1. Ekadhikena Purvena
– By one more than the previous one
For any integer ending with 5, the square always ends with 25 and begins
with the multiple of previous integer and one more than the integer. For
example:

Square of 25 is 2 x 3 .. 25 = 625.

Square of 85 is 8 x 9 .. 25 = 7225
2. Nikhilam Navatascaravam Dasatah
– All from 9 and the last from 10
For multiplication of any numbers near to less than multiple of 10,

Step 1: Subtract the numbers with their closest multiple of 10 and
multiply the results.

Step 2: Subtract the results with other numbers.

Step 3: Write the result of Step 2 in the beginning and result of Step 1 at
the end.
For example: 99 x 96 = ?
100 – 99 = 1 and 100 – 96 = 4
1 x 4 = 04
99 – 4 or 96 – 1 = 95
95 and 04
So, the answer is 9504
Example II: 94 x 83 =?
100 – 94 = 6 and 100 – 83 = 17
6 x 17 = 102
94 – 17 or 83 – 6 = 77
So, the answer is 7802
3. Urdva – Triyagbhyam
– Vertically and crosswise
For multiplication of any two two-digit numbers,

Step 1: Multiply the last digit

Step 2: Multiply numbers diagonally and add them.

Step 3: Place Step 1 at the end and Step 2 at the beginning.

Step 4: Multiply the first digit of both the number and put it at the most
beginning.

