- Pranav Nandurkar

# Do you know: Some Vedic maths tricks for quick calculations.

Actually, these are some Sutras and Sub-Sutras of Vedic mathematics wherein these tricks come in and which would in accurate and quick mathematical calculations.

### **1. Nikhilam Navatascaravam Dasatah. **

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.

**2. Urdva – Triyagbhyam **

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.

For 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

**3. Paraavartya Yojayet**

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

**4. 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.

For 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

**5. Anurupye – Sunyamanyat **

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.

For example 1: 3x + 7y = 2 4x + 21y = 6 Observe that the y-coefficients are in the ratio 7 : 21 i.e., 1 : 3, which is 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

**6. Sankalana – Vyavakalanabhyam **

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

**7. Puranapuranabhyam **

This is a method of completion of polynomials to find its factors.

For 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

**8. Chalana – Kalanabhyam **

**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’.

**9. Ekanyunena Purvena **

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.

For 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.

**10. 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.

For examples: 2x + 3y =6 3x + 4y = 3 Applying above formula: x = (9 – 24)/ (9 – 8) = -15 y = (18 – 6) (9 – 8) = 12 11 Sisyate Sesasamjnah 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 12 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. 43 x 47 = (4+1) x 4 | 3 x 7 = 5 x 4 | 21 = 20 | 21 = 2021.

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