vedic mathematics for simple calculation in mathematics

1. VEDIC METHEMATICS
BY
PROF.PAVAN KUMAR UIKEY

2. CONTENTS
Cube-Anurupyena
Cube-Ekadhikena
Cube-Vargank
Cube-Yavadhunum
CubeRoot-
Yavadhunum
Division-Nikhilum
Division- Paravartya
Equation-Vilokanam
Equation one
variable
Factorization 3
variables
HCF
Multiplication Adhva
Multiplication
Antyoradasakepi
Multiplication
Anurupeyna
 Multiplication Different
Sutra
 Multiplication Eknunena
 Multiplication Urdhva
Multiplication viloknum
Equation Quadratic
Equation Simple
Equation special
simultaneous
Square Dwandva
Square Ekadhikena
Square Sankalana
Square-
Yawadhunum
Square Root
Vilokanum
Vinculum Numbers

3. ANURUPYENA [CUBE]
Here we can find the cube of any two digits no. by anurupyena sutra. The answer is consists of four
parts from right to left starting from unit place digit , tenth place digit , hundredth place digit…….as
follows:
(I) Left hand most part containing cube of tenth place digit.
(II) Second part contain:
=3[(tenth place digit)^2 × (Unite place digit)]
(III) Third part contain :
=3[tenth place digit × (unit place digits)^2]
(IV) Right hand most part contains:
=(Unit place digits)^3
The above sutra is similar sutra is similar to the identity:-
( For calculation additional digits classified over to the left )

4. EKADHIKENA PURVENA.[CUBE]
By this sutra we can find the cube of two digits no. which are having the digits five in the end.
Procedure : The answer consists of four parts from left to right :
1) Square of second digit × Ekadhik
2) Second digit × Ekadhik × First digit
3) Second digit × square of first digit
4) Cube of first digit
Example: Find (35)^3 by the sutra
Solution : The given no. is 35, first digit= 5 , second digit = 3
Ekadhik 3 is 4
By the sutra we have
(35)^3 = 3^2 × 4 / 3×4×5/ 3×5^2/ 5^3
= 36 / 60 /75 /125
= 42/8/7/5
= 42875

5. ANURUPYENA+ YAVADUNUM
TAVADUNIKRITYA VARGANKA[CUBE]
By the sutra we can find the cube of no. which are nearest to the sub-base
, then
Example ; Find (205)^3 by the sutra
Solution :- the given no. is 205 , base = 100 , sub base =200
=
Deviation=(surplus) = 05
put the values in above sutra
= 860 /150 /125
=861/51/25
=8615125

6. YAVADUNUM TAVADUNI KRITYA
VARGANKA YOGYET[CUBE]
By this method we can find cube of no. which are nearest to the base. The answer will be in three parts as follow :
LHS / Middle / RHS.
LHS.= BASE + 3( DEVIATION ).
For calculation in RHS and middle part :-
1) The no. of digits must be equal to the no. zeroes of the base.
2) If the no. digits are less than the no. of zeroes of the base , then put zero or zeroes in the left side of digits.
3) If the no. of digits are more than the no. of zero or zeroes of the base , then additional digit or digits should be shifted from RHS part to
middle part or from middle part to LHS part.
4) Check the result by beejank method also.

7. CUBE ROOT
[VILOKANUM METHOD]
The cube root of perfect cubic number of six or less than six digits can be find by vilokanum method
TABLE – 1 TABLE – 2 TABLE - 3

8. NIKHILUM NAVATASH –
CHARMUM DASHTAH,[DIVISION]
The Nikhilum sutra states that “All from nine and last from ten”.
The process of division can be done by Nikhilum sutra to find the Quotient and
remainder ; when divisor is less than the base or nearest to the base :-
TABLEL FORMATION
Make a table of three parts from left to right by vertical lines also draw a horizontal lines
below it
FIRST PART:-
Write the deviser and its M.D. complement and converted digit below it
M.D. means modified – divisor or divisor’s complement or Nikhilum complement from 10 or
Difference from base. Also setup the nearest base of the divisor and its complement . The
complements of numbers 1,2,3,4,5 are 9,8,7,6,5 respectively.
SECOND AND THIRD PART:-
Split dividend into LHS part for the Quotient column and RHS part for
Remainder column as the number of zeroes of the base.