Step 5: For the result, more than 2 or more digits, add the beginning
digits to the beginning numbers.
Example : 45 x 87 = ?
5 x 7 = 35
(4 x 7) + (5 x 8) = 28 + 40 = 68
4 x 8 = 32
45 x 87 = 32 | 68 | 35 = 32 | 68 + 3 | 5 = 32 | 71 | 5 = 32 + 7 | 15 =
3915
Example II : 14 x 12 = ?
4x2=8
(1 x 2) + (4 x 1) = 6
1x1=1
14 x 12 = 168
Example III : 54 x 87 = ?
4 x 7 = 28
(5 x 7) + (4 x 8) = 35 + 32 = 67
5 x 8 = 40
54 x 87 = 40 | 67 | 28 = 4698
4. Paraavartya Yojayet
– Transpose and adjust
For dividing large numbers by number greater than 10. For example
3784 divided by 12.
Step 1: Write the negative of the last number of the divisor under the
dividend.
12
-2
Step 2: Separate the last digit of the dividend from the rest to calculate
the remainder.
378 4
Step 3: Multiply the first digit with the above result i.e., -2.
3 x (-2) = -6
Step 4: Add the second digit with the result and continue till the last.
378 4 = 3 (7-6) 8 4 = 31 (8-2) 4 = 316 (4-12) = 315 (14 – 10)
Result: Quotient = 315 and Remainder = 4
5. Sunyam Samya Samuccaya
– When the sum is the same, the sum is zero
This is related to equating with zero. For example, as x is common factor
in the equation “14x + 5x…… = 7x + 3x…..”, x will equal to 0.
Example : 9(x+3) = 4(x+3)
According to the definition, Since x+3 is a common factor, x + 3 =
0 that’s why, x = -3
Calculation with a simple algebraic method,
9x + 27 = 4x + 12
5x = -15
x = -3
6. Anurupye – Sunyamanyat
– If one is in ratio the other one is 0
We use this Sutra in solving a special type of simultaneous simple
equations in which the coefficients of ‘one’ variable are in the same ratio
to each other as the independent terms are to each other. In such a
context the Sutra says the ‘other’ variable is zero from which we get two
simple equations in the first variable (already considered) and of course
give the same value for the variable.
7. Sankalana – Vyavakalanabhyam
– By addition and By subtraction
In two general equation such as, ax + by = p and cx + dy = q, where x and
y are unknown values,
x = (bq – pd) / (bc – ad)
y = (cp – aq) / (bc – ad)
For example:
3x + 2y = 4 and 4x + 3y = 5
x = (10-12)/(8-9) = 2
y = (16-15)/(8-9) = -1
8. Puranapuranabhyam
– By the completion or non-completion
This is a method of completion of polynomials to find its factors.
Example:
x³ + 9x² + 24x + 16 = 0 i.e. x³ + 9x² = -24x -16
We know that (x+3)³ = x³+9x²+27x+27 = 3x + 11 (Substituting above
step).
i.e. (x+3)³ = 3(x+3) + 2 … (write 3x+11 in terms of LHS so that we
substitute a term by a single variable).
Put y = x+3
So, y³ = 3y + 2
i.e. y³ – 3y – 2 = 0
Solving using the methods discussed (coeff of odd power = coeff of even
power) before.
We get (y+1)² (y-2) = 0
So, y = -1 , 2
Hence, x = -4,-1
9. Chalana – Kalanabhyam
– Differences and Similarities
i. In the first instance, it is used to find the roots of a quadratic equation
7×2 – 11x – 7 = 0. Swamiji called the sutra as calculus formula. Its
application at that point is as follows. Now by calculus formula, we say:
14x–11 = ±√317
A Note follows saying every Quadratic can thus be broken down into two
binomial factors. An explanation in terms of the first differential,
discriminant with a sufficient number of examples are given in the
chapter ‘Quadratic Equations’.
ii. In 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’.
Further other Sutras 10 to 16 mentioned below are also used to get the
required results. Hence the sutra and its various applications will be
taken up at a later stage for the discussion.
10. Ekanyunena Purvena
– By one less than the previous one
For multiples of 9 as a multiplier, the first digit is 1 less than the first
digit of the multiplicand and the second digit is subtracting the lessened
digit from multiple of 9.
Examples:
5 x 9 = 45
5-1= 4, 9-4 = 5
14 x 99 = 1386
14-1=13, 99-13=86
For multiplier less than the multiplicand, the first digit is an additional 1
of the first digit of the multiplicand less than the latter and the second
digit is subtracting the second digit of the multiplicand by 10.
Examples:
12 x 9 = (12-2), (10-2)
= 108
28 x 9 = (28-3), (10-8)
= 252
11. Yaavadunam
– Whatever the extent of its deficiency
For the square root of a number close to multiple of 10,
Step 1: Subtract the number by multiple of 10
Step 2: First digit is the subtraction of the number from the result of Step
1
Step 3: the second digit is the square of result from Step 1
Examples:
i. Square of 9 = (9-(10-9), (10-9) x (10 – 9)
=81
ii. Square of 18 = 18 + 8, 8 x 8
= 26,64 = 2 (6+6) 4
= 324
12. Vyashtisamanstih
– Part and Whole
This solves the equation of polynomials to find its factors by
using Paravartya Sutra. For example:
i. x³ + 7x² + 14x + 8 = 0 i.e. x³ + 7x² = – 14x – 8
We know that (x+3)³ = x³+9x²+27x+27 = 2x² + 13x + 19 (Substituting
above step).
i.e. (x+3)³= 2x² + 13x + 19
Now we need factorize RHS in terms of (x+3). So apply Paravartya sutra.
Dividing 2x² + 13x + 19 by (x+3) gives
2x² + 13x + 19 = (x+3)(2x-7)-2
i.e. (x+3)³ = (x+3)(2x-7)-2
put y = x+3
So, y³ = y(2y+1) -2
Which gives y = 1,-1,2
Hence, x= -2, 4, 1, -1
13. Shesanyankena Charamena
– The remainders by the last digit
Example: 1/7
 As seen earlier successive remainders are 1, 3, 2, 6, 4 and 5.

We will write them as 3, 2, 6, 4, 5 and 1.

Multiply them with the last digit of divisor (7): 21, 14, 42, 28, 35 and 7

Now take their last digits and that’s the final answer: 0.142857.
(another interesting concept).
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Example: 1/13
 Successive remainders are 1, 10, 9, 12, 3 and 4.

We will write them as 10, 9, 12, 3, 4 and 1.

Multiply with the last digit of the divisor (3): 30, 27, 36, 9, 12 and 3.

