + 8 = x + 2
i.e.,, ( x + 2 )3 = ( x + 2 )
this is of the form y3 = y for y = x + 2
solution y = 0, y = 1, y = - 1
i.e.,, x + 2 = 0,1,-1
which gives x = -2,-1,-3
Example 2: x3 + 8x2 + 17x + 10 = 0
We know ( x + 3)3 = x3 + 9x2 + 27x + 27
So adding on the both sides, the term (x2 + 10x + 17 ), we get
x3 + 8x2 + 17x + x2 + 10x + 17 = x2 + 10x + 17
i.e.,, x3 + 9x2 + 27x + 27 = x2 + 6x + 9 + 4x + 8
i.e.,, ( x + 3 )3 = ( x + 3 )2 + 4 ( x + 3 ) – 4
67

70. y3 = y2 + 4y – 4 for y = x + 3
y = 1, 2, -2.
Hence x = -2, -1, -5
Thus purana is helpful in factorization.
Further purana can be applied in solving Biquadratic equations also.
Solve the following using purana – apuranabhyam.
1. x3 – 6x2 + 11x – 6 = 0
2. x3 + 9x2 + 23x + 15 = 0
3. x2 + 2x – 3 = 0
4. x4 + 4x3 + 6x2 + 4x – 15 = 0
9. Calana - Kalanabhyam
In the book on Vedic Mathematics Sri Bharati Krishna Tirthaji mentioned
the Sutra 'Calana - Kalanabhyam' at only two places. The Sutra means
'Sequential motion'.
i) In the first instance it is used to find the roots of a quadratic equation7x2 –
11x – 7 = 0. Swamiji called the sutra as calculus formula. Its application at
that point is as follows.Now by calculus formula we say: 14x–11 = ±√317
A Note follows saying every Quadratic can thus be broken down into two
binomial factors. An explanation in terms of first differential, discriminant with
sufficient number of examples are given under the chapter ‘Quadratic
ii) At the Second instance under the chapter ‘Factorization and Differential
Calculus’ for factorizing expressions of 3rd, 4th and 5th degree, the procedure
is mentioned as'Vedic Sutras relating to Calana – Kalana – Differential
Further other Sutras 10 to 16 mentioned below are also used to get the
required results. Hence the sutra and its various applications will be taken up
at a later stage for discussion.
But sutra – 14 is discussed immediately after this item.
68

71. Now the remaining sutras :
10. YĀVADŨNAM ( The deficiency )
11. VYAŞłISAMAŞłIH ( Whole as one and one as whole )
12. ŚEŞĀNYAŃ KENA CARAMEĥA ( Remainder by the last digit )
13. SOPĀNTYADVAYAMANTYAM ( Ultimate and twice the penultimate )
15. GUĥITASAMUCCAYAH ( The whole product is the same )
16. GUĥAKA SAMUCCAYAH ( Collectivity of multipliers )
The Sutras have their applications in solving different problems in different
contexts. Further they are used along with other Sutras. So it is a bit of
inconvenience to deal each Sutra under a separate heading exclusively and
also independently. Of course they will be mentioned and also be applied in
solving the problems in the forth coming chapter wherever necessary. This
decision has been taken because up to now, we have treated each Sutra
independently and have not continued with any other Sutra even if it is
necessary. When the need for combining Sutras for filling the gaps in the
process arises, we may opt for it. Now we shall deal the fourteenth Sutra, the
Sutra left so far untouched.
10. Ekanyunena Purvena
The Sutra Ekanyunena purvena comes as a Sub-sutra to Nikhilam which gives
the meaning 'One less than the previous' or 'One less than the one before'.
1) The use of this sutra in case of multiplication by 9,99,999.. is as follows .
Method :
a) The left hand side digit (digits) is ( are) obtained by applying the ekanyunena
purvena i.e. by deduction 1 from the left side digit (digits) .
e.g. ( i ) 7 x 9; 7 – 1 = 6 ( L.H.S. digit )
b) The right hand side digit is the complement or difference between the
multiplier and the left hand side digit (digits) . i.e. 7 X 9 R.H.S is 9 - 6 = 3.
c) The two numbers give the answer; i.e. 7 X 9 = 63.
Example 1: 8 x 9 Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit )
Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit )
Step ( c ) gives the answer 72
69

72. Example 2: 15 x 99 Step ( a ) : 15 – 1 = 14
Step ( b ) : 99 – 14 = 85 ( or 100 – 15 )
Step ( c ) : 15 x 99 = 1485
Example 3: 24 x 99
Answer :
Example 4: 356 x 999
Answer :
Example 5: 878 x 9999
Answer :
Note the process : The multiplicand has to be reduced by 1 to obtain the LHS
and the rightside is mechanically obtained by the subtraction of the L.H.S from
the multiplier which is practically a direct application of Nikhilam Sutra.
Now by Nikhilam
24 – 1 = 23 L.H.S.
x 99 – 23 = 76 R.H.S. (100–24)
_____________________________
23 / 76 = 2376
Reconsider the Example 4:
356 – 1 = 355 L.H.S.
x 999 – 355 = 644 R.H.S.
________________________
355 / 644 = 355644
70

73. and in Example 5: 878 x 9999 we write
0878 – 1 = 877 L.H.S.
x 9999 – 877 = 9122 R.H.S.
__________________________
877 / 9122 = 8779122
Algebraic proof :
As any two digit number is of the form ( 10x + y ), we proceed
( 10x + y ) x 99
= ( 10x + y ) x ( 100 – 1 )
= 10x . 102 – 10x + 102 .y – y
= x . 103 + y . 102 – ( 10x + y )
= x . 103 + ( y – 1 ) . 102 + [ 102 – ( 10x + y )]
Thus the answer is a four digit number whose 1000th place is x,100th place is
( y - 1 ) and the two digit number which makes up the 10th and unit place is the
number obtained by subtracting the multiplicand from 100.(or apply nikhilam).
Thus in 37 X 99. The 1000th place is x i.e. 3
100th place is ( y - 1 ) i.e. (7 - 1 ) = 6
Number in the last two places 100-37=63.
Hence answer is 3663.
Apply Ekanyunena purvena to find out the products
1. 64 x 99 2. 723 x 999 3. 3251 x 9999
4. 43 x 999 5. 256 x 9999 6. 1857 x 99999
We have dealt the cases
i) When the multiplicand and multiplier both have the same number of digits
ii) When the multiplier has more number of digits than the multiplicand.
In both the cases the same rule applies. But what happens when the multiplier
has lesser digits?
i.e. for problems like 42 X 9, 124 X 9, 26325 X 99 etc.,
71

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