9. • Nikhilum Procedure:-
• (1) Put the first digit of dividend from left under the second digit of quotient .
• (2) By adding the first digit of the dividend remains as the first digit of the quotient .
• (3) The first digit of the dividend is added to the second digit of dividend, to obtain the second digit of quotient .
• (4) Add second digit of the quotient, to the third digit , to obtain the next digit of the quotient .
• (5) Add the third digit of the quotient , to the last digit of the dividend thus sets the final remainder .
• NIKHILUM CALCULTION:-
• (1) If the number of digits are less than the number of zeroes of the base, than put zeroes at the left side of digits.
• (2) If the number of digits are greater than the number of zeroes of the base , than additional digit or digits should be
shifted from right hand side part to middle part or from middle part to LHS part.
• (3) The beejank of Dividend is always equal to the remainder.
• [A] WHEN THE REMAINDER IS LESS THAN THE DIVISOR :-
(1)Division of two digits number by 9
– Ex.01) Find 12÷9 by Nikhilum. Ex.2 )Find 512 ÷ 9 by NIKHILUM division.
– Solution;- Here Dividend = 12, Solution.;- Dividend = 512,divisor = 9 , base = 10 , complement
= 1,
– Divisor = 9 , base = 10 . by dividing 512 by 9 , we get
by dividing 12 by 9 , we get Q = 1 ,R = 3 Q=1, and R= 3,
– By NIKHILUM SUTRA we get :- by NIKHILUM DIVISION we have
– since base is 10 , one zero.
– Thus one digits in the Remainder column. Put LHS digits
– under the RHS digit we have .by adding column wise 5+1 = 1
– 2+6 = 8
Hence Q=56, R = 8,
BEEJANK of dividend 512 is 8 equal to Remainder 8 .
–
1 At it is 2+1=3
There for Q = 1, R=3.

10. [B]:- When the Remainder is equal or greater than Divisor. Ex, 5 )find 2342÷98 by NIKHILUM DIVISION ?
Ex. 3):- find 245 / 9 , base 10 by NIKHILUM DIVISION. Sol:- dividend = 2342, base = 100 , divisor =98 , complement = 02
SOL:- divisor = 9, complement = 1 by nikhilum division. By “Nikhilum division .”
4 + 2=6 2×02 =04
5 + 6=11
Remainder is greater than Divisor 3 × 02 = 06
Redividing 11 / 9 ,Q = 1 , R = 2 Hence Q = 23 , R = 88
26 + 1 = 27
,thus Q = 27 , R = 2
[c]When the Divisor is of two or more digits,
Ex. 4) :- Find 21422 ÷897 by NIKHILUM DIVISION ?
SOL:- Dividend = 21422
base 1000, divisor = 897. complement = 103.
2 × 103 = 206
3 × 103 = 306

11. PARAVARTYA YOJYET [DIVISION]
The sutra means “Transpose and apply.” or “change the sign and start the procedure”.
This method is suitable when the divisor is nearest but more than the base.
CALCULATION:-
(1) Multiply the first digits (1) of dividend 132 by -2 , we get
-2 × 1= -2. write the product(-2) bellow the second digits 3 and odd is
-2 × 1=-2 -2 + 3 =1.
-2+3 =1 (2) multiple the second digits sum (1) by -2 we get -2 × 1=-2
-2 × 1 =-2 and write the product (-2) under the third digits (2) of the third digits
2 – 2 = 0 2 of the dividend and odd we get -2+2=0
-2 × 0 = 0 (3) continue the process to the last digits , we get
Q=110 ,and R= 6.
Divisor column: divisor =12, converted =10
deviation =surplus =12-10= 2 . converted – digit with negative sign .= -2.
Remainder column : - split the dividend 1326 into two
Parts. 132/6 put the second part 6 in remainder column .here
number of digits must be equal to the number zero of base .
Quotient column : - write the first part or remaining digits
132 of the dividend

12. Ex.2):- find the 2596 ÷ 123 ,base = 100 paravatya yojyet.? Ex.3):- find the 13457 ÷ 1123 ,base =1000 by Paravartya Yojyet.?
SOI.- Her dividend = 2596 , divisor = 123 , base = 100,. Sol.- Here dividend = 13457 divisor = 1123, base = 1000 deviation (surplus
Deviation(surplus )=123 - 100 = 23 , converted digits with digits ) = 123 converted digit with negative sign = -1 -2 -3 .since base
negative sign = - 2 - 3 Since base has 2 zeroes so, last two digits 96 for has 3 zeroes so last 3 digits 457 should be in remainder column and
remainder column , and remaining two digits 25 for the Quotient. Remaining two digits 13 should be in Q Quotient column .
Calculation:-
1×(-1-2-3)= ̶ 1 – 2 – 3
2×(-1 -2 -3) = - 2 - 4 - 6
Here Remainder is -21
over come –negetive sign take
1 over from Quotient is 1123 to
remainder column thus 1123+21
= 1102, thus ,,
Q = 11 ...R = 1102
Calculation:-
2 × (-2-3)=-4,-6.
5 – 4 = 1
1(-2,-3) = - 2 – 3.,
thus Q = 21, R = 13.

13. VILOKANUM.
[EQUATION]
The sutra states that
“By overving ” or VILOKNAM.

14. SHUNYAM SAAMYA SAMUCHAYE.
[EQUATION OF ONE VARIABLE.]
The sutra states that ‘’when the sum is same, that sum is zero”. Or when samuchaya is same , the numerical
value of samuchaye is zero . Here we are discussing special equation of one variable having the power one.
First INTER PRETATION ;-
When both sides of eqathion having the same common factor x.
Ex. 1 ) solve 10x+5x=10x+4x.
Sol. The given equation has the same common factor x in all its term . Thus shunyam samuchaye:-
x=0
Ex.2 ) solve the equation 7(x+1) = 6 ( x+1 )
Sol:- The given equation has (x+1) as a common factor in both sides. Thus shunyam samuchaye is :-
x+1=0
x = -1
Ex.3. solvethe equation
Sol:- (2x - 3) is common.
Shunyam Samuchaya is :-
2x – 3 =0
x = 3/2
SECOND INTERPRETATION:-
‘‘As a product of independent terms in expression like (x+p) (x+q)
Ex.4.solve (x+3)(x+4)=(x+2)(x+6)
Sol:-The product of independent terms on both sides are equal . Hence samuchaye is :
3 × 4 = 2 × 6
since the answer on both sides are same . Shumyam Samuchay is :-
X = 0

