Download the Handout set ( format)

1. Multiplication
Urdhva-Tiryak Sutra
"Vertically and Crosswise": That's all you need to remember to multiply two numbers!
Line diagrams
2 digit numbers
32 x 54 = ?
32
54
CARRY
2
1728
3 digit numbers
749 x 322 = ?
749
322
CARRY
1 2 5 3
241178
4 digit numbers
6724 x 1823 = ?
6724
1283
CARRY
1 3
4 8 7 2
8626892
Multiplying a number by 11
To multiply any 2 digit number by 11, we just put their total between the 2 figures.
26 x 11 = 2 | 2+6 | 6 = 286
48 x 11 = 4 | 4+8 | 8 = 4 |1 2 | 8 = 528
89 x 11 = 8 | 8+9 | 9 = 8 |1 7 | 9 = 979
Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE)
2. Squaring & Cubing of 2 digit numbers
Nikhilam Sutra for Squaring 2 digit numbers
1. For a given two digit number always take the base as 10 and the working base as
the multiple of ten nearest to the given number.
2. Find the difference between the given number and the working base and add this
to the given number.
3. The number now obtained is multiplied by the number of times the working base
is a multiple of 10. This forms the LHS of the answer.
4. Square the difference of the number and the working base. This is the RHS of the
answer.
This method is much simpler than it may appear at first and is best understood through
examples.
642:
 Base = 10, Working Base = 60(10 x 6).
 64 – 60 = 4. Adding 4 to 64 we get 68.
 68 x 6 = 408 and 42 = 16
 ANS = 408 | 16 = 4096
2
78 : WB = 80 (10 x 8), 78 – 80 = –2, 78 + (–2) =
76, 76 x 8 = 608, (–2)2 = 4, ANS =6084
Anurupya Sutra for Cubing 2 digit numbers
Remember cubes of the first 9 numbers
13 = 1
23 = 8
33 = 27
3
3
6 = 216
7 = 343
83 = 512
43 = 64
93 = 729
1. The formula for a 2 digit no. with digits a and b is
2. Below the a2b and ab2 terms write twice their values.
3. Add ‘vedically’ to get the cube of the number!
53 = 125
a3
a2b
ab2
b3
Examples:
1
5
6
143 = 1
4
8
7
16
32
4
2
4
5
2
233 = 8
12
24
1
18
36
6
12
64
=
2744
=
12167
4
27
7
Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE)
3. Simultaneous Equations in 2 variables
Paravartya Sutra
This method is essentially similar to what we know today as “Cramer’s Rule”.
This can be understood with the help of the following example
Solve
and
2x + 3y = 8
4x + 5y = 14
To find ‘x’: Consider the next column to the ‘x’ column, i.e. that containing y and cross
multiply forwards (with – sign in between) to get numerator.
Therefore, numerator = (3 x 14) – (8 x 5) = 2
Cross multiply backwards for the denominator. Denominator = (3 x 4) – (2 x 5) = 2
Hence, x = 2/2 = 1
To find ‘y’: Consider the next column to ‘y’ column, i.e. the constant column and cross
multiply forwards with – sign in between. Therefore, numerator = (8 x 4) – (2 x 14) = 4
Denominator of ‘y’ term is same as that of ‘x’ term = 2 (note this point to avoid
confusion). Hence, y = 4/2 = 2
ANS: x = 1, y = 2
Evidently this method can be done mentally after just a few practice examples!
Examples:
Solve x – y = 7
and
5x + 2y = 42
ANS: x = – 42 – 14 = – 56 = 8 ;
–5 – 2
–7
y = 35 – 42 = 1
–7
Solve 11x + 6y = 28
and
7x – 4y = 10
ANS: x = 60 – (–112) = 172 = 2 ;
42 – (–44)
86
y = 196 – 110 = 1
86
Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE)
4. Factorisation of Simple Quadratics
Aadyam-aadyena Aantam-antyena Sutra
Instead of following the cumbersome method of splitting and taking common terms to
factorise a quadratic expression, the Vedic method can be employed effectively.
Procedure:
For a quadratic of the form ax2 + bx + c
1. Split middle coefficient (i.e. b) into 2 parts such that:
ratio of 1st coeff (a) to 1st part = ratio of 2nd part to constant term (c)
This ratio gives first factor
2. To find second factor: divide first term of quadratic by first term of first factor
and divide last term of quadratic by last term of first factor
After going through just a few examples, this can be done mentally!