Taking just last digits gives final answer: 0.076923.
14. Sopaantyadvayamantyam
– The ultimate and twice the penultimate
For the equation in the format 1/AB + 1/AC = 1/AD + 1/BC, the result is
2C(penultimate) + D(ultimate) = 0.
Example:
1/(x+2)(x+3) + 1/(x+2)(x+4) = 1/(x+2)(x+5) + 1/(x+3)(x+4)
Appying the formula, 2(x+4) + (x+5) = 0
or, x = -13/3
15. Gunitasamuchyah
– The product of the sum is equal to the sum of the product
For the quadratic equation, in order to verify the result, the product of
the sum of the coefficients of ‘x’ in the factors is equal to the sum of the
coefficients of ‘x’ in the product.
Examples:
(x + 3) (x + 2) = x² + 5x + 6
or, (1+3) (1+2) = 1 + 5 + 6
or, 12 = 12 ; thus verified.
(x – 4) (2x + 5) = 2x² – 3x – 20
or, (1 – 4) (2 + 5) = 2 – 3 – 20
or -21 = – 21 ; thus verified.
16. Gunakasamuchyah
– The factors of the sum is equal to the sum of the factors
For quadratic equation, the factor of the sum of the coefficients of ‘x’ in
the product is equal to the sum of the coefficients of ‘x’ in the factors.
The 13 sub-sutras of Vedic Maths
1. Anurupyena – Proportionality
This Sutra is highly useful to find products of two numbers when both of
them are near the Common bases i.e powers of base 10. For example:
i) 46 x 43
We take 50 as a base point because it is the nearest powers of base 10.
46 = 50 – 4
43 = 50 – 7
Multiplying the difference from 50 to the numbers,
4 x 7 = 28 …… (a)
From cross subtraction,
43 – 4 = 39 or 46 – 7 = 39……. (b)
Adding (a) with the base i.e., 28+50 = 78 (goes on right hand side)
Dividing 39 by 2 give 19 with remainder 1. (goes on left hand side)
Thus, 46 x 43 = 1978
ii) 56 x 53
56 = 50 + 6
53 = 50 + 3
56 x 53 = (56 + 3) or (53 + 6), (6 x 3) + 50
= 59,18 = 59/2, 18 = 2968
2. Adyamadyenantya – mantyena
– The first by the first and the last by the last
Example: 2x² + 5x -3
Divide the first term’s coeff (2) of the equation by 1st term of factor(1)
and divide the last term of equation (-3) by 2nd term of factor (3). So,
2nd factor: 2x-1
3. Yavadunam Tavadunikrtya Varganca Yojayet
– Whatever the deficiency subtract that deficit and write
alongside the square of
To find squares of numbers close to base 10, we subtract the number
from base 10 and take a square of the result. Then we subtract the result
from the number and cross the results.
Examples:
i. Square of 8
10 – 8 = 2, square of 2 is 4
8–2=6
Thus, Square of 8 = 64
ii. Square of 6
10 – 6 = 4, square of 4 is 16
6–4=2
Thus, Square of 6 = 2(1)6=36
4. Antyayor Desakepi
This is the multiplication of two numbers in the same structure of
numbers to make their sum being the multiple of 10. For example: 43 x
47, 116 x 114, 1125 x 1125, etc.
For this you apply ekadhikena for the first digits of left hand side leaving
the last digit and multiply with the first number of right hand side. The
you multiply the last digit of the left hand side with the last digit of right
hand side. Practically explained as below.
i. 43 x 47 = (4+1) x 4 | 3 x 7
= 5 x 4 | 21
= 20 | 21
= 2021
ii. 116 x 114 = (11 + 1) x 11 | 6 x 4
= 132 | 24
= 13224
iii. 1125 x 1125 = 113 x 112 | 5 x 5
= 12656 | 25
= 1265625
5. Antyayoreva – Only the last terms
For the equation in the format (AC + D) / (BC + E) = A/B, the result is
A/B = D/E. For
Example:
i) (x²+x+1) / (x²+3x+3) = (x+1) / (x+3)
or, {x(x+1)+1} / {x(x+3)+1} = (x+1) / (x+3)
or, (x+1) / (x+3) = 1 / 3
or, x = 0
6. Lopana Sthapanabhyam
– By alternate elimination and retention
This is to factorize complex equation by eliminating any of the variable.
For example:
i. Factorize 2x² + 6y² + 3z² + 7xy + 11yz + 7zx
We have 3 variables x,y,z.
Remove any of the variable like z by putting z=0.
Hence the given expression
E = 2×2 + 6y2 + 7xy = (x+2y) (2x+3y) … (Combination of Anurupyena &
Adyamadyenantyamantya).
Similarly, if y=0, then
E = 2×2 + 3z2 + 7zx = (x+3z) (2x+z)
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As x and 2x are present separately and uniquely. Hence we may map to
get factors.
E = (x+2y+3x) (2x+3y+z)
7. Vilokanam – By mere observation
Generally, we come across problems that can be solved by mere
observation. But we follow the same conventional procedure and obtain
the solution. But the hint behind the Sutra enables us to observe the
problem completely and find the pattern and finally solve the problem by
just observation.
Example:
x + 1/x = 5/2
x + 1/x = 2 + 1/2
Hence, x = 2 and 1/x = 1/2
8. Gunitahsamuccayah Samuccayagunitah
For the quadratic equation, in order to verify the result, the product of
the sum of the coefficients of ‘x’ in the factors is equal to the sum of the
coefficients of ‘x’ in the product.
Example:
i) (x + 3) (x + 2) = x² + 5x + 6
or, (1+3) (1+2) = 1 + 5 + 6
or, 12 = 12 ; thus verified.
ii) (x – 4) (2x + 5) = 2x² – 3x – 20
or, (1 – 4) (2 + 5) = 2 – 3 – 20
or -21 = – 21 ; thus verified.
9. Sisyate Sesasamjnah
– The remainder remains constant
For multiplication of any numbers near to more than multiple of 10,