15. THIRD INTERPRETATION :-
Samuchaye ‘‘As the sum of the
denominators of two fraction , having the same numerators.”
Ex.5.Solve
Sol. Since the numerators are same .i.e. 1 . Thus According to
Shunyam Samuchaye:-
Sum of the denominators = 0
(3x-1)+(2x-1)=0
x = 2/5
FOURTH INTERPRETATION:-
‘‘Samuchaye as combination or
total”.If the sum of the numeraters and sum of the denominaters
be same, then that sum = 0.
Ex.6.Solve
Sol. Let
N1 = 3x+5 , N2 = 3x+ 6.
D1 = 3x + 6, D2 = 3x+ 5.
According to Shunyam Samuchaye ,
Sum of the numerators :-
N1 +N2 = 3x+5+3x+6 =6x+11
Sum of the denominators:-
D1+D2 = 3x+6+3x+5 = 6x+11
Then N1 +N2 = D1+D2 gives
6x+11=0
x= -11/6
Now consider the examples like (N1+N 2 ) = k (D1+D2 )
Where k is any Constant k, Procead as above.
FFIFTH INTERPRETATION :-
‘‘Samuchaye as Combination or total”
leading to Quadratic Equations.
If N1+N2= D1+D2 and N1+D1 = N2-D2 Then both
are equated to Zero, thus solution gives the two
values of x.
Ex. 7. Solve
Sol. Let,
N1= 2x+7 , N2=3x+2,D1=3x+2,D2=2x+7.
According to Shunyam Samuchaye , we have
N1+N2=(2x+7)+(3x+20)=5x+9
D1+D2=(3x+2)+( 2x+ 7)=5x+9
Then N1+N2=D1+D2, gives
5x+9=0
x =-9/5
Father
N1 ̶ D1=(2x-7)-(3x-2)
= -(x-5)
and
N2-N2 = (3x+2)-(2x+7)
= x-5
According to Shunyam Samuchaye ,
(N1-D1)=(N2-D2) = -(x-5)=0
x = 5

16. SIXTH INTERPRETATION :-
‘‘Samuchaye as Combination or total”
With different application.
Ex. 8. Solve Ex.11. Solve:
=
Sol. Let, Sol. This can be solve by the idendtity . Now
D1= x-2 , D2= x-5,D3= x-3,D4= x-4. Sum of the terms within the cube on
According to Shunyam Samuchaye , we have LHS = x-149+x+147
D1+D2=(x-2)+(x-5)=2x-7 = 2(x-1)
D3+D4=(x-3)+( x-4)=2x-7 Sum of the terms within the cube on
Then D1+D2=D3+D4, gives RHS =2 (x-1).
2x-7=0 By ShunYam Samuchaye:
x =7/2 2(x-1)=0
x = 1
. Ex.9. Solve: Ex.12.Solve
Sol. This can be solve by the idendtity . Now Sol. Let N1 = x+3, N2=x+1
Sum of the terms within the cube on D1 =x+5, D2=x+7
LHS = x-2a+x-2b Within the cube.
= 2(x-a-b) Sum of numerator and denominator on
Sum of the terms within the cube on LHS , N1+D1 =(x+3)+(x+5) = 2(x+4)
RHS =2 (x-a-b) Sum of the numerator and denominator on
By ShunYam Samuchaye: RHS , N1+D2 = (x+1)+(x+7) = 2(x+4)
2(x-a-b)=0 By ShunYam Samuchaye
x = a+b. 2(x+4)=0
x = -2
.

17. The sutra state that ‘‘ By altarnate elimination and
retention”.
Consider the homogeneous Quadratic
equation :-
Of second degree in three variables x,y,z,
make the process of factorization easy by the
following steps .By Adyamadhyena sutra we
have
(1) Eliminate z by putting z=0, retain x and y
obtaind a quadratic equation:-
and factorize .
(2) Eliminate y , by putting y=0 retain x and z
obtain the quadratic equation :-
and factorize
(3) By three two sets of factors , fill in the gaps
caused by the eleminatio process of y and z
resp.
Ex.1. Factorization
Consider the homogeneous Quadratic
equation :-
by sutra.
Sol. By acrranging the equation
steps,
(1) Eleminate z by putting z=0, retain x and y
obtaind a quadratic equation:-
(x-y)(3x-y)----------(1)
(2) Eliminate y , by putting y=0 retain x and z
obtain the quadratic equation :-
(x-z)(3x+2z)---------(2)
(3) Fill in the gaps caused by adding we get
(x-y)(3x-y) + (x-z) (3x-2z)
(x-y-z)(3x-y+2z)
Note.(i) when (x-y) +(x-z) is taken we get (x-y-z)
where x is taken once
(ii) adding (3x-y)+(3x+2y),where 3x is taken
once

