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Vedic Mathematics - Methods
Preface

------------------------------------------------------------------------------------------------ 1

I
...
Vedic Mathematical Formulae -------------------------------------------------------------------------- 5
1
...
Nikhilam navatascaramam Dasatah ----------------------------------------------------------------- 18
3
...
Paravartya Yojayet -------------------------------------------------------------------------------------- 41
5
...
Anurupye - Sunyamanyat ------------------------------------------------------------------------------ 64
7
...
Puranapuranabhyam ------------------------------------------------------------------------------------ 67
9
...
Ekanyunena Purvena ---------------------------------------------------------------------------------- 69
11
...
Adyamadyenantya - mantyena ---------------------------------------------------------------------- 82
13
...
Antyayor Dasakepi ------------------------------------------------------------------------------------- 93
15
...
Lopana Sthapanabhyam ---------------------------------------------------------------------------- 101
17
...
Gunita Samuccayah : Samuccaya Gunitah ------------------------------------------------------ 113
III Vedic Mathematics - A briefing ---------------------------------------------------------------------- 115
1
...
Addition and Subtraction ------------------------------------------------------------------------------ 130

3
...
Division ----------------------------------------------------------------------------------------------- 144
5
...
The word Veda covers all Veda-sakhas known to humanity
...

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
...

According to him, there has been considerable literature on Mathematics in
the Veda-sakhas
...
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
...
They
apply even to complex problems involving a large number of mathematical
operations
...
Though
the solutions appear like magic, the application of the Sutras is perfectly
logical and rational
...
The Sutras provide not only
methods of calculation, but also ways of thinking for their application
...
The explanations offered make the processes clear to the
learners
...

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
...
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
...
The innovation in
the presentation is the algebraic proof for every elucidation of the Sutra or
the Sub-Sutra concerned
...
Why Vedic Mathematics?
Many Indian Secondary School students consider Mathematics a very difficult
subject
...
Some students feel it difficult to manipulate symbols and balance
equations
...

Many such difficulties in learning Mathematics enter into a long list if prepared
by an experienced teacher of Mathematics
...
Learning Mathematics is an unpleasant experience to some
students mainly because it involves mental exercise
...

Dr
...
A few of his opinions are stated
hereunder:
i) Mathematics, derived from the Veda, provides one line, mental and superfast methods along with quick cross checking systems
...

iii) Vedic Mathematics offers a new and entirely different approach to the
study of Mathematics based on pattern recognition
...

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
...

v) Vedic Mathematics with its special features has the inbuilt potential to
solve the psychological problem of Mathematics - anxiety
...
T
...


3

A
...
Nicholas (1984) puts the Vedic Mathematics system as 'one of the most
delightful chapters of the 20th century mathematical history'
...
R
...
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'
...
J
...
Kapur says 'Vedic Mathematics can be used to remove mathphobia, and can be taught to (school) children as enrichment material along
with other high speed methods'
...
Michael Weinless, Chairman of the Department of Mathematics at the
M
...
U, Iowa says thus: 'Vedic Mathematics is easier to learn, faster to use and
less prone to error than conventional methods
...
'
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
...


4

II
...
In the text authored by the Swamiji, nowhere has
the list of the Mathematical formulae (Sutras) been given
...
The
list so compiled contains Sixteen Sutras and Thirteen Sub - Sutras as stated
hereunder
...
So are
also the words Upa-sutra, Sub-sutra, Sub-formula, corollary used
...
Explanation,
methods, further short-cuts, algebraic proof, etc follow
...


6

1
...

i) Squares of numbers ending in 5 :
Now we relate the sutra to the ‘squaring of numbers ending in 5’
...

Here the number is 25
...
For the
number 25, the last digit is 5 and the 'previous' digit is 2
...
The Sutra, in this context, gives the
procedure'to multiply the previous digit 2 by one more than itself, that is, by 3'
...
H
...
The R
...
S
(right hand side) of the result is52, that is, 25
...

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
...
x2+ 2abx + b2
...
102 + 2
...
5 + 52
= a2
...
102+ 52
= (a2+ a )
...
102 + 25
...
In such a case the number (10a + 5)2 is of the form whose L
...
S is a (a + 1)
and R
...
S is 25, that is, a (a + 1) / 25
...

Example:

45 = (40 + 5)2, It is of the form (ax+b)2 for a = 4, x=10
and b = 5
...


b) Any three digit number is of the form ax2+bx+c for x =10, a ≠ 0, a, b, c Є
W
...
x3+(b2+ 2ca)x2+2bc
...

This identity for x = 10, c = 5becomes (a
...
10 + 5) 2

= a2
...
a
...
103 + (b2+ 2
...
a)102+ 2
...
5
...
104+ 2
...
b
...
102+ 52
= a2
...
103+ b2
...
103 + b 102+ 52

= a2
...
103 + (b2+ b)102 +52

= [ a2
...
10 + a
...
102 + 25

8

= P (P+1) 102+ 25, where P = 10a+b
...


Example :

1652= (1
...
10 + 5) 2
...
It gives the
answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’
...


Apply Ekadhikena purvena to find the squares of the numbers 95, 225,
375, 635, 745, 915, 1105, 2545
...
In the conversion
of such vulgar fractions into recurring decimals, Ekadhika process can be
effectively used both in division and multiplication
...

The numbers of decimal places before repetition is the difference of numerator
and denominator, i
...
,, 19 -1=18 places
...

Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2
...
Now the method of division is as
follows:
Step
...


i
...
,, 1 / 20 = 0
...
10 ( 0 times, 1 remainder)
Step
...
e
...
005( 5 times, 0 remainder )

Step
...
e
...
0512 ( 2 times, 1 remainder )
Step
...
e
...
e
...
0526 ( 6 times, No remainder )
Step
...
e
...
05263 ( 3 times, No remainder )
Step
...
e
...
0526311(1 time, 1 remainder )
Step
...
e
...
e
...
05263115 (5 times, 1 remainder )
Step
...
e
...
e
...
052631517 ( 7 times, 1 remainder )
Step
...
e
...
e
...
05263157 18 (8 times, 1 remainder )
Step
...
e
...
e
...
0526315789 (9 times, No remainder )
Step
...
e
...
0526315789 14 (4 times, 1 remainder )
Step
...
e
...
e
...
052631578947 ( 7 times, No remainder )

Step
...
e
...
05263157894713 ( 3 times, 1 remainder )
Step
...
e
...
e
...
052631578947316 ( 6 times, 1 remainder )
Step
...
e
...
e
...
052631578947368 (8 times, No remainder )
Step
...
e
...
0526315789473684 ( 4 times, No remainder )
Step
...
e
...
05263157894736842 ( 2 times, No remainder )
Step
...
e
...
052631578947368421 ( 1 time, No remainder )
Now from step 19, i
...
,, dividing 1 by 2, Step 2 to Step
...

1 / 19 =0
...
052631578947368421


...
Nowhere
the division by 19 occurs
...
Here the
last digit is 9
...
And one more than of it is 1 + 1 = 2
...
We write the last digit in the
numerator as 1 and follow the steps leftwards
...
1 :

1
11

Step
...
3 :

421(multiply 2 by 2, put to left)

Step
...
5 :
Step
...
7 :

7368421 ( 3 X 2, = 6 +1 [Carryover]
= 7, put to left)

Step
...
9 :

147368421

(as in the same process)

947368421 ( Do – continue to step 18)

Step
...
11 :

178947368421

Step
...
13 :

11578947368421

Step
...
15 :

631578947368421

Step
...
17 :

52631578947368421

Step
...

Thus 1 / 19 = 0
...

Observations :
a) For any fraction of the form 1 / a9 i
...
,, 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
...

c) Starting from right most digit and counting from the right, we see ( in the given example 1 /
19)
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
...
052631517 and next step
...
052631578
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
...
,, 0
...
8 :

147368421

13

Step
...
e
...


0
...

d) When we get (Denominator – Numerator) as the product in the multiplicative
process, half the work is done
...

e) Either division or multiplication process of giving the answer can be put in a
single line form
...

Now consider the problem of 1 / 19
...
5) + 10-2 (0
...
125)+ ---------= 0
...
0025 + 0
...
00000625+ - - - - - - = 0
...
Find 1 / 49 by ekadhikena process
...
‘One more than the previous’ is 4 + 1 = 5
...

1 / 49 =
...
02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder )
=
...
0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder )
=
...
020408 - - - - - (divide 8 by 5, 1 time, 3 remainder )
=
...
02040811 6 - - - - - - -

continue
15

=
...
e
...

i
...
The remaining half can be obtained as
complements from 9
...

Thus 1 / 49 = 0
...

979591836734693877551
Now finding 1 / 49 by process of multiplication left ward from right by 5, we
get
1 / 49 = ----------------------------------------------1
= ---------------------------------------------51
= -------------------------------------------2551
= ------------------------------------------27551
= ---- 483947294594118333617233446943383727551
i
...
,,Denominator – Numerator = 49 – 1 = 48 is obtained as 5X9+3
( Carry over ) = 45 + 3 = 48
...
The remaining
half is automatically obtained as complements of 9
...

= 0
...

979591836734693877551

16

Example 2: Find 1 / 39 by Ekadhika process
...
Now the rule
predicting the completion of half of the computation does not hold
...

Now continue and obtain the result
...


Find the recurring decimal form of the fractions 1 / 29, 1 / 59,
1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible
...


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
...
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
...
, to the powers of 10
...
The numbers taken can be either less
or more than the base considered
...

Deviation may be positive or negative
...
Now observe the following table
...

Eg : 94
...
e
...

b) The two numbers under consideration are written one below the other
...

Eg : Multiply 7 by 8
...
Since it is near to both the numbers,
we write the numbers one below the other
...
A vertical or a slant linei
...
, a slash may be drawn for the
demarcation of the two parts i
...
,

(or)
d) The R
...
S
...
It
shall contain the number of digits equal to number of zeroes in the base
...
e
...

e) L
...
S of the answer is the sum of one number with the deviation of the
other
...


i) Cross-subtract deviation 2 on the second row from the original number7 in
the first row i
...
, 7-2 = 5
...
e
...

i
...
, (7 + 8) – 10 = 5
iv) Subtract the sum of the two deviations from the base
...
e
...

_
3
_
8 2
‾‾‾‾‾‾‾‾‾‾‾‾
5/

Thus

7

Now (d) and (e) together give the solution
_
7 3
7
_
8 2 i
...
, X 8
‾‾‾‾‾‾‾
‾‾‾‾‾‾
5/ 6
56
f) If R
...
S
...
H
...
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
...
S of the answer
...
ThenN1 X N2 can be
represented as

Case (i) : Both the numbers are lower than the base
...

20

Now let us solve some more examples by taking bases 100 and 1000
respectively
...
1: Find 97 X 94
...
Now following the rules, the working is
as follows:

Ex
...


Ex
...
Base is 100
...
4: 986 X 989
...


Ex
...
Base is 1000
...
6: 750X995
...

