Vedic Maths – Sutra 1. Ekadhikena Purvena (By one more than the previous one)


I Square of numbers ending in 5

75 x 75 = (7 x (7+1) ) 25 = (7×8) 25 = 5625

35 x 35 = (3 x (3+1) ) 25 = (3×4) 25 = 1225

125 x 125 = (12 x (12+1) 25 = (12 x 13) 25 = 15625

II When sum of the last digits is the base(10) and previous parts are the same

56 x 54 = (5 x (5+1)) (6×4) = 3024

III 1 divided by 19, 29, 39,…

take 1 / 19.In the divisor(19), previous one or the number before 9 is 1. By sutra,Eka adhika or by adding 1 more to the previous one, we get 2. Lets call the previous one+1 (here 2) as "x".In this method,we start from the end. There will be (divisor-1) terms in the answer. That is 18. Now, Assign last number to be 1.Now,multiply it with "x". ie , 2 1
Now go on multiplying with "x" for (divisor-1)/2 times (here,(19 − 1) / 2 = 9) ,ie,

Result : 9 4 7 3 6 8 4 2 1
Process:(2*x+1)|(7*x)|(3*x+1)|[6*x+1(carry of last
multiplication)]|(8*x)|(4*x)|(2*x)|(1*x)

In the next step, we write the compliment of 9 from the last number onwards,(divisor-1)/2 times

Result : 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1
Process : (9-i)|(9-h)|(9-g)|(9-f)|(9-e)|(9-d)|(9-c)|(9-b)|(9-a)|i|h|g|f|e|d|c|b|a

Now, prefix 0.,and this is your final answer, more accurate than a value that your calculator or computer can give.

1 / 19 = 0.0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1

IV 1 / 7 =7 / 49 previous digit is 4 so multiply 7 by 4+1 (=5,x) Result : 0 . 1 4 2 8 5 7
Process:0.(4*x+1)|(2*x+4)|(8*x+2)|(5*x+3(carry of last multiplication))|(7*x) (7 − 1 = 6 terms)

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