Hey guys, lets look into some cool faster mathematics tricks, that we generally call as Vedic Mathematics. We will discuss some of them, which have application almost every time with examples.

**Trick 1: 1/(<any_number>9) = ?**

The condition here is the denominator ends with 9. Increase the digits preceding by 1 and divide 1 with the result.

Example: 1/19 = Dividing 1 by 1+1=2 gives ) with remainder 1. Divide 10 by 2 gives 5. Dividing 5 gives 2 with 1 as remainder and so on. The final answer is thus formed with all the quotients, which in this case is .052 (up to 3 digits).

**Trick 2: Multiplication with Base 10**: Always remember that RHS has just the same number of the digits as the number of 0’s in the base.

Lets us multiply 13 * 17. 13 is +3 than 10. Similarly, 17 is +7 than 10.

13 +3

17 +7

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(13+7 or 17+3)20 | (+7*+3)21 = (20+2) | 1 = 221

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**Trick 3: Multiplication with base 100**:

Let us multiply 93 * 111

93 -7

111 +11

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(93+11 or 111-7) 104 | -77 = (104-1) | (100-77) = 10323

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**Trick 4: Trick 2 and Trick 3 can be extended to any number with bases multiples of 10**.

**Trick 5: Multiplication with something half of multiple of 10**.

For illustration, let us say 50 as the base and we are multiplying 41 * 41

41 -9

41 -9

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(41 -9)32 | 81 = 16 | 81 = 1681.

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**Trick 6: Squaring**

(Base is 10 here ) Square of 12 = (12+2)|(square of 2)= 14|4= 144.

(Base is 100) Square of 94 = (94-6) | (square of -6) = 88|36=8836

**Trick 7: Multiply 2-digit number**

ax+b and cx+d, where x =10. The result is (a*c)(x^2)|((a*d)+(b*c))x|(b*d)

21 * 43 = (2*4)|((2*3)+(1*4))|(1*3) = 8|10|3=(8+1)|0|3=903

**Trick 8: Multiply 3-digit numbers**

a(x^2)+bx+c and d(x^2)+ex+f, where x 10. The result is (a*d)(x^4)+((a*e)+(b*d))(x^3)+((a*f)+(b*e)+(c*d))(x^2)+((b*f)+(c*e))x+(c*f).

**Trick 9: Squares of number ending in 5**.

25 = (2*(2+1))|(square of 5)=(2*3)|25=625

55 = (5*6)|25 = 3025

The above mentioned 9 tricks should help in faster calculation in all competitive exams.