The method and rules follow as they are
...
Instead of cross – subtract, we follow cross – add
...
7: 13X12
...
8: 18X14
...


Ex
...
Base is 100
...
10: 1275X1004
...

22

( rule -f )

1275 275
1004 004
‾‾‾‾‾‾‾‾‾‾‾‾
1279/ 275x4 = 1279 / 1100
____________ = 1280100

( rule -f )

Case ( iii ): One number is more and the other is less than the base
...
So the
product of deviations becomes negative
...
To have a clear representation
and understanding a vinculum is used
...

Ex
...
Base is 10

Note : Conversion of common number into vinculum number and vice versa
...
etc
The procedure can be explained in detail using Nikhilam Navatascaram Dasatah,
Ekadhikenapurvena, Ekanyunena purvena in the foregoing pages of this book
...
12: 108 X 94
...


23

Ex
...
Base is 1000
...

N1 = (x-a), N2 = (x-b)
...
Observe that x is a multiple of 10
...
x – x
...
x + ab

= x (x – a – b ) + 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]
...
e
...
b yields a product consisting of more than the required digits
...

Case ( ii ) :
When both the numbers exceed the selected base, we have N1 = x + a,N2 = x +
b, x being the base
...
b holds
good, of course with relevant details mentioned in case(i)
...
and the procedure is evident from the examples
given
...

1) 7 X 4

2) 93 X 85

3) 875 X 994

4) 1234 X 1002

5) 1003 X 997

7) 1234 X 1002

6) 11112 X 9998

8) 118 X 105

Nikhilam in Division
Consider some two digit numbers (dividends) and same divisor 9
...

i) 13 ÷ 9 The quotient (Q) is 1, Remainder (R) is 4
...


iii)

60 ÷ 9, Q is 6, R is 6
...


Now we have another type of representation for the above examples as given
hereunder:
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
...


13 as 1 / 3,

34 as 3 / 4 ,

80 as 8 / 0
...


25

Eg
...
Put the
same digit under the right hand part for the remainder, add the two and place
the sumi
...
, sum of the digits of the numbers as the remainder
...

1/3
1
______ ,
1/4

3/4
8/0
3
8
______ , ______
3/7
8/8

Now the problem is over
...
e
...
e

Q=2, R=3

2
‾‾‾‾‾‾
2 / 3
b) 43 ÷ 9 as
9) 4 / 3
4
‾‾‾‾‾‾
4 / 7

i
...


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
...

i)
9 ) 104 ( 11
99

9 ) 10 / 4
1 / 1

26

‾‾‾‾‾‾
5

as

‾‾‾‾‾‾‾
11 / 5

ii)
9 ) 212 ( 23
207
‾‾‾‾‾
5

as

9 ) 21 / 2
2 / 3
‾‾‾‾‾‾‾
23 / 5

iii)
9 ) 401 (44
396
‾‾‾‾‾
5

as

9 ) 40 / 1
4 / 4
‾‾‾‾‾‾‾‾
44 / 5

Note that the remainder is the sum of the digits of the dividend
...
This digit added to the third
digit sets the remainder
...

Consider

511 ÷ 9

Add the first digit 5 to second digit 1 getting 5 + 1 = 6
...

Now second digit of 56 i
...
, 6 is added to third digit 1 of dividend to get the
remainder i
...
, 1 + 6 = 7
Thus
9)

51 / 1
5/ 6
‾‾‾‾‾‾‾
56 / 7

Q is 56, R is 7
...

Eg : 1204 ÷ 9
i) Add first digit 1 to the second digit 2
...
i
...
, 3 to third digit ‘0’ and obtain the
Quotient
...
e
...
R = 3 + 4 = 7, Q = 133
In symbolic form

9 ) 120 / 4
13 / 3
‾‾‾‾‾‾‾‾
133 / 7

Another example
...
What
about the case when the remainder is equal or greater than the divisor?
Eg
...

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
...


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

‾‾‾‾‾‾‾‾
27 / 3
Now consider the divisors of two or more digits whose last digit is 9,when
divisor is 89
...


Representing in the previous form of procedure, we have
89 ) 1 / 13
/ 11
‾‾‾‾‾‾‾
1 / 24

89 ) 100 / 15
12 / 32
‾‾‾‾‾‾‾‾‾‾
112 / 47

But how to get these? What is the procedure?
Now Nikhilam rule comes to rescue us
...
Now if you want to find 113 ÷ 89, 10015 ÷ 89, you have to apply
nikhilam formula on 89 and get the complement 11
...

89 ) 1 / 13
/ 11
‾‾‾‾‾‾‾‾
1 / 24

89 ) 100 /15
‾‾
11
11 /

first digit 1 x 11

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
...
While carrying the
added numbers to the place below the next digit, multiply by 02
...
e
...


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
...
e
...
In case of more digited numbers we apply
Nikhilam and proceed
...


* Guess the logic in the process of division by 9
...

1) 311 ÷ 9
4) 2342 ÷ 98

2) 120012 ÷ 9

3) 1135 ÷ 97

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
...
The answer lies in Vedic Methods
...
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
...

Ex
...
e
...
e
...
The product 4 X 2 = 8 forms the right hand most part of the answer
...
e
...
e
...
e
...
e
...
e
...
It gives the next, i
...
, second
digit of the answer
...


iii) Now, multiply the second digit of the multiplicand i
...
, 1 and second digit of
the multiplieri
...
, 1 vertically, i
...
, 1 X 1 = 1
...


Thus the answer is 16 8
...


Step i) :
31

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)

Ex
...


32

What happens when one of the results i
...
, either in the last digit or in the
middle digit of the result, contains more than 1 digit ? Answer is simple
...
e
...
The digits carried
over may be written as in Ex
...


Ex
...


Here 6 is to be retained
...


Step (iii) :

3 X 2 = 6
...

i
...
, 6 + 1 = 7
...


33

Note that the carried over digit from the result (3X4) + (2X2) = 12+4 = 16
i
...
,1 is placed under the previous digit 3 X 2 = 6 and added
...


Ex
...


Step (i) :
8 X 5 = 40
...

Step (ii) : 2 X 5 = 10; 8 X 3 = 24; 10 + 24 = 34; add the carried over 4 to
34
...
Now 8 is retained as the second digit of the
answer and3 is carried over
...
The result 6 + 3 = 9 is the
third or final digit from right to left of the answer
...

Ex
...


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
...


Algebraic proof :
a) Let the two 2 digit numbers be (ax+b) and (cx+d)
...
Now
consider the product
34

(ax + b) (cx + d) = ac
...
d

= ac
...
d

Observe that
i) The first term i
...
, the coefficient of x2 (i
...
, 100, hence the digit in the100th
place) is obtained by vertical multiplication of a and c i
...
,the digits in10th place
(coefficient of x) of both the numbers;
ii) The middle term, i
...
, the coefficient of x (i
...
, 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
...


Let the two numbers be(ax2 + bx + c) and (dx2 + ex + f)
...
x4+bd
...
x2+ae
...
x2+ce
...
x2+bf
...
x4 + (bd + ae)
...
x2 + (ce + bf)x + cf

35

Note the following points :
i) The coefficient of x4 , i
...
, ad is obtained by the vertical multiplication of the
firstcoefficient from the left side :

ii)The coefficient of x3 , i
...
, (ae + bd) is obtained by the cross –wise
multiplication of the first two coefficients and by the addition of the two
products;

iii) The coefficient of x2 is obtained by the multiplication of the first coefficient
of the multiplicand(ax2+bx +c) i
...
, a; by the last coefficient of the multiplier
(dx2 +ex +f)i
...
,f ; of the middle one i
...
, b of the multiplicand by the middle
one i
...
, e of the multiplier and of the last onei
...
, c of the multiplicand by the
first one i
...
, d of the multiplier and by the addition of all the three productsi
...
,
af + be +cd :

iv) The coefficient of x is obtained by the cross wise multiplication of the second
coefficienti
...
, b of the multiplicand by the third one i
...
, f of the multiplier, and
conversely the third coefficienti
...
, c of the multiplicand by the second
coefficient i
...
, e of the multiplier and by addition of the two products,i
...
, bf +
ce ;

36

v) And finally the last (independent of x) term is obtained by the vertical
multiplication of the last coefficients c and f i
...
, cf

Thus the process can be put symbolically as (from left to right)

Consider the following example

124 X 132
...
First digit = 8
ii) (2 X 2) + (3 X 4) = 4 + 12 = 16
...
Second digit = 6
...
The carried over 1 of above
step is added i
...
, 12 + 1 = 13
...
Thus third digit = 3
...
the carried over 1 of above step is added
i
...
, 5 + 1 = 6
...
Thus fourth digit = 6
v) ( 1 X 1 ) = 1
...
Thus fifth digit = 1
124 X 132 = 16368
...

234
x 316
‾‾‾‾‾‾‾
61724
1222
‾‾‾‾‾‾‾
73944
i) 4 X 6 = 24 : 2, the carried over digit is placed below the second 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
...

v) ( 2 X 3 ) = 6
...

Note :
1
...

2
...

3
...


38

Example 1 : Find the product of (a+2b) and (3a+b)
...
e
...
e
...
e
...
e
...
x2 + 4x + 7
0
...
e
...
e
...
e
...

1) 25 X 16

2) 32 X 48

3) 56 X 56

4) 137 X 214

5) 321 X 213

6) 452 X 348
8) (5a2 + 1) (3a2 + 4)

7) (2x + 3y) (4x + 5y)
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
...
x3 + 5x2 + 3x + 7
It is the first term of the Quotient
...
But 5x2 in the dividend hints7x2 more since 7x2 – 2x2 =
5x2
...
Hence
second term of Q is 7x
...
But the 3rd term in the dividend is 3x for
which ‘17x more’ is required since 17x – 14x =3x
...
Hence third term of quotient is 17

40

Thus
x3 + 5x2 + 3x + 7
_________________
x–2

gives Q= x2 + 7x +17

iv) Now last term of Q, i
...
, 17 multiplied by –2 gives 17X–2 = -34 but the
relevant term in dividend is 7
...
As there no
more terms left in dividend, 41 remains as the remainder
...


Find the Q and R in the following divisions by using the converse
process of urdhva – tiryagbhyam method :
1) 3x2 – x – 6
‾‾‾‾‾‾‾‾‾
3x – 7
3) x3+ 2x2 +3x + 5
‾‾‾‾‾‾‾‾‾‾‾‾‾‾
x-3

2) 16x2 + 24x +9
‾‾‾‾‾‾‾‾‾‾‾‾
4x+3
4) 12x4 – 3x2 – 3x + 12
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
x2 + 1

4
...

Example 1 : Divide 1225 by 12
...

12
-2
‾‾‾‾
Step 2 : Write down the dividend to the right
...


41

i
...
,,

12
-2

122

5

Step 3 : Write the 1st digit below the horizontal line drawn under
thedividend
...

i
...
,,

12
122 5
-2
-2
‾‾‾‾‾ ‾‾‾‾
10

Since 1 x –2 = -2and 2 + (-2) = 0
Step 4 : We get second digits’ sum as ‘0’
...

