Tricks

__//**The vedic mathematics **//__

The vedic mathematics has many tricks to make the tricks easy.

An interesting sub-application of this formula is in computing squares of numbers ending in five. Examples: 35×35 = ((3×3)+3),25 = 12,25 and 125×125 = ((12×12)+12),25 = 156,25

or by the sūtra, multiply "by one more than the previous one." 35×35 = ((3×4),25 = 12,25 and 125×125 = ((12×13),25 = 156,25

The latter portion is multiplied by itself (5 by 5) and the previous portion is square of first digit or first two digit (3×3) or (12×12) and adding the same digit in that figure (3or12) resulting in the answer 1225.

(Proof) This is a simple application of (a + b)2 = a2 + 2ab + b2 when a = 10c and b = 5, i.e. (10c + 5)2 = 100c2 + 100c + 25 = 100c(c + 1)+25

It can also be applied in multiplications when the last digit is not 5 but the sum of the last digits is the base (10) and the previous parts are the same. Examples: 37 × 33 = (3 × 4),7 × 3 = 12,21 29 × 21 = (2 × 3),9 × 1 = 6,09 ?

This uses (a + b)(a − b) = a2 − b2 twice combined with the previous result to produce: (10c + 5 + d)(10c + 5 − d) = (10c + 5)2 − d2 = 100c(c + 1) + 25 − d2 = 100c(c + 1) + (5 + d)(5 − d).

Another trick

1 21 421 8421 68421 (carry 1) – we got 16, so we keep 6 and carry 1 368421 (carry 1) – we get 6*2 + carry 1 = 13, so we keep 3 and carry one

do this to eighteen digits (19–1. If you picked up 1/29, you'll have to do it till 28 digits). You'll get the following 1/19 = 052631578947368421 10100111101011000