Step 1: Subtract the closest multiple of 10 from the number and
multiply the results.

Step 2: Add the results with other numbers.

Step 3: Write the result of Step 2 in the beginning and result of Step 1 at
the end.
For example: 104 x 101 = ?
104 – 100 = 4 and 101 – 100 = 1
4 x 1 = 04
101 + 4 or 104 + 1 = 105
105 and 04
So, the answer is 10504
10. Gunakasamuccayah
For the quadratic equation, the factor of the sum of the coefficients of ‘x’
in the product is equal to the sum of the coefficients of ‘x’ in the factors.
11. Vestanam – By Osculation
Let us check whether 21 is divisible by 7.
For 7, Ekadhika(positive osculator) is 5
So as per the mentioned process, multiply 5 with 1 and add 2 to the
product.

21; 1×5+2 = 7 (Divisible by 7)

91; 1×5+9 = 14 (Divisible by 7). Can be continued further as
14; 4×5 + 1 = 21; and
21;1×5+2 = 7

112; 2×5+11= 21. (seen earlier)

2107; 7×5 + 210 = 245
245; 5×5+24= 49 (Divisible by 7 or continue further).
12. Yavadunam Tavadunam
Consider following 2 general equations
ax + by = p
cx + dy = q
Solving,
x = (bq – pd) / (bc – ad)
y = (cp – aq) / (bc – ad)
Notice that for calculation of numerators (x any y) cyclic method is used
and Denominators remains same for both x and y.
Examples:
2x + 3y =6
3x + 4y = 3
Applying above formula:
x = (9 – 24)/ (9 – 8) = -15
y = (18 – 6) (9 – 8) = 12
-3x + 5y = 2
4x + 3y = -5
Applying above formula:
x = (-25 -6) / (20+9) = -31/29
y = (8-15) / (20+9) = -7/29
13. Kevalaih Saptakam Gunyat
For 7 the Multiplicand is 143 (Kevala: 143, Sapta: 7).
On the basis of 1/7, without any multiplication, we can calculate 2/7, 3/7,
4/7, 5/7 and 6/7. For that 1/7=0.142857 is to be remembered. But since
remembering 0. 142857 is difficult we remember Kevala(143). This is the
only use of this sutra (for remembrance).
1/7 = 0.142857
and By Ekanyuna, 143 x 999 = 142857
We need to remember 142857 for following reasons:

2/7 = 0.285714

3/7 = 0.428571

4/7 = 0.571428

5/7 = 0.714285

6/7 = 0.857142
All the values are in cyclic order.
So if we can remember 1/7 then we can obtain 2/7, 3/7 and etc as

2/7: the last digit of answer must be 4 (2*7=14)

3/7: the last digit of answer must be 1 (3*7=21)

4/7: the last digit of answer must be 8 (4*7=28)

5/7: the last digit of answer must be 5 (5*7=35)

6/7: the last digit of answer must be 2 (6*7=42)
We are aware that this attempt is only to make you familiar with a few
special methods of Vedic Mathematics. The methods discussed, and
organization of the content here are intended for any reader with some
basic mathematical background. That is why the serious mathematical
issues, higher-level mathematical problems are not taken up in this
article, even though many aspects like four fundamental operations,
squaring, cubing, linear equations, simultaneous equations.
factorization, H.C.F, recurring decimals, etc are dealt with. Many more
concepts and aspects are omitted unavoidably, keeping in view the scope
and limitations of the present volume.