12
-2
‾‾‾‾

122
5
-20
‾‾‾‾‾‾‾‾‾‾
102
5

Step 5 : Continue the process to the last digit
...
e
...

Thus Q = 102 andR = 1
Example 2 :

Divide
14
-4
‾‾‾‾

1697 by 14
...

Example 3 :

Divide

2598 by 123
...
So we have to set up the last two
42

digits of the dividend for the remainder
...

Hence Q = 21 and R = 15
...
The divisor has 5 digits
...

1 1213
-1-2-1-3
‾‾‾‾‾‾‾‾

23 9 479
-2 -4-2-6
with 2
-1-2-1-3
with 1
‾‾‾‾‾‾‾‾‾‾‾‾‾
21
40 0 6

Hence Q = 21, R = 4006
...
e
...
To
over come this situation, take 1 over from the quotient column, i
...
,, 1123
over to the right side, subtract the remainder portion 20 to get the actual
remainder
...


43

Find the Quotient and Remainder for the problems using
paravartya – yojayet method
...

Example 1 :

Divide
X- 1
‾‾‾‾‾‾
1

Example 2 :

Divide
X- 5
‾‾‾‾‾
5

6x2 + 5x + 4 by x – 1
6x2 + 5x + 4
6+ 11
‾‾‾‾‾‾‾‾‾‾‾‾
6x + 11 + 15

ThusQ = 6x+11,R=15
...

The procedure as a mental exercise comes as follows :
i) x3 / xgives x2 i
...
,, 1 the first coefficient in the Quotient
...
It gives 1 X( +5) =
+5, adding to the next coefficient, i
...
,, –3 + 5 = 2
...

iii) Continue the process : multiply 2 by +5, i
...
,, 2 X +5 =10, add to the
next coefficient 10 + 10 = 20
...
Thus
Quotient is x2 + 2x + 20
iv) Nowmultiply 20 by + 5i
...
,, 20 x 5 = 100
...
e
...


44

Example 3:
x4 –3x3 + 7x2 + 5x + 7
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
x+4
Now thinking the method as in example ( 1 ), we proceed as follows
...

or we proceed orally as follows:
x4 / x gives 1 as first coefficient
...

ii) – 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q
...

iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R
...

Note :
1
...

2
...

Now consider the divisors of second degree or more as in the following
example
...


Here x2 term is missing in the dividend
...
x2 or 0
...
x
...
x2 - 3x + 2
x2+ 0
...
x + 4 = 4
...


We treat the dividend as 2x5 –5x4 + 0
...
x + 3 and proceed as follows :
x3 – 2x2 + 0
...
x3 + 3x2 – 4x + 7
4

0
-2

-6
0 + 3
-4
0 + 6
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
2 - 1
-2
- 7 - 1 +13

Thus Q = 2x2 – x – 2, R = - 7x2 – x + 13
...
And yet paravartya goes much farther and is capable of
numerous applications in other directions also
...

1)
2)
3)
4)
5)

(4x2 + 3x + 5)÷(x+1)
(x3 – 4x2 + 7x + 6) ÷ (x – 2)
(x4 – x3 + x2 + 2x + 4) ÷(x2 - x – 1)
(2x5 +x3 – 3x + 7) ÷ (x3 + 2x – 3)
(7x6 + 6x5 – 5x4 + 4x3 –3x2 + 2x – 1)÷ (x-1)

46

Paravartya in solving simple equations :
Recall that 'paravartya yojayet' means 'transpose and apply'
...
i
...
, + becomes - and conversely ; and X becomes ÷ and conversely
...

Type ( i ) :
Consider the problem 7x – 5 = 5x + 1
7x – 5x = 1 + 5
i
...
,, 2x = 6

x = 6 ÷ 2 = 3
...

‾‾‾‾‾‾‾‾
a-c

In this example a = 7, b = - 5, c = 5, d = 1
Hence
x =

Example 2:

1 – (- 5)
_______
7–5

1+5
= ____
7-5

=

6
__
2

=

3

2
__
1

=

Solve for x,3x + 4 = 2x + 6

x =

d-b
_____
a-c

6-4
= _____
3-2

=

2

Type ( ii ) :Consider problems of the type (x + a) (x+b) = (x+c) (x+d)
...
)

Example 1 : (x – 3) (x – 2 ) = (x + 1 ) (x + 2 )
...


Now

cd - ab
___________
a+b–c–d

x=

=

(3) (-4) – (7) (-6)
________________
7 + (-6) – 3 - (-4)

=

- 12 + 42
____________
7–6–3+4

=

30
___
2

=

15

Note that if cd - ab = 0 i
...
,, cd = ab, i
...
,, if the product of the absolute
terms be the same on both sides, the numerator becomes zero giving x = 0
...
= 12
Type ( iii) :
Consider the problems of the type

ax + b
______
cx + d

48

m
= __
n

By cross – multiplication,
n ( ax + b) = m (cx + d)
nax + nb = mcx + md
nax - mcx = md – nb
x( na – mc ) = md – nb

x

=

md - nb
________
na - mc
...
C
...

m(x+b) + n (x+a)
______________
(x + a) (x +b)

=

0

mx + mb + nx + na
________________
x + a)(x + b)

=

0

(m + n)x + mb + na

=

0

(m + n)x = - mb - na

-mb - na
________
(m + n)

x =

Thus the problem

m
____
x+a

+

n
____ =
x+b

0,

by paravartya process

gives directly
x

Example 1 :

gives

=

-mb - na
________
(m + n)

3
4
____ + ____
x+4
x–6
-mb - na
x = ________
(m + n)

=

=

0

Note that m = 3, n = 4, a = 4, b = - 6

-(3)(-6) – (4) (4)
_______________
( 3 + 4)

=

50

18 - 16
______
7

2
= __
7

Example 2 :
5
____
x+1
gives
x

=

6
_____ =
x – 21

+

-(5) (-21) - (6) (1)
________________
5+6

=

0

105 - 6
______
11

99
= __
11

= 9

I
...

1) 3x + 5 = 5x – 3

6) (x + 1) ( x + 2) = ( x – 3) (x – 4)

2) (2x/3) + 1=x - 1
3) 7x + 2
______
3x- 5

=

7) (x – 7) (x – 9)= (x – 3) (x – 22)

5
__
8

8) (x + 7) (x + 9)= (x + 3 ) (x + 21)

4) x + 1 / 3
_______ = 1
3x - 1
5)

5
____ +
x+3

2
____
x–4

=

0

II)
1
...


2
...
x + m (c - a)
...
x + n(b - c)
...


By paravartya rule we can easily remember the formula
...
Sunyam Samya Samuccaye
The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that
Samuccaya is Zero
...
e
...
The term 'Samuccaya'
has several meanings under different contexts
...

Example 1: The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all
its terms
...
e
...

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
follows
...

iii) We interpret ' Samuccaya 'as the sum of the denominators of two fractions
having the same numerical numerator
...

1
____
3x-2

1
+ ____
2x-1

=

0

for this we proceed by takingL
...
M
...
e
...
e
...

If the sum of the numerators and the sum of the denominators be the
same, then that sum = 0
...


55

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
x = __
6

-3
= __
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
...

3x + 4 = 0

3x = -4

x = - 4 / 3
...
e
...
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
...
e
...

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
8x = -10
x = - 10 / 8
=-5/4

3( x + 1 ) = 0
x+1=0
x = -1

vi)‘Samuccaya’ with the same sense but with a different context and
application
...

x–6+x-4
___________
(x–4) (x–6)

=

x–8+x-2
___________
(x–2) (x-8)

58

2x-10
_________
x2–10x+24

=

2x-10
_________
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 sumtotal of the denominators on the L
...
S
...
H
...
be the
same, that total is zero
...
But a little work regarding transposition
makes the above as follows
...
e
...

x = - 16 / 2 = - 8
...

1
...


( x + 6 ) ( x + 3 ) = ( x – 9 ) ( x – 2)

3
...


1
______
4x-3

5
...


7
...


4x - 3
______ =
2x+ 3

9
...


1
____ x-7

x+ 4
_____
3x - 2

=

1
____ =
x-6

1
1
____ + _____
x-3
x-4

1
1
_____ - _____
x - 10
x-9

Sunyam Samya Samuccaye in Certain Cubes:
Consider the problem ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5)3
...
Taking out the numerical factor 2,
we have ( x – 5 ) = 0, which is the factor under the cube on R
...
S
...
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
...
And x – 1
...
H
...


61

cube, it is enough to state that x – 1 = 0 by the ‘sutra’
...
No cubing or any other mathematical operations
...


( x – 3 )3 + ( x – 9 )3 = 2 ( x – 6 )3

2
...


( x + a + b – c )3 + ( x + b + c – a )3 = 2 ( x + b )3

62

Example :
(x + 2)3
x+1
______ = _____
(x + 3)3
x+4
with the text book procedures we proceed as follows
x3 + 6x2 + 12x +8
_______________
x3 + 9x2 + 27x +27

x+1
= _____
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
...
S
...

By vedic formula we have 2x + 5 = 0

x = - 5 / 2
...


(x + 3)3
______ =
(x + 5)3

x+1
____
x+7

63

2
...
Anurupye - Sunyamanyat
The Sutra Anurupye Sunyamanyat says : 'If one is in ratio, the other one is
zero'
...
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
variable
...
e
...
e
...
e
...
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
...

y = 0 and 323 x = 1615 or 969 x = 4845 gives x = 5
...

1
...


64

3x + 7y = 24

16x + 96y =16
3
...


ax + by = bm
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
x + 5xy + 4y + 4x - 2y = 20
2

2

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
...
,, 8y= 20 y = 10/4 = 2½
Thus x = 0 and y = 2½ is the solution
...
Sankalana - Vyavakalanabhyam
This Sutra means 'by addition and by subtraction'
...

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
...
It is difficult process to think of
...
e
...
e
...
e
...
e
...

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
subtract - 2y = - 6

x=2
y=3

Solve the following problems usingSankalana – Vyavakalanabhyam
...


3x + 2y = 18
2x + 3y = 17

2
...


659x + 956y = 4186
956x + 659y = 3889

66

8
...
Purana is well known in the present
system
...

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
...
,, purana ) on the L
...
S
...

Example 1
...

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
...
,, x3 + 6x2 + 12x + 8 = x + 2
i
...
,, ( x + 2 )3 = ( x + 2 )
this is of the form y3 = y for y = x + 2
solution y = 0, y = 1, y = - 1
i
...
,, 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
...
,, x3 + 9x2 + 27x + 27 = x2 + 6x + 9 + 4x + 8
i
...
,, ( x + 3 )3 = ( x + 3 )2 + 4 ( x + 3 ) – 4

67

y3 = y2 + 4y – 4 for y = x + 3
y = 1, 2, -2
...

Further purana can be applied in solving Biquadratic equations also
...

1
...

