25+ Vedic Maths Tricks In Simplified Version

This is a simplified and summarized version of Vedic Maths – a book written by the Indian Hindu Priest Bharati Krishna Tirthaji and first published in 1965. The Vedic maths contains 16 sutras (formulae) and 13 sub-sutras.

There are many Vedic Maths Tricks you can find when you search for easy calculation methods using Vedic Maths. 

How will Vedic Math Trick help you?

Vedic Mathematics is a unique collection of techniques drawn from ancient Indian mathematics, attributed to the Vedic period and significantly enhanced by the notable Indian mathematician, Krishna Tirthaji Maharaja, also known as Tirtha Ji or Tirthaji Maharaj.

Vedic Mathematics has its roots in the Atharva Veda, one of the oldest Indian scriptures and encompasses a wide range of mathematical skills, offering beautiful methods for performing complex calculations, especially arithmetical computation and algebraic operations.

The Vedic Mathematics system centers around 16 Vedic sutras or aphorisms, including ‘Ekadhikena Purvena’ and other Vedic Mathematics tricks, which can be utilized for solving mathematical problems more quickly and efficiently than conventional methods.

These methods empower students to enhance their calculation speed, thereby significantly reducing calculation time. Mathematical operations with two or even three-digit numbers become simpler using these sutras.

The system enables the development of mental mathematics skills, enabling the execution of difficult calculations without relying on pen and paper, or calculators. This mental agility with numbers is particularly advantageous for competitive exams, including entrance exams and banking exams where calculation speed can give an edge over others. Below are 16 Vedic Maths Tricks for you.

The 16 Sutras of Vedic Maths

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 the 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 the multiplication of any numbers near to less than the 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 at the beginning and the 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 the 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 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 =?

4 x 2 = 8
(1 x 2) + (4 x 1) = 6
1 x 1 = 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 numbers 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 the common factor in the equation “14x + 5x…… = 7x + 3x…..”, x will equal 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.

Example 1:

3x + 7y = 24x + 21y = 6

Observe that the y-coefficients are in the ratio 7 : 21 i.e., 1 : 3, which is the same as the ratio of independent terms i.e., 2 : 6 i.e., 1 : 3. Hence the other variable x = 0 and 7y = 2 or 21y = 6 gives y = 2 / 7

Example 2:

323x + 147y = 1615

969x + 321y = 4845

7. Sankalana – Vyavakalanabhyam

– By addition and By subtraction

In two general equations 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 the 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 a 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 is 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 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 a 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 the multiple of 10,

Step 1: Subtract the number by a multiple of 10
Step 2: The first digit is the subtraction of the number from the result of Step 1
Step 3: the second digit is the square of the 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 the above step).

i.e. (x+3)³= 2x² + 13x + 19

Now we need to factorize RHS in terms of (x+3). So apply the 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 remainder 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).

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 the last digits gives the 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)
Applying 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 a 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.

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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 power 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 gives 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, the square of 2 is 4
8 – 2 = 6
Thus, a 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 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 the left-hand side leaving the last digit and multiply with the first number of the right-hand side. Then you multiply the last digit of the left-hand side with the last digit of the right-hand side. Practically explained 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 equations by eliminating any of the variables. For example:

i. Factorize 2x² + 6y² + 3z² + 7xy + 11yz + 7zx

We have 3 variables x,y,z.

Remove any of the variables 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)

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

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 the 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 at the beginning and the 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 by 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 the following 2 general equations
ax + by = p
cx + dy = q

Solving,
x = (bq – pd) / (bc – ad)
y = (cp – aq) / (bc – ad)

Notice that for the calculation of numerators (x and y), cyclic method is used and Denominators remains the 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.

Thus the present article serves as only an ‘introduction’.

While it is often debated how much of Vedic Mathematics comes directly from the Vedic period, there’s no denying the fact that these techniques, refined and taught by Tirthaji Maharaj, have provided countless students with a practical, efficient tool for numerical calculations.

The principles of Vedic Mathematics, passed down from generation to generation, continue to be a valuable part of the history of mathematics. The introductory volume of Vedic Mathematics and subsequent works by proponents like Gaurav Tekriwal have further extended the reach of these ancient techniques, proving beneficial to a broad community of students and educators worldwide.