18. Ex.2 Factorization the expression
Sol. Steps
(1) Eliminate y , z by putting y=0, z=0
= by solving
= (x-4)(3x+5)-----------(i)
(2) Eliminate x, z by putting x=0 , z=0 .
= by solving
= (y+4)(4y+5)-----------(ii)
(3) Eliminate x , y by putting x=0, y=0
= by solving
= (-z+4)(z+5--------------(iii)
By adding (i)(ii)(iii) we get
= (x+y-z+4)(3x+4y+z+5)
NOTE-
(a) By adding (x+4)(y+4)(-z+4) we get (x+y-z+4) where 4 is taken once .
(b) By adding (3x+5)(4y+5)(z+5) we get (3x+4y+z+5) where 5 is taken once.

19. HIGHEST COMMON FACTOR (HCP)
The HCF is also known as Greatest Common
Multiple (GCM).By Lapana Sthapana and
Adhayamadhaya Sutra. We can find the HCF
as follows :-
Here Lapana – Sthapana Means
Samkalana(addition),
Vyavakalana (subtraction).

20. Ex.1. Find the H.C.F. Of
Sol. On multiplying by 2 the expression
we get
------------- (1)
By Vyavakalana (subtraction ) .
By Lapana we get
=
3 is omitted because it is common
By Sanskalana (addition),
+
6
=
here 6x is omitted becouse it is common.
Thus HCF is
= .
Ex.2. Find the H.C.F. Of
and
Sol.
By Sanskalana (addition),
+
By Lapana we get
=
2 is omitted because it is common
By Vyavakalana (subtraction ) .
=
Lapana.
here 4 is omitted becouse it is common.
Thus HCF is
= .

21. ADHAYAMADHYENTYAMANTYENA
[MULTIPLICATION]
Sutra Adhya madhyenantua mantyena means ‘‘The first
by the first and last by the last.”
Ex.1 In an electric bill the expenditure is 9 Rs. 10 P. Ex.2. The speed of a man is 4 km. 900 m.per hr.How much
Per hour. Find the expenditure in 6 hours 20 min. Distance will be travel in 6 hr. 20.min.
Sol. 9 Rs. 10 paise. Sol. 4 km. 900 m.
× 6 hrs. 20 min. × 6 hr. 20 m.
By adding there, we get By adding we get
54+0.6+3+0.03 24+1.33+54+0.3
Ans:- = 57.63 Rs . Hr Ans:- = 31.03

22. Ex.3. Find the area of a rectangular plate whose
length and Breath are reap. 7 ft. 4 inch and Both ratio are same Thus it gives first factor of
6 ft. 8 inch. Quadratic equation ,
Sol. 7’4’’ can be written as (7x+4) = (x+3)
6’8’’ can be written as (6x+8)
Where x = 1 .ft. = 12 inch. And
is sq.ft. Now
(7x+4)(6x+8) is
Thus second factor is
=(3x+5)
Ex.4 Fctorize the Quadratic equation Hance the factors of Quadratic equation
Sol. (1) Multiply the Co.-ef.of andConstant =(3x+5)(x+3).
3×15=45
(2) Find the factors of 45
9×5
(3) By adding or subtraction 9 and 5 to get
the Co.-ef. Of x is
9+5=14
(14) Thus first is 9 and second part is 5

23. ANTYORDASAKEPI.
[MULTIPLICATION.]
The sutra ‘‘Antyordasakepi” means, ‘The sum
of last digits is ten’.The sutra indicats that
numbers of which the last digits are added up
to give 10,100,100,........We can solve the
problems by EKADHIKENA PURVENA and
ANTYORDASAKEOPI.

24. Ex.1 Find the product of 401×409 by the sutra.
Sol.
(1) Sum of the last digits of both the number digits
=1+9
(2) Product of last digits of both numbers
= 1×9
(3) Same part or. Left part of the digits of both the
numbers = 40
(4) Ekadhikena of same part of both number
=(same part+1)
= 40+1
= 41
LHS = Part of same part and its Ekadhikena
= 40×41
RHS = Product of last digits
= 1×9
Acording to the sutra
LHS / RHS
= 40×41/1×9
= 1640 /09
= 164009
Ex.1 Find the product of 694×606 by the sutra.
Sol.
(1) Sum of the last digits of both the number digits
=94+06
(2) Product of last digits of both numbers
= 94×06
(3) Same part or. Left part of the digits of both the
numbers = 06
(4) Ekadhikena of same part of both number
=(same part+1)
= 06+1
= 07
LHS = Part of same part and its Ekadhikena
= 06×07
RHS = Product of last digits
= 94×06
Acording to the sutra
LHS / RHS
= 40×41/94×06
= 42/0564
= 420564