3
...


x3
x3
x2
x4

– 6x2 + 11x – 6 = 0
+ 9x2 + 23x + 15 = 0
+ 2x – 3 = 0
+ 4x3 + 6x2 + 4x – 15 = 0

9
...
The Sutra means
'Sequential motion'
...
Swamiji called the sutra as calculus formula
...
Now by calculus formula we say: 14x–11 = ±√317
A Note follows saying every Quadratic can thus be broken down into two
binomial factors
...

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
Calculus'
...
Hence the sutra and its various applications will be taken up
at a later stage for discussion
...


68

Now the remaining sutras :
10
...

12
...

15
...


YĀVADŨNAM ( The deficiency )
VYAŞłISAMAŞłIH ( Whole as one and one as whole )
ŚEŞĀNYAŃ KENA CARAMEĥA ( Remainder by the last digit )
SOPĀNTYADVAYAMANTYAM ( Ultimate and twice the penultimate )
GUĥITASAMUCCAYAH ( The whole product is the same )
GUĥAKA SAMUCCAYAH ( Collectivity of multipliers )

The Sutras have their applications in solving different problems in different
contexts
...
So it is a bit of
inconvenience to deal each Sutra under a separate heading exclusively and
also independently
...
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
...
Now we shall deal the fourteenth Sutra, the
Sutra left so far untouched
...
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'
...
is as follows
...
e
...

e
...
( i ) 7 x 9; 7 – 1 = 6 ( L
...
S
...
i
...
7 X 9 R
...
S is 9 - 6 = 3
...
e
...

Example 1:

8 x 9 Step ( a ) gives 8 – 1 = 7 ( L
...
S
...
H
...
Digit )
Step ( c ) gives the answer 72
69

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
...
S from
the multiplier which is practically a direct application of Nikhilam Sutra
...
H
...

x 99 – 23 = 76 R
...
S
...
H
...

x 999 – 355 = 644 R
...
S
...
H
...

x 9999 – 877 = 9122 R
...
S
...
102 – 10x + 102
...
103 + y
...
103 + ( y – 1 )
...
(or apply nikhilam)
...
The 1000th place is x i
...
3
100th place is ( y - 1 ) i
...
(7 - 1 ) = 6
Number in the last two places 100-37=63
...


Apply Ekanyunena purvena to find out the products
1
...
723 x 999

3
...
43 x 999

5
...
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
...
But what happens when the multiplier
has lesser digits?
i
...
for problems like 42 X 9, 124 X 9, 26325 X 99 etc
...

Multiplication table of 9
...

a
b
11 x 99 = 10 89 = (11–1) / 99 – (11–1) = 1089
12 x 99 = 11 88 = (12–1) / 99 – (12–1) = 1188
13 x 99 = 12 87 = (13–1) / 99 – (13–1) = 1287
------------------------------------------------18 x 99 = 17 82 ---------------------------19 x 99 = 18 81
20 x 99 = 19 80 = (20–1) / 99 – (20–1) = 1980
The rule mentioned in the case of above table also holds good here
Further we can state that the rule applies to all cases, where the multiplicand
and the multiplier have the same number of digits
...

(i)
a b
11 x 9 = 9 9
12 x 9 = 10 8
13 x 9 = 11 7
---------------------18 x 9 = 16 2
19 x 9 = 17 1
20 x 9 = 18 0
72

(ii)
21 x 9 = 18 9
22 x 9 = 19 8
23 x 9 = 20 7
----------------------28 x 9 = 25 2
29 x 9 = 26 1
30 x 9 = 27 0
(iii)
35 x 9 = 31 5
46 x 9 = 41 4
53 x 9 = 47 7
67 x 9 = 60 3
-------------------------so on
...
Here
L
...
S of products are uniformly 2 less than the multiplicands
...
Here L
...
S of products are uniformly 3 less
than the multiplicands
...
e
...
H
...
H
...

4) The right hand side of the product in all the tables and cases is obtained by
subtracting the R
...
S
...

Keeping these points in view we solve the problems:
Example1 : 42 X 9
i) Divide the multiplicand (42) of by a Vertical line or by theSign : into a right
hand portion consisting of as many digits as the multiplier
...
e
...
i
...
left portion of multiplicand is 4
...


73

We have to subtract this from multiplicand
i
...
write it as
4 : 2
:-5
--------------3 : 7
This gives the L
...
S part of the product
...
e
...

iii) Subtract the R
...
S
...

i
...
R
...
S of multiplicand is 2
its nikhilam is 8
It gives the R
...
S of the product
i
...
answer is 3 : 7 : 8 = 378
...

Example 2 :

124 X 9

Here Multiplier has one digit only
...
e
...
H
...
of multiplicand is 4
...

1
...


62 x 9

3
...


832 x 9

5
...
111011 x 99

Ekanyunena Sutra is also useful in Recurring Decimals
...

Thus we have a glimpse of majority of the Sutras
...
Any how we now proceed into the use of SubSutras
...

But some approaches in the Vedic Mathematics book prompted some serious
research workers in this field to mention some other Upa-Sutras
...


11
...
This Sutra is highly useful
to find products of two numbers when both of them are near the Common bases
i
...
It is very clear that in such cases the expected
'Simplicity ' in doing problems is absent
...

43 -57
‾‾‾‾‾‾‾‾
75

Now by ‘anurupyena’ we consider a working base In three ways
...

Method 1: Take the nearest higher multiple of 10
...

Treat it as 100 / 2 = 50
...

i
...
, working base is 100 / 2 = 50
ii) Write the numbers one below the other
i
...


4 6
4 3
‾‾‾‾‾‾‾

iii) Write the differences of the two numbers respectively from 50 against each
number on right side
i
...


46 -04
43 -07
‾‾‾‾‾‾‾‾‾

iv) Write cross-subtraction or cross- addition as the case may be under the line
drawn
...

46 -04
43 -07
____________
39 / -4 x –7
= 28
vi) Since base is 100 / 2 = 50 , 39 in the answer represents 39X50
...
This 1 as
Reminder gives one 50 making the L
...
S of the answer 28 + 50 = 78(or
Remainder ½ x 100 + 28 )
i
...
R
...
S 19 and L
...
S 78 together give the answer1978 We represent it as
46 -04
43 -07
‾‾‾‾‾‾‾‾‾
2) 39 / 28
‾‾‾‾‾‾‾‾‾
19½ / 28
= 19 / 78 = 1978
Example 2: 42 X 48
...
We take the same working base 50
...
i
...
we operate with 10 but not with 100 as in method
now

(195 + 2) / 8 = 1978
[Since we operate with 10, the R
...
S portion shall have only unit place

...
The L
...
S portion of
the answer shall be multiplied by 5, since we have taken 50 = 5 X 10
...


Since 10 is in operation 1 is carried out digit in 18
...
H
...
e
...
H
...
of answer as 1978
...

We proceed in the same method for 42 X 48

Let us see the all the three methods for a problem at a glance
Example 3: 24 X 23

78

Method - 1:

Working base = 100 / 5 = 20

24 04
23 03
‾‾‾‾‾‾‾‾
5) 27 / 12
‾‾‾‾‾‾‾‾
5 2/5 / 12 = 5 / 52 = 552
[Since 2 / 5 of 100 is 2 / 5 x 100 = 40 and 40 + 12 = 52]
Method - 2: Working base 2 X 10 = 20

Now as 20 itself is nearest lower multiple of 10 for the problem under
consideration, the case of method – 3 shall not arise
...

Example 4: 492 X 404
Method - 1 : working base = 1000 / 2 = 500
492 -008
404 -096
‾‾‾‾‾‾‾‾‾‾‾
2) 396 / 768
‾‾‾‾‾‾‾‾‾‾‾
198 / 768

since 1000 is in operation
= 198768

79

Method 2: working base = 5 x 100 = 500

Method - 3
...

Thus

No need to repeat that practice in these methods finally takes us to work
out all these mentally and getting the answers straight away in a single line
...
e
...
H
...
i
...
the answer is

A simpler example for better understanding
...

Use anurupyena by selecting appropriate working base and method
...

1
...
57 x 57

3
...


18 x 18

5
...
229 x 230

7
...
87965 x 99996

10
...
49x499

12
...

Suppose we are asked to find out the area of a rectangular card board whose
length and breadth are respectively 6ft
...
8 inches
...

Area = Length X Breath
= 6’ 4" X 5’ 8" Since 1’ = 12", conversion
= ( 6 X 12 + 4) ( 5 X 12 + 8) in to single unit
= 76" 68" = 5168 Sq
...

Since 1 sq
...
=12 X 12 = 144sq
...
e
...
ft 128 Sq
...
e
...
= 12 in;x2 is sq
...

Now ( 6x + 4 )(5x + 8 )
=
=
=
=
=
=
=
=

30x2 + 6
...
x + 4
...
x + 32
30x2 + 48x + 20x + 32
30x2 + 68
...
x + 32 Writing 68 = 5 x 12 + 8
35x2 + 8
...
ft
...
in + 32 Sq
...
ft
...
in + 32 Sq
...
ft
...
in

82

It is interesting to know that a mathematically untrained and even uneducated
carpenter simply works in this way by mental argumentation
...
e
...
ft
...
e
...
in
...

Adjust as many '12' s as possible towards left as 'units' i
...
68 = 5 X 12 +8 , 5
twelve's as 5 square feet make the first 30+5 = 35 sq
...
e
...
in
...
in
...
units
...
e
...
ft 128 sq
...
ft
...
inches
...
in = 12 X 12 = 1 sq
...
ft
...


83

I
...

1)
...


l = 12’ 5" , b = 5’ 7"

3)
...
b = 2 yards 5 ft
...


l = 6 yard 6 ft
...


II
...

Recall area =½ h (a + b) where a, b are parallel sides and h is the
distance between them
...


a = 3’ 7", b = 2’ 4", h = 1’ 5"

2)
...


a = 8’ 4", b = 4’ 6", h = 5’ 1"
...
The steps are as
follows
...
Split the middle coefficient in to two such parts that the ratio of the first
coefficient to the first part is the same as the ratio of the second part to the last
coefficient
...
e
...
e
...
e
...
It is clear that 3:6 = 2:4
...
Now the ratio 3: 6 = 2: 4 = 1:2 gives one factor x+2
...
Second factor is obtained by dividing the first coefficient of the quadratic by
the fist coefficient of the factor already found and the last coefficient of the
quadratic by the last coefficient of the factor
...
e
...
1: 4x2 + 12x + 5
i) Split 12 into 2 and 10 so that as per rule 4 : 2 = 10 : 5 = 2 : 1i
...
,, 2x + 1 is
first factor
...


Eg
...
Both are
same
...
e
...

ii) Now
15x2
____
3x

8y2
+ ___
-4y

= 5x + 2y is second factor
...

It is evident that we have applied two sub-sutras ‘anurupyena’
i
...
‘proportionality’ and ‘adyamadyenantyamantyena’ i
...
‘the first by the
first and the last by the last’ to obtain the above results
...
3x2 + 14x + 15
2)
...
8x2 – 22x + 5
4)
...
Yavadunam Tavadunikrtya Varganca Yojayet
The meaning of the Sutra is 'what ever the deficiency subtract that deficit from
the number and write along side the square of that deficit'
...