Examples:
2x2 + 5x + 2
:
Step 1  Split 5 = 4 + 1 since 2 : 4 :: 1 : 2
Ratio is 1:2 … therefore 1st factor is (x + 2)
Step 2  2x2 ÷ x = 2x and 2 ÷ 2 = 1 … 2nd factor is (2x + 1)
ANS: 2x2 + 5x + 2 = (x + 2)(2x + 1)
6x2 + 11x + 3 :
Step 1  Split 11 = 9 + 2 since 6 : 9 :: 2 : 3
Ratio is 2:3 … therefore 1st factor is (2x + 3)
Step 2  6 x2 ÷ 2 = 3x and 3 ÷ 3 = 1 … 2nd factor is (3x + 1)
ANS : 6x2 + 11x + 3 = (2x + 3)(3x + 1)
8x2 – 22x + 5 :
Step 1  Split -22 = -2 - 20 since 8 : -2 :: -20 : 5 =- 4:-1
Ratio is 4:-1 … therefore 1st factor is (4x - 1)
Step 2  8x2 ÷ 4 = 2x and 5 ÷ -1 = -5 … 2nd factor (2x - 5)
ANS: 8x2 – 22x + 5 = (4x - 1)(2x - 5)
12x2 – 23xy + 10y2: Step 1  Split -23 = -8 - 15 since 12 : -8 :: -15 : 10 =- 3:-2
Ratio is 3:-2 … therefore 1st factor is (3x – 2y)
Step 2  12x2 ÷ 3 = 4x and 10y2 ÷ -2y = -5y … 2nd factor (4x – 5y)
ANS: 12x2 – 23xy + 10y2 = (3x – 2y)(4x – 5y)
Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE)
5. Quadratic Equations: A few types
A general formula for solving any quadratic equation is
First Differential of Quadratic = Square root of Discriminant
Consider x2 + 4x - 3 = 0
Then its first differential is 2x + 4
Discriminant is 42 – [4 x 1 x (-3)] = 28
Therefore, the solution is given by
2x + 4 =  (281/2 )
Hence x = (1/2) – 2
Numerous categories have been developed to further simplify the solution of quadratic
equations.
One such category is the “reciprocals type”
Examples:
1.
Solve
x + 1 = 17 .
x
4
=>
x + 1 = 4 + 1
x
4
2.
Solve
=>
x – 1 = 35– 25
x
2
3
3.
Solve
=>
x – 1 = 49 – 4 = 7 – 2
x
14
14
2 7
Hence, directly x = 4 or x = 1/4
x – 1 = 55
x
6
x – 1 = 45
x
14
Hence, directly x = 3/2 or x = -2/3
[14 = 7 x 2 and 45 = 72 – 22]
Hence, directly x = 7/2 or x = -2/7
[Express numerator as a difference of squares of factors of denominator]
Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE)
6. Method of Auxiliary Fractions
Ekadhika Purva Sutra
TYPE 1: Dividing by a number ending with 9
Consider 7 .,
149
To write the auxiliary fraction (AF), first note the number of 9s at the end of denominator.
Then shift the decimal point of the numerator by this number and in the denominator
increase the digit immediately preceding the 9s by 1.
AF = .7 .,
15
0.7 ., 
15 R 
Divide the numerator successively, each time prefixing the
‘remainder’ with the ‘quotient’ arrived at after dividing.
This is continued till the required number of decimal places.
0.0
7
4
6
10
9
14
11
Another example:
Consider 83 .
AF = 0.83
399
4
0.83 .,  0 . 20 80
20
4
R
3
0
0
7
ANS = 0.04697
14
05
ANS = 0.20802005
0
Note: This method can also be applied for numbers ending in 3, by multiplying
numerator and denominator by 3 to get denominator ending in 9.
TYPE 2: Dividing by a number ending with 1
Drop the 1 from the denominator (eg. 31 becomes 30, 291 becomes 290), and subtract 1
from numerator. The denominator will always end with 0; get rid of this by dividing
numerator and denominator by 10. This is the auxiliary fraction (AF). Successively
divide in similar fashion to Type 1 but each time replace the ‘quotient’ by its nine’s
complement (9 – the number, for eg. 9’s complement of 3 = 6, 9’s complement of 1 = 8)
Consider 39 .,
171
3.8 .,
17
C

R
AF = 38 . = 3.8
170
17
7
0.2
4
7
2
13
1
8
1
9
0
11
2
9
7 0 1
0
[9’s COMPLEMENT]
ANS = 0.2280701
2 12
Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE)
7. Miscellaneous Topics
Difference of Squares
Did you know that any number can be expressed as a difference of two squares?
Procedure:
1. Express the number as a product of either two even or two odd factors, say ‘a’ and ‘b’.
(if not possible then express as any two factors)
2. Then the number can be written as a + b 2
a–b 2
2
2
Examples:
45 = 9 x 5
=> 45 = 72 - 22
316 = 158 x 2 => 316 = 802 - 782
( )( )
Special type of Linear Equations
If the denominators contain linear expressions in x and the numerical values of the
numerators are same with RHS = 0, then we can add denominators directly
Examples:
3 + 3 = 0
2x + 4 x – 3
=> (2x + 4) + (x – 3) = 0
i.e. 3x + 1 = 0
x = –1/3
1
+ 1
= 0
x–4
x+3
=> (x – 4) + (x + 3) = 0
i.e. 2x – 1 = 0
x = 1/2
Equation of a Straight Line passing through two known points
Say the points are (p,q) and (r,s)
The equation will be of the form ax – by = c, where a, b and c have to be found
Procedure:
1. To find a, i.e. coefficient of x, find the difference of y coordinates [q – s]
2. To find b, i.e. coefficient of y, find the difference of x coordinates [p – r]
3. To find c, i.e. constant term, find middle product – extreme product [qr – ps]
Examples:
(11,6) and (4,7)
a = -1, b = 7, c = 24-77 = -53
ANS: –x – 7y = –53
(17,9) and (13,-8)
a = 17, b = 4, c = 117 – (-136) = 253
ANS: 17x – 4y = 253
Seminar on Vedic Mathematics by Madhav Nayak (SE Computer Engineering, SPCE)