25. Ex.1 Find the product of 198×102 by the sutra.
Sol.
(1) Sum of the last digits of both the number digits
=98+02=100
(2) Product of last digits of both numbers
= 98×02=0196
(3) Same part or. Left part of the digits of both the
numbers = 1
(4) Ekadhikena of same part of both number
=(same part+1)
= 1+1
= 2
LHS = Part of same part and its Ekadhikena
= 1×2=2
RHS = Product of last digits
= 98×02
Acording to the sutra
LHS / RHS
= 1×2/98×02
= 2/0196
= 20196
Ex.1 Find the product of 392×308 by the sutra.
Sol.
(1) Sum of the last digits of both the number digits
=92+08
(2) Product of last digits of both numbers
= 92×08=0736.
(3) Same part or. Left part of the digits of both the
numbers = 3
(4) Ekadhikena of same part of both number
=(same part+1)
= 3+1
= 4
LHS = Part of same part and its Ekadhikena
= 3×4
RHS = Product of last digits
= 92×08=0736
Acording to the sutra
LHS / RHS
= 3×4/92×08
= 12/0736
= 120736.

26. ANU RUPYENA.
[MULTIPLICATION]
SutraAnurupen means ‘‘PRAPORTIONALITY’’.
For finding the products of two numbers, when
both of them are near to the common base, Here
,Deviation means surplus digits or digits or deficit
digits.
SUTRA:- [LHS / RSH].
LHS = (one number +deviation of second number )
× ratio.
RHS = (Product of deviation ).

27. Ex.1. Solve 63 × 68 Anurupyena Sutra.
Sol. Heae working base is 10 and sub base is 60, = L.H.S. Part /R.H.S. Part
because 60 is near to both numbers Thus = (45-07)or(43-5×5)/-5×-7
ratio = 38×5/35
Deviation :- = 190/35 { base =10
= 63/03 = 190+3/ 5
= 68/08 = 193/5
According to the sutra ; =1935.
= L.H.S. Part / R.H.S. Part Method (ii), When working base is 100 sub bade is 50:-
= [(63+8)×6/3×8 Then
= 71×6 / 24
= 426 / 24 Dificit digit of 45 is -5 ,and Dificit digits of 43 is -7.
Since working base is 10, then there one digits Then ,
on R.H.S. Is = L.H.S. Part / R.H.S. Part
= 426+2/4 ( one number +deviation of second number )
= 428 / 4 ×Ratio/Product of deviation.
= 4284 = (45-7)or(43-5)× / -5×-7
= 19/35 { base 100
Ex.2. Solve 45×43 by Anurupyena Sutra. =1935
Sol. Method (i) Here working base is 10 and sub base NOTE : (i), If RHS side contains more number of digits than
is 50 because 60 is near to both numbers -7 the number of zeroes in the base , then the excase digits or
digits are to be added to LHS of the asnswer.
Division :- (ii) If RHS contain less number of zeroes in the base,
= 45/- 05. then remaining digits are to be filled with zero on the left
= 43/- 07 side of RHS.
According to the Sutra ;

28. MULTIPLICATION BY DIFFERENT
VEDIC SUTRA
Multiplication by vedic sutra can
be done by any of the following
methods. Arurupyena ,Ekadhikena-
purvena ,Nikhilam Navatas caramun
Desatah. Antyordasa kepi,Urdhva-
Triyagbhyam,Vinculum,Yavadunum
Tavaduni-kritya vargaca Yojyet
Algebraic expressions.

29. Ex.1. Find the squaring of 185.
Sol. (i) By conventional method. = 2+340/25
= 342/25
(ii) By Nikhilum Navatas caramum Desath and = 34225
Anurupyena ,, (v) By Antyayar dasakep[i and Ekadhi kena
Working base =200
is .. 2 × 100 =342/25
185-15 = 34225
× 185-15 (vi) By Urdhav Triyagbhyam
(185-15)/15 × 15 Grup Analysis:-
= 2 × 170 / 225 (2 × 170).
= 342/25 = 340+2
= 34225 =342
(iii) By Ekadhikena Purvena..
Since the number ends up in 5 , the answer
is split up into parts . LHS part is 18 and RHS part 5.
1 8+8 5 + 5 +64 40+40 25
+ 2 16 74 80 +2 5
=342 / 25 +8 +8 2
=34225. 3 4 2
(iv) By Yavadunam tavaduni Kritya Varganca Yojyet
and Anurupyena. ⇛ 34225
Here working is 200 is 2 × 100 deficit is 15.
= (185-15)/
= 170/25
= 340/25