Method-1 : Numbers near and less than the bases of powers of 10
...

The answer is separated in to two parts by a’/’
Note that deficit is 10 - 9 = 1
Multiply the deficit by itself or square it
12 = 1
...
e
...

Now put 8 on the left and 1 on the right side of the vertical line or slash
i
...
, 8/1
...

Eg
...

Since deficit is 100-96=4 and square of it is 16 and the deficiency
subtracted from the number 96 gives 96-4 = 92, we get the answer 92 / 16
Thus 962 = 9216
...
3: 9942 Base is 1000
Deficit is 1000 - 994 = 6
...

Deficiency subtracted from 994 gives 994 - 6 = 988
Answer is 988 / 036
Eg
...


Deficit = 10000 - 9988 = 12
...

Deficiency subtracted from number = 9988 - 12 = 9976
...
5:

882

[since base is 10,000]
...


Deficit = 100 - 88 = 12
...

Deficiency subtracted from number =88 - 12 = 76
...

Thus
(1) 9 = (10 -1) (2) 96 = ( 100-4) (3) 994 = (1000-6)
(4) 9988 = (10000-12 ) (v) 88 = (100-12)
( x – y )2 =x2 – 2xy + y2
= x ( x – 2y ) + y2
= x ( x – y – y ) + y2
= Base ( number – deficiency ) + ( deficit )2
Thus
9852 = ( 1000 – 15)2
= 1000 ( 985 – 15 ) + (15)2
= 1000 ( 970 ) + 225
= 970000 + 225
= 970225
...

gives

a2 = (a + b) ( a - b) +b2

Thus for a = 985 and b = 15;
a2= (a + b) ( a - b) + b2
9852 = ( 985 + 15 ) ( 985 - 15 ) + (15)2
87

= 1000 ( 970 ) + 225
= 970225
...
2 : Numbers near and greater than the bases of powers of 10
...
(1): 132
...

for 132 , base is 10, surplus is 3
...

Square of surplus = 32 = 9
Answer is 16 / 9 = 169
...
(2):

1122

Base = 100, Surplus = 12,
Square of surplus = 122 = 144
add surplus to number = 112 + 12 = 124
...

(x + y)2 =x2 + 2xy + y2
= x ( x + 2y ) + y2
= x ( x + y + y ) + y2
= Base ( Number + surplus ) + ( surplus)2
gives
1122 =100 ( 112 + 12 ) +122
= 100 ( 124 ) + 144
88

= 12400 + 144
= 12544
...
3:

100252
= ( 10025 + 25 ) / 252
= 10050 / 0625

[ since base is 10,000 ]

= 100500625
...
etc
...

Example 1:

3882 Nearest base = 400
...
As the number is less than the base we proceed
as follows
Number 388, deficit = 400 - 388 = 12
Since it is less than base, deduct the deficit
i
...
388 - 12 = 376
...

376 x 4 = 1504
Square of deficit = 122 = 144
...

Example 2: 4852 Nearest base = 500
...

1
...
982

3
...
142

5
...
10122

7
...
4752

11
...
60142
...
7962

10
...
Now with a slight modification yavadunam can also be applied for
finding the cubes of numbers
...


i)

We proceed as follows:
For 106, Base is 100
...

Here we add double of the surplus i
...
106+12 = 118
...

i
...
answer proceeds like 118 / - - - - -

ii) Put down the new surplus i
...
118-100=18 multiplied by the initial surplus
i
...
6=108
...
e
...

iii) Write down the cube of initial surplus i
...
63 = 216 as the last portion
i
...
right hand side last portion of the answer
...

Answer is 118 / 108 / 216
Now proceeding from right to left and adjusting the carried over, we get the
answer
119 / 10 / 16 = 1191016
...
(1):

1023 = (102 + 4) / 6 X 2 / 23
=

106

=

12 = 08

= 1061208
...


91

Eg
...
Here it is deficit contrary to the
above examples
...
Twice of it -6 X 2 = -12
add it to the number = 94 -12 =82
...

Product of new deficit x initial deficit = -18 x -6 = 108
iii) deficit3 = (-6)3 = -216
...
Adjusting the carried over in
order, we get the answer
( 82 + 1 ) /( 08 – 03 ) / ( 100 – 16 )
= 83 / = 05
/ = 84
= 830584
__
16 becomes 84 after taking1 from middle most portion i
...
100
...

_
Now 08 - 01 = 07 remains in the middle portion, and2 or 2 carried to it
makes the middle as 07 - 02 = 05
...

Eg
...

9983 = (998 – 2 x 2) / (- 6 x – 2) / (- 2)3
=

994

/

= 012

=

994 / 011 / 1000 - 008

=

994 / 011 / 992

=

/ = -008

994011992
...

1
...
96

2
...
993

3
...
9991
92

4
...
1000008

5
...
999992
...
Antyayor Dasakepi
The Sutra signifies numbers of which the last digits added up give 10
...
e
...
Note that in each case the sum of the last digit of first
number to the last digit of second number is 10
...
At that instant use
Ekadhikena on left hand side digits
...

Example 1 : 47 X 43
See the end digits sum 7 + 3 = 10 ; then by the sutras antyayor dasakepi and
ekadhikena we have the answer
...

Example 2: 62 x 68
2 + 8 = 10, L
...
S
...
e
...

Ekadhikena of 6 gives 7
62 x 68 = ( 6 x 7 )/ ( 2 x 8 )
= 42 / 16
= 4216
...

Example 4: 65 x 65
We have already worked on this type
...

We have 65 x 65 = 6 x 7/ 5 x 5
= 4225
...


Use Vedic sutras to find the products
1
...
34 x 36

3
...
401 x 409

5
...
1404 x 1406

It is further interesting to note that the same rule works when the sum of the
last 2, last 3, last 4 - - - digits added respectively equal to 100, 1000, 10000 -- -
...
Your can observe that this is more convenient
while working with the product of 3 digit numbers
...
1: 292 x 208
Here 92 + 08 = 100, L
...
S portion is same i
...
2
292 x 208 = ( 2 x 3 )/ 92 x 8
60 / =736 ( for 100 raise the L
...
S
...

Eg
...
H
...

Now R
...
S product 48 X 52 can be obtained by ‘anurupyena’ mentally
...

[Since L
...
S product is to be multiplied by 10 and 2 to be carried over as the
base is 100]
...
3:

693 x 607
693 x 607 = 6 x 7/ 93 x 7
= 420 / 651
= 420651
...
318 x 312

2
...
796 x 744

4
...
397 x 393

6
...
Antyayoreva
'Atyayoreva' means 'only the last terms'
...

The type of equations are those whose numerator and denominator on the
L
...
S
...
H
...
stand to each
other
...

Example 1:
x2 + 2x + 7
__________
x2 + 3x + 5

x+2
= _____
x+3

In the conventional method we proceed as
x2 + 2x + 7
__________
x2 + 3x + 5

x+2
= _____
x+3

(x + 3) (x2 + 2x + 7) = (x + 2) (x2 + 3x + 5)
x + 2x + 7x +3x2 + 6x + 21 = x3 + 3x2 + 5x + 2x2 +6x + 10
x3 + 5x2 + 13x + 21 = x3 + 5x2 + 11x+ 10
3

2

Canceling like terms on both sides
13x + 21 = 11x + 10
13x – 11x = 10 – 21
2x = -11
x = -11 / 2
Now we solve the problem using anatyayoreva
...
Hence from the sutra
x+2
_____
x+3

7
= __
5

5x + 10 = 7x + 21
7x – 5x = -21 + 10
2x = -11
x = -11 / 2
Algebraic Proof:
Consider the equation
AC + D
______
BC + E

A
= ___
B

------------- (i)

This satisfies the condition in the sutra since
AC
___
BC

A
= ___
B

Now cross–multiply the equation (i)
B (AC + D) = A (BC + E)
BAC + BD = ABC + AE
BD = AE which gives
A
__
B

D
= __
E

--------(ii)

97

i
...
, the result obtained in solving equation (i) is same as the result obtained in
solving equation (ii)
...

2x + 3
_____
3x+4

=

10
__
14

Cross–multiplying
28x + 42 = 30x + 40
28x – 30x = 40 – 42
-2x = -2
x = -2 / -2 = 1
...

Example 3: (x + 1) (x + 2) (x + 9) = (x + 3) (x + 4) (x + 5)
Re–arranging the equation, we have
(x + 1) (x + 2)
____________
(x + 4) (x + 5)

=

x+3
_____
x+9

i
...
,
x2 + 3x + 2x + 3
= ______________
x2 + 9x + 20x + 9
Now
x2 +3x
x (x + 3)
______ = _______
x2 +9x
x (x + 9)

x+3
= _____
x+9

gives the solution by antyayoreva

98

Solution is obtained from
x+3
____
x+9

=

2
__
20

20x + 60 = 2x + 18
20x – 2x = 18 – 60
18x = -42

x = -42 / 18 = -7 / 3
...
In such a case the equation can be adjusted into the form suitable
for application of antyayoreva
...
H
...
= 3x + 16
Sum of the binomials on R
...
S
...
Hence antyayoreva can be applied
...

3x2 + 5x + 8
__________
5x2 + 6x +12
2
...


(x + 3) (x +4) (x + 6) = (x + 5) (x + 1) (x + 7)

4
...


2x2 + 3x + 9
__________
4x2 +5x+17

=

2x + 3
______
4x + 5

100

16
...

Consider the case of factorization of quadratic equation of type ax2 +by2 + cz2 +
dxy + eyz + fzx This is a homogeneous equation of second degree in three
variables x, y, z
...
The steps are as follows:
i) Eliminate z by putting z = 0 and retain x and y and factorize thus obtained a
quadratic in x and y by means of ‘adyamadyena’ sutra
...

iii) With these two sets of factors, fill in the gaps caused by the elimination
process of z and y respectively
...

Example 1: 3x2 + 7xy + 2y2 + 11xz + 7yz + 6z2
...


Eliminate z i
...
, z = 0; factorize
12x2 + 11xy + 2y2 = (3x + 2y) (4x + y)

Step (ii): Eliminate y i
...
, y = 0; factorize
12x2 - 13xz + 3z2 = (4x -3z) (3x – z)
Step (iii): Fill in the gaps; the given expression
= (4x + y – 3z) (3x + 2y – z)
Example 3:

3x2+6y2+2z2+11xy+7yz+6xz+19x+22y+13z+20

Step (i): Eliminate y and z, retain x and independent term
i
...
, y = 0, z = 0 in the expression (E)
...
e
...

Then E = 6y2 + 22y + 20 = (2y + 4) (3y + 5)
Step (iii): Eliminate x and y, retain z and independent term
i
...
, x = 0, y = 0 in the expression
...


In this way either homogeneous equations of second degree or general
equations of second degree in three variables can be very easily solved by
applying ‘adyamadyena’ and ‘lopanasthapanabhyam’ sutras
...


x2 + 2y2 + 3xy + 2xz + 3yz +z2
...


3x2 + y2 - 4xy - yz -2z2 - zx
...


2p2 + 2q2 + 5pq + 2p – 5q - 12
...


u2 + v2 – 4u + 6v – 12
...


x2 - 2y2 + 3xy + 4x - y + 2
...


3x2 + 4y2 + 7xy - 2xz - 3yz -z2 + 17x + 21y – z + 20
...
e
...
C
...
of algebraic expressions, the
factorization method and process of continuous division are in practice in the
conventional system
...
C
...

Example 1:

Find the H
...
F
...


1
...
C
...
is ( x + 1 )
...
Continuous division process
...
e
...
C
...


3
...
e
...
e
...
C
...
C
...
of 2x2 – x – 3 and 2x2 + x – 6

x3 – 7x – 6 and x3 +8x2 + 17x + 10
...


(or)

Example 5:

2x3 + x2 – 9 andx4 + 2x2 + 9
...

÷ x2 gives

x2 + 2x + 3 ------ (i)

Subtract after multiplying the first by x and the second by 2
...

104

Thus ( x2 + 2x + 3 ) is the H
...
F
...

Algebraic Proof:
Let P and Q be two expressions and H is their H
...
F
...
C
...

P
Q
i
...
, __ = A and __ = B
H
H

which gives P = A
...
H

P + Q = AH + BH and P – Q = AH –BH
= (A+B)
...
H
Thus we can write P ± Q = (A ± B)
...
AH and NQ = N
...
C
...
of P and Q is also the H
...
F
...

i
...
we have to select M and N in such a way that highest powers and lowest
powers (or independent terms) are removed and H
...
F appears as we have
seen in the examples
...
C
...
in each of the following cases using Vedic sutras:
1
...


x3 – 3x2 – 4x + 12,x3 – 7x2 + 16x - 12

3
...


6x4 – 11x3 +16x2 – 22x + 8,
6x4 – 11x3 – 8x2 + 22x – 8
...
Vilokanam
The Sutra 'Vilokanam' means 'Observation'
...
But we follow the same conventional
procedure and obtain the solution
...

Let us take the equation x + ( 1/x ) = 5/2 Without noticing the logic in the
problem, the conventional process tends us to solve the problem in the
following way
...
e
...


Consider some examples
...
C
...

simplification and factorization
...


Example 3:
5x + 9
5x – 9
82
_____ + _____ = 2 ___
5x - 9
5x + 9
319

107

At first sight it seems to a difficult problem
...
e
...


We follow in the conventional way that
(x – y)2 = (x + y)2 – 4xy =92 – 4 (14) = 81 - 56 = 25
x – y = √ 25 = ± 5

x + y = 9 gives 7 + y = 9
y = 9 – 7 = 2
...

But by Vilokanam, xy = 14 gives x = 2, y = 7 or x = 7, y = 2 and these two
sets satisfy x + y = 9 since 2 + 7 = 9 or 7 + 2 = 9
...

Example 2:

5x – y = 7 and xy = 6
...

Observe that x = 6, y = 1; x = 1, y = 6: are not solutions because they do not
satisfy the equation 5x – y = 7
...

Hence x = 2, y = 3 is a solution
...

Hence it is not a solution
...
Thus one set of the
solutions i
...
, x = 2, y = 3 can be found
...

i
...
, x = -3 / 5, y = -10
...


2x + 7
____________
(x + 3)(x + 4)

=

A
______
(x + 3)

B
+ ______
(x + 4)

A (x + 4) + B (x + 3)
= __________________
(x + 3) (x + 4)

109

2x + 7

≡ A (x + 4) + B (x+ 3)
...
(i) X 3
Independent of x : 4A + 3B = 7
...
e
...

2 (-3) + 7 ≡ A (-3 + 4) + B (-3 + 3)
1 = A (1)

...

2 (- 4) + 7 = A (-4 + 4) + B (-4 + 3)
-1 = B(-1)
...

2x + 7
____________
(x + 3) (x + 4)

=

1
1
_____ + _____
(x +3)
(x + 4)

2x + 7
But by Vilokanam ____________
(x + 3) (x + 4)
(x + 3) + (x + 4) =2x +7,

can be resolved as

directly we write the answer
...
Solve the following by mere observation i
...
vilokanam
1
...

1
25
x + __ = __
x
12

1
x - __ =
x

3
...

x+7
x+9
32
____ - ____ = ___
x+9
x+7
63

111

5
__
6

II
...

1
...


x + y = 7, xy = 10

3
...


x + y = 4,x2 + xy + 4x = 24
...
Resolve the following into partial fractions
...

2x - 5
____________
(x – 2) (x – 3)
2
...

x – 13
__________
x2 - 2x - 15
4
...
Gunita Samuccayah : Samuccaya Gunitah
In connection with factorization of quadratic expressions a sub-Sutra, viz
...
It is intended for the
purpose of verifying the correctness of obtained answers in multiplications,
divisions and factorizations
...

Example 2:

(x – 4) (2x + 5) = 2x2 – 3x – 20

Sc of the product 2 – 3 – 20 = - 21
Product of the Sc = (1 – 4) (2 + 5) = (-3) (7) = - 21
...

In case of cubics, biquadratics also the same rule applies
...
Verified
...
e
...
e
...


We apply and interpret So and Sc as sum of the coefficients of the odd powers
and sum of the coefficients of the even powers and derive that So = Sc gives (x
+ 1) is a factor for thee concerned expression in the variable x
...


113

Verify whether the following factorization of the expressions are
correct or not by the Vedic check:
i
...
Gunita
...


(2x + 3) (x – 2) = 2x2 – x - 6

2
...


12x2 + 13x – 4 = ( 3x – 4 ) ( 4x + 1 )

4
...


( x + 2 ) ( x + 3 ) ( x + 8 ) = x3 +13x2 + 44x + 48

So far we have considered a majority of the upa-sutras as mentioned in the
Vedic mathematics book
...
They are
2) S’ISYATE S’ESASAMJ ÑAH
4) KEVALAIH SAPTAKAMGUNYAT
5) VESTANAM
6) YAVADŨNAM TAVADŨNAM and
10) SAMUCCAYAGUNITAH already find place in respective places
...
They are mentioned in the
Vedic Mathematics text also
...
We shall also discuss them at appropriate places, with these
three included, the total number of upa-Sutras comes to sixteen
...
Further we think how 'the element of choice in the Vedic
system, even of innovation, together with mental approach, brings a new
dimension to the study and practice of Mathematics
...

K
...
Williams, London)
...
In this approach we
have missed to note some points and merits of one method over the other
methods at some instances
...
You may question why this book first
gives examples and methods and then once again try to proceed as if an
introduction to the Vedic Mathematics has been just started
...
This is clear from the examples given
...

Question some body showing the symbol Π
...

Very few may state it is a function or so
...
A girl thinking of set notation simply
says that it is Cartesian product of the sets A and B
...

Another may conclude that it is a product of two matrices A and B
...

Some other may go deep in to elementary number theory and may take ' X ' to
be the symbol ' X ' (does not divide) and conclude 'A does not divide B'
Now the question arises does a student fail to understand and apply contextual
meaning and representation of symbols and such forms in mathematical
writings? certainly not
...

Again a careful observation brings all of us to a conclusion that even though the
Sutras are not like mathematical formulae so as to fit in any context under
consideration but they are intended to recognize the pattern in the problems
and suggest procedures to solve
...

115

Terms and Operations
a) Ekadhika means ‘one more’
e
...
g:

Ekanyuna of 1 2 3
...
14
...

is
0 1 2
...
13
...


c) Purak means ‘ complement’
e
...
8
...
2 1

d) Rekhank means ‘a digit with a bar on its top’
...

_
e
...
It is called rekhank 7 or bar 7
...

_
_
e
...

__
-21 can be shown as 21
...

Adding a bar-digit i
...
Rekhank to a digit means the digit is subtracted
...
g: 3 + 1 = 2, 5 + 2 = 3, 4 + 4 = 0

116

Subtracting a bar - digit i
...
Rekhank to a digit means the digit is added
...
g: 4 - 1 = 5, 6 - 2 = 8, 3 - 3 = 6
1
...
g: 3 + 4 = 7
_
_ _
(-2) + (-5) = 2 + 5 = 7 or -7
f) Multiplication and Division using rekhank
...
Product of two positive digits or two negative digits ( Rekhanks )
_
_
e
...
e
...
Product of one positive digit and one Rekhank
_
_
_
__
e
...
e
...

3
...

_
_
e
...
e
...
Division of a positive by a Rekhank or vice versa
...
g: 10 ÷ 5 = 2, 6 ÷ 2 = 3 i
...
always negative or Rekhank
...
If the
addition is a two digit number, Thenthese two digits are also to be added up to
get a single digit
...
g: Beejank of 27 is 2 + 7 = 9
...
i
...
6 is Beejank
...
e
...


117

19

1+9

1

ii) Easy way of finding Beejank:
Beejank is unaffected if 9 is added to or subtracted from the number
...

eg 1: Find the Beejank of 632174
...

Hence remaining 1 + 4
5 is the beejank of 632174
...


But we can cancel 1& 8, 2& 7, 5 & 4 because in each case the sum is 9
...

h) Check by Beejank method:
The Vedic sutra - Gunita Samuccayah gives ‘the whole product is same’
...
It means that the operations carried
out with the numbers have same effect when the same operations are carried
out with their Beejanks
...

i) 42 + 39
Beejanks of 42 and 39 are respectively 4 + 2 = 6 and 3 + 9 = 12 and 1+2=3
Now 6 + 3 = 9 is the Beejank of the sum of the two numbers
Further 42 + 39 = 81
...

we have checked the correctness
...

64
125

6+4

10

1+2+5

1+0
8
118

1

Sum of these Beejanks 8 + 1 = 9
Note that
64 + 125 = 189

1+8+9

18

1+8

9

we have checked the correctness
...

v) 24 X 16 = 384
Multiplication of Beejanks of
24 and 16 is 6 X 7 = 42
Beejank of 384

3+8+4

4+2

6

15

1+5

6

12

1+2

3

Hence verified
...


Beejank of 18

1+8

9

Product of the Beejanks = 3 X 9
Beejank of 4266

27

4+2+6+6

2+7
18

9

1+8

9

Hence verified
...

viii) 3562 = 126736
Beejank of 356

3+5+6

Square of it = 52 = 25

2+ 5

Beejank of result 126736
(

5
7

1 + 2 + 6 +7 + 3 + 6

1+6

7

7 + 2 = 9; 6 + 3 = 9) hence verified
...

Further the relationship between them is P = ( Q X D ) + R
eg 1: 187 ÷ 5
we know that 187 = ( 37 X 5 ) + 2 now the Beejank check
...

187

1+8+7

(37 X 5) + 2
5+2

7(

1 + 8 = 9)

Beejank [(3 + 7) X 5] + 2
7
120

Hence verified
...