30. EKANYUNENA PURVENA.
[MULTIPLICATION]
The Ekanunena purvena sutra states that ‘‘ By one less (case -1) When the multiplier and multiplicand having the
than the previus one’’ eg.Ekanyunena of 4 is 3 or 4-1 =3. same number of digits .
Ex.1 . Find the answere of P×Q by Ekanunane Purvena . Ex.2. Solve 6×9 by the sutra .
1. Methed :- Sol. (i) LHS part of the answere gives 6-1= 5
(i) Here P is multiplicand digits,or Left hand side LHS digits. (ii) RHS part of the answere gives ,
Where Q is multiplier digits or Right hand side RHS digits . = 9- LHS part
(ii) LHS part of the answere , = 9-5=4
= deducting 1 from multiplicand. Thus LHS part / RHS part
= (P-1). = 5 / 4
(iii) RHS part of the answere = 54 .
= different between multiplier and LHS part Same Qutions:-
= Multiplier LHS part 1) 54 × 99 2) 355 × 999 .
= Q-(P-1) Ans.5346 Ans .354645.
Summation of two part given the sutra
⇛ LHS part / RHS part
⇛ (P-1) / [Q-(P-1)]
the following are multiplication by
9,99,999,9999,........................
Case (ii) When multiplier has more number of digits
than
the multiplicand .
EX.3. Solve 672 9999 by Ekanyunena Sutra.
Sol. Here P=678 and Q=9999 then the Sutra
= (P-1) / [Q-(P-1)]
= (678-1) / [9999-(678-1)]
= 677 / 9322.
= 6779322.

31. URDHAVA TRIYAGBHYAM.
[MULTIPLICATION].
The Urdhav Triyagbhyam sutra state that ‘‘ Vertically and cross multiplication “.Urdhva means verticle
multiplication and Triyagbham means cross wise or diagonal multiplication .Multiplication process may be
left to right to left to right .The sutra is use ful in all case of multiplication, Division and Algebraic
expressions.
(1) Multiplication of 2 digits number say ab × pq.
where ab is multiplicand and pq is multiplier .
Where b is the first right hand most digit of multiplicand ,
and a is the seconde digit of multiplicand..And q is the first
right and most digit of multiplier .Pis the second digit
of multiplier. Multiplication of ,
a b
× p q
Group Analysis:-
a a b b
p p q q
(a×p) (a×q) +(b × p) (b × q)

32. Ex.2 Multiplication of 3 digits number say
a b c
× p q r
Group analysis:-
a a b a b c b c c
p p q p q r q r r
(a×p) (a ×q) +(b×p) (a×r)+(b×q )+(c×p) (b ×r)+(c ×q) (c ×r)
Ex.3. Multiplication of 4 digit number say a b c d × p q r s
Group analysis :-
a a b a b c a b c d b c d c d d
p p q p q r p q r s q r s r s s
(a ×p) (a ×q) + (a×r)+(b×q) (a×s)+(b×r)+ (b×s)+(c×r) (c×s) + (d×r) (d×s)
(b×p) ( c×p) (c×q)+(d×p) + (d×q)

33. MULTIPLICATION
METHODS ARE US FOLLOW :-
1. VILOKNAM SUTRA.
2. EKANUYUNENA PURYENA
3. EKADHIKENA PURVENA ANTIYORDESA KEPI
4. NIKHILUM NAVA TASAH- CHARAMUM DESHATAH
5. URDHAVA TRIYAGBHYAM
6. ANURUPYENA
7. ADHYA MADHYENANT- YAMANTI YE NA
8. MULTIPLICATION BY DIFFERENT METHODS

34. VILOKNUM SUTRA
[MULTIPLICATION]
We can find the answers simply by observing the
problems follows
1,. Multiply any numbers by 10, 100,1000,10000,..........
Put the same numbers of zeroes to the right side of
multiplicand.
Ex. Multiply 27 by 10.
Solution Here 27 is multiplicand and 10 is multiplies , .
Then
27× 10 = 270
2. Multiply any number by 5.
Solutiion Then put one zero at the right side of
mutiplicand and then half of it.
Ex. Multiply 82 by 5 .
Solution ; 82 × 5 = 820/2 = 410
3. Multiply any number by 50.
Solution. Put two zeroes at the right side of
multiplicand and then half of it.
Ex.Multiply 148 by 50.
Solution 148×50=14800/2 =7400

35. PURNAQ PURNABHAYAM
[QUADRATIC EQUATON]
The sutra states that’‘By completing the non completion ‘’ For solving the roots of general form of
quadratic equation and factorisation.By quadratic equation also
Ex. 1 solve
Sol We know the general formula
The givenb equaton is
This is non completon form
By substracting (x+2) on both sides for completion.
⇒
⇒
⇒
⇒
⇒
⇒
⇒ Hence

36. Ex.2. Solve
Sol.We know that
Thus given equation is
It is non compltion form for complete the
cube add (4x+12) on both sides of the given
equation we get
LHS is complete

37. ANTYA YOREVA.
[SIMPLE EQUATIONS].
The sutra states that ‘‘Only by the last term.’’
The equation whose nu mberator and
Dinominator on LHS , whose contant terms
are in the same ratio , as the numerator and
the denominator of RHS stand to esch
order:-
Let the equation :
according to sutra it satisfies -
By cross multiplying equ.(1)
Q(PR+S) = P(QR+T)
⇒ PQR +QS = PQR +PT
QS=PT
The result in solving equation (1)is same as
the result in equation (2).
Ex.1 solve the equation.
Sol. Observe that the left part of the given
equation , solving as,
But by the sutra we have
Thus ,
Ex.2 Solve the equation
Sol. By taking the left part of the given
equation and solving we get
according to the sutra,
By solving we get
x=-2.