Check the following results by Beejank method
1
...
4381 - 3216 = 1165

3
...
(1234)2 = 1522756

5
...
2556 ÷ 127 gives Q =20, R = 16

i) Vinculum : The numbers which by presentation contains both positive and
negative digits are called vinculum numbers
...

We obtain them by converting the digits which are 5 and above 5 or less than 5
without changing the value of that number
...
(Note it is greater than 5)
...
It is 2 in this case and add 1 to the left (i
...
tens place) of
8
...

The number 1 contains both positive and negative digits
121

_
_
i
...
1 and 2
...

_
Thus 12 = 10 - 2 = 8
Conveniently we can think and write in the following way

General Number

Conversion

6

10 - 4

97

100 - 3

289

300 - 11

Vinculum
number
_
14
_
103
__
311

etc
...

eg 1: 289, Edadhika of 2 is 3
_
Nikhilam from 9 : 8 - 9 = -1 or 1
_
Charmam from 10 :9 -10 = -1 or 1
__
i
...
289 in vinculum form311
eg 2: 47768
‘Ekadhika’ of 4 is 5
___
‘Nikhilam’ from 9 (of 776) 223
_
‘Charmam from 10 (of 8) 2
____
Vinculum of 47168 is 5 2232
eg 3: 11276
...
We need not convert
...
e
...

The conversion can also be done by the sutra sankalana vyavakalanabhyam as
follows
...

sankalanam (addition) = 315+315 = 630
...

Samkalanam = 47768 + 47768 = 95536
Vyavakalanam = 95536 - 47768
...

eg 5: 12637
1
...
gives, 12637 + 12637 = 25274
__
25274 – 12637 = (2 – 1) / (5 – 2) / (2 – 6) / (7 – 3) / (4 – 7) = 13443

123

2
...

As in the number 1 2 6 3 7, the smaller and bigger digits (i
...
less than 5 and;
5, greater than 5) are mixed up, we split up in to groups and conversion is
made up as given below
...
" Now by ‘Ekadhika and Nikhilam
...

315 Since digits of 31 are less than 5,
We apply the sutras on 15 only as
Ekadhika of 1 is 2 and Charman of 5 is 5
...

eg 6:

24173

As both bigger and smaller numbers are mixed up we split the number 24173
as 24 and 173 and write their vinculums by Ekadhika and Nikhilam sutras as
_
__
24 = 36 and 173 = 227
_ __
Thus 24173 = 36227

Convert the following numbers into viniculum number by
i
...
Sankalana vyavakalana sutra
...

1
...
289

3
...
2879

5
...
123456799

8
...
823672

124

ii) Conversion of vinculum number into general numbers
...
Rekhanks are
converted by Nikhilam where as other digits by ‘Ekanyunena’ sutra
...
2 = 10 – 2
__
ii) 326 = (3 – 1) / (9 – 2) / (10 – 6)
= 274
_ _
iii) 3344 = (3 – 1) / (10 – 3) / (4 – 1) / (10 – 4)
= 2736
(note the split)
__ __
iv) 20340121 = 2/(0–1)/(9–3)/(10–4)/(0–1)/(9–1)/(10–2)/1
_ _
= 21661881
_
_
= 21 / 6 / 61 / 881
...

iii) Single to many conversions
...

eg 1: 86 can be expressed in following many ways
__
86 = 90 - 4 =94
__
= 100 – 14 = 114
125

___
= 1000 – 914 = 1914
_
__
___
____
Thus 86 = 94 = 114 = 1914 = 19914 = …………
...
etc
...

1) 274

2) 4898

3) 60725

4) 876129
...

Consider a number and find its Beejank
...
If both are same the conversion is correct
...

_
196 = 216
...


1+6

7

Thus verified
...
e
...
Then it is to be converted to +ve number by adding 9 to Rekhank (

126

already we have practised) and hence 9 is taken as zero, or vice versa in finding
Beejank
...


Use Vedic check method of the verification of the following result
...
24 = 36
2
...

__
_ _
3
...
23213 = 17187

Addition and subtraction using vinculum numbers
...
e
...

i) Change the numbers as vinculum numbers as per rules already discussed
...


iii) add the digits of the next higher level i
...
,, 1 + 1 = 2
_
13
_
14
127

____
_
27
iv) the obtained answer is to be normalized as per rules already explained
...

_
i
...
, 27 = (2 - 1) (10- 7) = 13 Thus we get 7 + 6 = 13
...

_
973 = 1 0 3 3
_ __
866 = 1 1 3 4

_
1033
___
1134
______
___
2161

___
But 2161 = 2000 - 161 = 1839
...

Observe that in this representation the need to carry over from the previous
digit to the next higher level is almost not required
...

i
...
,, 4247
-1828
______
____
Step (i) : write –1828 in Bar form i
...
,, 1828
____
(ii) : Now we can add 4247 and 1828 i
...
,,
4247
____
+1828
_______
_ _
3621
_ _
_
_ _
_
since 7 + 8 = 1, 4 + 2 = 2, 2 + 8 =6, 4 + 1 = 3
_ _
(iii) Changing the answer 3621 into normal form using Nikhilam, we get
_ _
_
_
128

3621 = 36 / 21 split
= (3 –1) / (10 – 6) / (2 – 1) / (10 – 1) = 2419
4247 – 1828 = 2419

Find the following results using Vedic methods and check them
1) 284 + 257

2) 5224 + 6127

3) 582 - 464

4) 3804 - 2612

129

2
...

234 + 403 + 564 + 721
write as 234
403
564
721
Step (i): 4 + 3 + 4 + 1 = 12

2 retained and 1 is carried over to left
...
e
...

step (iii): 2 + 4 + 5 + 7 = 18

carried over ‘1’ is added

i
...
, 18 + 1 = 19
...

thus

234
403
564
+721
_____
1922 is the answer

we follow sudhikaran process Recall ‘sudha’ i
...
, dot (
...

(a) 1 + 4 = 5, 5 + 3 = 8, 8 + 4 = 12 The final result is more than 9
...
e
...
We say at this stage sudha and a
dot is above the top 4
...

4
3
4
1
__
2
b) Before coming to column (2) addition, the number of dots are to be counted, This shall
be added to the bottom number of column (2) and we proceed as above
...

3
0
6
2
__
2

dot=1,

1+2=3
3+6=9
9+0=9
9 + 3 = 12

2 retained and ‘
...



...

v) 3+4=7,7+2=9 Retain 9 in 3rd digit i
...
,in 100th place
...
Here it is 1 only and the number is carried out left
side ie
...

Thus

234
403


...

Though it appears to follow the conventional procedure, a careful observation and practice
gives its special use
...

437


...

586
+162
______
1809
Steps 1:
i) 2 + 6 = 8, 8 + 4 = 12 so a dot on 4 and 2 + 7 = 9 the answer retained under column (i)
ii) One dot from column (i) treated as 1, is carried over to column (ii),
thus 1 + 6 = 7, 7 + 8 = 15 A' dot’; is placed on 8 for the 1 in 15 and the 5 in 15 is added to
2 above
...
e
...

(iii) The number of dots counted in column (iii) are 2
...
Thus 8 is placed under column (iii)
...
It is ‘1’ only
...
As there is no fourth column 1 is the answer for 4th column
...

Example 3:

Check the result verify these steps with the procedure mentioned above
...
e
...


Step 2: 2+2 (

2 dots) = 4; 4+9 = 13: 1 dot and 3+0= 3; 3+8 = 11;

1 dot and 1 answer under second column - total 2 dots
...

Step 4: 4 + 2 (

2 dots ) = 6; 6 + 5 =11:

1 dot and 1+3 = 4; 4+2 = 6
...

Step 5: 1 dot in the fourth column carried over to 5th column (No digits in it) as 1
Thus answer is from Step5 to Step1;16512
Example 2:

Steps
(i): 8 + 9 = 17; 7 + 4 = 11; 1 + 1 = (2)

(2dots)

(ii): 7 + 2 = 9; 9 + 1 = 10; 0 + 8 = 8, 8 + 9 = 17, (7)

(2dots)

(iii): 2 + 2 = 4; 4 + 6 = 10; 0 + 0 = 0; 0 + 7 = (7)

(1 dot)

(iv): 3 + 1 = 4; 4 + 4 = 8; 8 + 3 = 11; 1 + 1 = (2)

(1 dot)

(v): 1

133

Thus answer is 12772
...

1
...

5432
3691
4808
+6787
‾‾‾‾‾‾
‾‾‾‾‾‾

3
...
If not write the correct process
...


SUBTRACTION:
The ‘Sudha’ Sutra is applicable where the larger digit is to be subtracted from the smaller
digit
...

Procedure:
i) If the digit to be subtracted is larger, a dot ( sudha ) is given to its left
...


134

Example (i): 34 - 18
34


...

Steps: (i): Since 8>4, a dot is put on its left i
...
1
(ii) Purak of 8 i
...
2 is added to the upper digit i
...
4
_
2 + 4 = 6
...
e
...
e
...

Now at the tens place a dot (means1) makes the ‘1’ in the number into 1+1=2
...
i
...
3 - 2 = 1
...

-18
_____
16
Example 2:
63


...

Steps: (i) 7>3
...
e
...
e
...
e
...

This is unit place of the answer
...

Example (3) :
3274


...
No sudha
...

(ii) 9 > 7 sudha required
...
e
...
e
...

(iv) Now means 8 + 1 = 9
...

(v) As 9 > 2, once again the same process: dot on left of i
...
,1
(vi) purak of 9 i
...
1, added to upper 2 gives1 + 2 = 3, the third digit
...

(vii) Now 1 means 1+1 = 2
(viii) As 2 < 3, we have 3-2 = 1, the fourth digit
Thus answer is 1382
Vedic Check :
Eg (i) in addition : 437 + 624 + 586 + 162 = 1809
...
(3) in subtraction :
3274 – 1892 = 1382
now beejanks
3274

3+2+7+4

3+4

1892

1+8+9+2

2

3292-1892
1382

7-2

7

5

1+3+8+2

5 Hence verified
...

In the conventional method we first add all the +ve terms
423 + 847 + 204 = 1474
Next we add all negative terms
- 654 - 126 = -780
At the end their difference is taken
1474 - 780 = 694
Thus in 3 steps we complete the problem
But in Vedic method using Rekhank we write and directly find the answer
...


Example (2):
6371 – 2647 + 8096 – 7381 + 1234
____
____

137

= 6371 + 2647 + 8096 + 7381 + 1234
_
_
_
_
_
_
_
_
=(6+2+8+7+1)/(3+6+0+3+2)/(7+4+9+8+3)/(1+7+6+1+4)
_
=6/4/7/3
= (6 – 1) / (10 – 4)/ 73
= 5673

* Find the results in the following cases using Vedic methods
...


138

3
...
Let us recall
them
...

Example (i) : Find the square of 195
...

The right side part is 52 where as the left side part 19 X (19+1) (Ekhadhikena)
Thus 1952 = 19 X 20/52 = 380/25 = 38025
(iii) By Nikhilam Navatascaramam Dasatah; as the number is far from base 100, we
combine the sutra with the upa-sutra ‘anurupyena’ and proceed by taking working base
200
...

Now 1952 = 195 X 195

iv) By the sutras "yavadunam tavadunikritya vargamca yojayet" and "anurupyena"
1952, base 200 treated as 2 X 100 deficit is 5
...

vi) Now "urdhva-tiryagbhyam" gives

By the carryovers the answer is 38025

Example 2 : 98 X 92
i) ‘Nikhilam’ sutra
98 -2
x 92 -8
______________
90 / 16 = 9016
ii) ‘Antyayordasakepi’ and ‘Ekadhikena’ sutras
...

98 X 92 = 9 X ( 9 + 1) / 8X2 = 90/16 = 9016
...

1) ‘Nikhilam’ Method and ‘Anurupyena’:
a) Working base is 500, treated as 5 X 100

b) Working base is 500, treated as 1000 / 2
493

-7

141

497 -3
_________
2) 490 / 021
_________
245 / 021

= 245021

2) ‘Urdhva tiryak’ sutra
...
Further Ekadhikena Sutra is also applicable
...

_
493 = 500 –07 = 507
_
497 = 500 –03 = 503
...

99
X 99
_______
8121
168
_______
9801
2) By vinculum method
_
99 = 100 - 1 = 101
Now 99 X 99 is
_
101
_
x 101
______
_
10201

= 9801

3) By Nikhilam method
99 -1
99 -1
_________
98 / 01 = 9801
...

In the above examples we have observed how in more than one way problems can be
solved and also the variety
...
Not only
that which method suits well for easier and quicker calculations
...


143

4
...

Quotient
_______
Divisor ) Dividend
------------------_________
Remainder

or

Divisor ) Dividend ( Quotient
------------------_________
Remainder

But in the Vedic process, the format is
Divisor ) Dividend
--------------__________________
Quotient / Remainder
The conventional method is always the same irrespective of the divisor
...

Example 1: Consider the division 1235 ÷ 89
...

_____
78
ii) Nikhilam method:
This method is useful when the divisor is nearer and less than the base
...
Let us recall the nikhilam division already dealt
...
Obtain the modified divisor
(M
...
) applying the Nikhilam formula
...
D
...

Thus for the divisor 89, the M
...
obtained by using Nikhilam is 11 in the last from 10 and
the rest from 9
...
H
...
D
...
D
...
D
...
Here it is 1
...
e
...
Write this product place wise under the 2nd and 3rd columns of the dividend
...
D
...
e
...
Write the digits of this result column wise as shown below, under 3rd and 4th
columns
...
e
...

Example 2: Find Q and R for 121134 ÷ 8988
...

iii) Paravartya method: Recall that this method is suitable when the divisor is nearer but
more than the base
...

The divisor has 4 digits
...


Now the remainder contains -19, -12 i
...
negative quantities
...

Take 1 over from the quotient column i
...
1x1028 = 1028 over to the right side and
proceed thus: 32 - 1 = 31 becomes the Q and R = 1028+200 - 190 - 12 =1028-2 =1026
...

The same problem can be presented or thought of in any one of the following forms
...
The word Dhvjanka means " on the top of the flag"
Example 4: 43852 ÷ 54
...
Separate the dividend into two parts where the right part
has one digit
...
The representation is as follows
...
Now Q= 8 and R = 3
...
e
...

Now 38 becomes the gross-dividend ( G
...
) for the next step
...
D
...
e
...
This is the net - dividend (N
...

Step3: Now N
...
6 ÷ 5, Q = 1, R = 1
...

4:43 8 5:2
5 :
3 1
________________
: 8 1
Step4: Now G
...
D=15-4=11 divide N
...
The representation is
4:43 8 5:2
5 :
3 1 :1
________________
: 8 1 2:
Step5: Now the R
...
S part has to be considered
...
e
...
Thus the division ends into
4:43 8 5:2
5 :
3 1 :1
________________
: 8 1 2:4
Thus 43852 ÷ 54 gives Q = 812 and R = 4
...
The divisor 54 can be represented by
5x+4, where x=10
The dividend 43852 can be written algebraically as 43x3 + 8x2 + 5x + 2
since x3 = 103 = 1000, x2 = 102 = 100
...

5x + 4 ) 43x3+ 8x2 + 5x + 2 ( 8x2 + x + 2
43x3+ 32x2
_________________
3x3 – 24x2
= 6x2 + 5x
( 3x3 = 3
...
x2
2
5x + 4x
= 3
...
x = 10x )
10x + 8
__________________
x–6
= 10 – 6
= 4
...
43x3 ÷ 5x gives first quotient term 8x2 , remainder =3x3 - 24x2 which really mean 30x2 +
8x2 -32x2 = 6x2
...
D = 6
...
6x2 ÷ 5x gives second quotient term x, remainder = x2 + x which really mean 10x + x =
11x
...
D =11
...
11x ÷ 5x gives third quotient term 2, remainder = x - 6 , which really mean the final
remainder 10-6=4
...
We now seperate the dividend into two parts where the RHS part
contains two digits for Remainder
...
D = 37
...
D is obtained as

= 37 – ( 8 + 0)
= 29
...
D ÷ Operator = 29 ÷ 5 gives Q = 5, R = 4 and G
...

ii) N
...

i
...
,
24 : 2 3 7 9 : 63
5 :
3 4 :
_________________
: 4 5
:
Step 4:
i) N
...
D = 363
...

24 : 2 3 7 9 : 63
5 :
3 4 :3
_________________
: 4 5 4 :
Step 5: We find the final remainder as follows
...


150

i
...
,,

Note that 2, 4 are two falg digits: 5, 4 are two last quotient digits:

represents the last flag - digit and last quotient digit
...

Thus the Vedic process of division which is also called as Straight division is a simple
application of urdhva-tiryak together with dhvajanka
...


5
...
Straight Squaring:
We have already noticed methods useful to find out squares of numbers
...
Now we go to a more general formula
...

They are i) by squaring ii) by cross-multiplying
...
We define for a single digit ‘a’, D =a2
...
If it is a 3 digit number like ‘abc’,
D =2( a x c ) + b2
...
i
...
if the digit is single
central digit, D represents ‘square’: and for the case of an even number of digits equidistant
from the two ends D represent the double of the cross- product
...
Thus in this process, we take extra dots to the left one less than the number of digits
in the given numbers
...
Thus

...

Examples:2 2342 Number of digits = 3
...
234
_____
42546
1221
_____
54756

for 4, D = 42 = 16
for 34, D = 2 x 3 x 4 = 24
for 234, D = 2 x 2 x 4 + 32 = 25
for
...
0
...
2
...
234, D = 2
...
4 + 2
...
3 + 22 = 4

Examples:3 14262
...
e

...
2
...
4
...
1
...
4
...
1426, D = 2
...
6 + 2
...
2 + 42 = 20

...
0
...
0
...
1
...
1426, D = 12 = 1
Thus 14262 = 2033476
...

Algebraic Proof:
Consider the first example 622
Now 622 = (6 x 10 + 2)2 =(10a + b)2 where a = 6, b = 2
= 100a2 + 2
...
b + b2
= a2 (100) + 2ab (10) + b2
i
...
b2 in the unit place, 2ab in the 10th place and a2 in the100th place i
...
22 = 4 in units
place, 2
...
2 = 24 in the10th place (4 in the 10th place and with carried over to 100th place)
...

Thus the answer 3844
...

Applying the Vedic sutra Dwanda yoga
...
CUBING
Take a two digit number say 14
...
e
...
e
...
e
...
e
...


iv) Write twice the values of 2nd and 3rd terms under the terms respectively in second row
...
e
...

1

4 16 64 Since 16 + 32 + 6 (carryover) = 54
8 32
4 written and 5 (carryover) + 4 + 8 = 17
______________
2 7
4
4
7 written and 1 (carryover) + 1 = 2
...

Consider the row a3
a2b ab2
b3
3
the first isa and the numbers are in the ratio a:b
since a3:a2b=a2b:b3=a:b
Now twice of a2b, ab2 are 2a2b, 2ab2
a3 + a2b + ab2 + b3
2a2b + 2ab2

________________________________

a3 +3a2b + 3ab2 + b3 = (a + b)3
...
Now cubing can be done by using the vedic sutra ‘Yavadunam’
...


i) The base is 100 and excess is 6
...

i
...
106 + 12 = 118
...

i
...
1063 = 118 / - - - ii) Multiply the new excess by the initial excess
i
...
18 x 6 = 108 (excess of 118 is 18)
Now this forms the middle portion of the product of course 1 is carried over, 08 in the
middle
...
e
...

i
...
63 = 216
...

i
...
1063 = 118 / 081 /16 = 1191016
1
2
Example 4: Find 10023
...
Excess = 2
...


155

ii) New excess x initial excess = 6 x 2 = 12
...

iii) Cube of initial excess = 23 = 8
...

Thus 10023 = 1006 / 012 / 008 = 1006012008
...

i) Base = 100, deficit = -6
...

ii) New deficit x initial deficit = -(100-82)x(-6)=-18x-6=108
Thus middle potion of the cube = 08 and 1 is carried over
...

Find the cubes of the following numbers using Vedic sutras
...


3
...

1
...

It is passing through (x1, y1) theny1 = mx1 + c
...

Solving these two simultaneous equations, we get ‘m’ and ‘c’ and so the equation
...
The formula
(y2 - y1)
y – y1 = ________ (x – x1)
(x2- x1)

156

and substitution
...
But the paravartya sutra enables us to arrive at
the conclusion in a more easy way and convenient to work mentally
...

Step1: Put the difference of the y - coordinates as the x - coefficient and vice - versa
...
e
...

Thus L
...
S of equation is 5x - 4y
...
H
...
H
...
e
...
H
...

Thus the equation is 5x - 4y = 17
...

Step 1 : x[-3-(-7)] –y[2-4] = 4x + 2y
...

Step 3 : Equation is 4x + 2y =2 or 2x +y = 1
...

Step 1 : x[9 - (-7)] – y(7 - 3) = 16x - 4y
...

1
...
(5,-2), (5,-4)

3
...


(a, o) , (o,b)

157

IV Conclusion

After going through the content presented in this book, you may, perhaps, have noted a
number of applications of methods of Vedic Mathematics
...
The methods discussed, and
organization of the content here are intended for any reader with some basic
mathematical background
...

factorization, H
...
F, recurring decimals, etc are dealt with
...

Thus the present volume serves as only an 'introduction'
...
Still more steps are needed to touch
the latest developments in Vedic Mathematics
...
Sri
Sathya Sai Veda Pratisthan intends to bring about more volumes covering the aspects
now left over, and also elaborating the content of Vedic Mathematics
...
Further it has given rationale
and proof for the methods
...
An impartial reader can easily experience the beauty, charm and
resourcefulness in Vedic Mathematics systems
...
The reader can also compare and contrast both the
methods
...


158



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