38. Ex.3 solve the eqution
(x+1)(x+5)(x+7)=(x+3)(x+4)(x+6)
Sol. Sum of the constant term on LHS and
RHS is 13 Now the given equation can
be arranged like ,
Here also the constant terms on LHS
and RHS are equal.
On LHS multipling the constant term
we get
according to sutra.
By solving we get
Second method :-
From the equation (i) we have,.
L.H.S. Part
Thus,
Hence.

39. ANURUPYEN SHUNYAMANYAT.
[SPECIAL SIMULTANEOUS EQUATION].
The sutra states that ‘‘ If is in
ratio , then the
Other is zero.”
Simultaneous
equations,in which the co-
efficient (co-ef.) of one
variable are in the same
ratio , as the independent
terms are to each order, The
other variable is zero.
Ex. 1 solve :-
6x + 7y = 8
19x+21y = 24
Sopl. The ratio of y co-ef=7/21=1/3
The ratio of independent terms
= 8/24 =1/3
The other variable x = 0
Put x = 0 in any equation we get
y = 8/7.
Ex. 2 solve :-
23x + 147y = 161.
69x+321y = 483.
Sopl. The ratio of x co-ef =23/69=1/3
The ratio of independent terms
= 161/483 =1/3
The other variable y = 0
Put y = 0 in any equ ation we get
x = 7.

40. SANSKALANA VYAKALANABHYAM.
[SPECIAL SIMULTINEONS EQUATION].
Sutra states that ‘‘ The x co-ef.
And y co-ef. Are
interchanged.”or ‘‘Co-ef. Of
one variable in one equation
is variable in one equation is
the same as the co-ef. Of
other variable in second
equation ”.
Then solve the
simultaneous equation by
addition and subtraction..
Ex. 1 solve :-
5x - 21y = 26-------(i)
21x - 5y = 26-------(ii)
Sopl. By adding (i) and (ii)
(5x-21y)+(21x-5y) = 26+26
26x – 26y = 52
x – y = 2 -----(iii)
By subtracting
(5x-21y) – (21x-5y) = 26-26
x + y = 0 ------(iv)
By solving (iii)and (iv) we get
x = 1, y = -1
Ex. 2 solve :-
12x + 17y = 53-------(i)
17x + 12y = 63-------(ii)
Sopl. By adding (i) and (ii)
(12x+17y)+(17x+12y) = 53+63
29(x+y )= 116
x + y = 4 -----(iii)
By subtracting
(12x+17y) – (17x+12y) = 53-63
x -y = 2 ------(iv)
By solving (iii)and (iv) we get
x = 3, y = 1

41. DWANDA YOG.
DUPLEX COBINATION PROCESS.
[SQUARE].
The sutra is used in two ways by
Squaring and by cross multiplication .
Let D represents Dwandva Yog
( D uplex Group Number ).
Calculation For Dwandva Yog.
(1):- For single digits number say 2 ,
D=square of that number = =4
(2):- For two digits number say 12 ,
D = Twice the product of 2 digits.
= 2×1×2=4
(3):- For three digits number say 123,
D= twice the product of first and
last digit +square of the middle
term.
= 2×1×3+ = 10
(4) For four digits number say 1234,
D = twice the product of first and
last digit + twice the product of
second and third digit.
= 2×1×4+2×2×3=20
(5):- For Five digits number say 12345,
D=twice the product of first and last
digit+twice the product of second and
fourth digits + square of the third
digits.= 2×1×5+2×2×4+ =35
(2):- For n digits number say 12 ,
D = square of the the number contains
2n digits.
(3):- For even number of digits,
equidistance from the two ends,
represent double of the cross product.
NOTE:-
(1) Make the group numbers increasing
and then by decreasing way from left
to right.
(2) The total number of squares and
Dwandva Yog is one less than the
twice the total number.

42. Ex.1 Find by Dwandva yog sutra.
Solv. The group numbers of 75 are 7 , 75, 5
(i) = 49
(ii) Dwandva Yog of 75 is
2×7×5=70
(iii) =25
Thus
= 49/70/25
= 49+7/0+2/5
= 5625
Ex.2 Find by Dwandva yog sutra.
Solv. The group numbers of 123 are 1,12,123,23,3,
Thus
= /2×1×2/2×1×3+ /2×2×3/
= 1/4/10/12/9
= 1/ 4+1/0+2/9
= 1/5/1/2/9
= 15129.
Ex.3 Find by Dwandva yog sutra.
Solv. The group numbers of 123 are
1,12,123,1234,234,34,4.
Thus
= /2×1×2/2 ×1×3+
/2×1×4+2×2×3/2×2×4+ /2×3×4/
= 1/4/10/20/25/24/16
= 1/ 4+1/0+2/0+2/5+2/4+1/6
= 1/5/2/2/7/5/6
= 1522756.

43. EKADHIKENA PURVENA .
[SQUARE].
Sutra Ekadhikena Purvena mean ‘‘By one more
than the privious one.”
Square of number
= Ekadhikena of Purvena / square of last digits
.
= purvena(purvena+1)/ square of last digits
= Tenth digits ×(Tenth digits +1)/square of last
digits
(1) Square of numbers ending in 5.
Ex.1 Find the square of number 25 by
Ekadhikenaq Purvena.
Sol:- The given number is 25, Purvena or privious
or tenth place digits is 2 and last digits is 5 by
putting these valuse in –
the sutra we get,
= 6 / 25
= 625
Ex.2 Find the square of number 325 by
Ekadhikenaq Purvena.
Sol:- The given number is 325, Purvena or
privious digitsor tenth place digit or left over
digits other than 5 is 32 and last digits is 5 by
the above sutra we have –
= 32(32+1) /
= 1056/25
= 105625 .
Ex.3. Find the square of 1105 by Ekadhikena
Purvena.
Sol. Given number is 1105 Previous digits is 110
and last digits = 5. by the sutra .
= 110 (110+1) /
= 12210 / 25
= 1221025.

44. SANKALANA VYAVAKALANABHYAME.
[SQUARE].
SUTRA STATES THAT ‘‘ By adding- and subtracting”.
We can find the squares of two numbers
which are having one more digits than the
Shunyat (known) number.
Let .
x = sunyat or known number
x+1 = one more than the sunyat number
x-1 = one less than the sunyet number.
According to the defination of sutra the
square can be given by the following ,
equation
apply the Algebraic formula
Also check the result of both sides of
the above two equation by Beejank method.
we can find the square of sunyat number by
Ekadhikena Purvena sutra.
Ex.1 Find the square of 41 and 39 if
Sol.
Here x=40 x+1=41 and x-1=39
=1681.
= 1521.

45. SQUARE RO0OT
[VILOKNUM METHOD].
We can find the square root of a perfect square by Viloknum(observing) method.
We knoe that the square root of one digits number is less than 100. The
Square root of two digits number will be of 3 or 4 digits. The square of 3 digits number
will be of 5 or 6 digits . The square root of 4 digits number will be of 6 or 7 digits .
Thus how many digits , will be in the square root, we can know simply by observing it ,
here we are finding the square root of six digits number.
TABLE NO.1 TABLE NO.2 TABLE NO. 3
SQUARE
ROOT
SQUARE BEEJANK
1 1 1
2 4 4
3 9 9
4 16 7
5 25 7
6 36 9
7 49 4
8 64 1
9 81 9
10 100 1
Charmank of
square number
Charamank of
SQUARE root.
1 1 or 9
4 2 or 8
5 5
6 4 or 6
9 3 or 7
NUMBER TENTH PLACE
DIGIT
100 TO 399 1
400 TO 899 2
900 TO 1599 3
1600 TO 2499 4
2500 TO 3599 5
3600 TO 4899 6
4900 TO 6399 7
6400 TO 8099 8
8100 TO 9999 9

46. CONDITION FOR PERFECT SQUARE.
1. There should be even number of zeroes at the end.
2. The unit place digits dosen’t have 2,3,7,8.
3. There should not be 6 at unit place and 0,2,4,6,8,.
4. There should not be 1,4,5,9, at unit place and 0,2,4,6,8. at tenth place.
5. If perfect square is even , then last 2 digits should be divisible by 4.
6. Perfect square have 5 at unit place and 2 at tenth place.
‘‘ METHODS”
1. CHARMANK IS THAT DIGITS FROM WHERE THE PROCESS OF CHANGE IS OCCUR. GENERALLY
THEUNIT PLACE DIGITS IS KNOWN AS CHARMANK.
2. SPLIT THE GIVEN NUMBER INTO TWO GROUP.
3. THE GROUP WHICH CONTAINS THE CHARMANK WIILL BE THE GROUP AND REST WILL BE THE
SECOND GROUP .
4. THE UNIT PLACE DIGITS OF A SQUARE ROOT IS THE CHARMANK.
5. THE TENTH PLACE DIGIT OF SQARE ROOT IS OBTAINED BY THE SECOND GROUP.
6. FIND THE MULTIPLICATION OF TENTH DIGITS OF THE SQUARE ROOT AND ITS EKADHIK (ONE MORE
THAN THE PRIVIOUS ).
7. IF THE SECOND GROUP IS LESS THAN MULTIPLICATION , THEN LOWER DIGIT.IF THE SECOND GROUP
IS GREATER THAN THE MULTIPLICATION , THEN PUT THE UPPER DIGITS FOR THE UNIT PLACE.

47. Ex1 Find he square root of perfect square number
2116 by viloknum method.
Sol.
1. split the given number 2110 into two group
from right to left
2. The first group is 16 and second group is 21
3. From first group 16 the unit place digit is 6
the charmank will be 6, thus from table no.2
THE Charmank of square root will be 4 or 6.
4. From table no.1 the second group 21 lies
between 16 and 25 Thus tenth place digits of
square root is 4.
5. Combining two groups the square root may
be 44 or 46 .
6. Now Ekadhik of 4 is 5. (4+1=5)
7. Multiplying the tenth place digits of square
root and its Ekadhik 4×5 = 20 , new
multiplying it by 100, we get 2000.
8. The given number 2116 is greater than
2000.thus 44 is not the square root bet 46 is
correct square root .
Ex2 Find he square root of perfect square number
12544 by viloknum method.
Sol.
1. Split the given number 12544 into two group
from right to left.