| Ekadhikena
Purvena or "By one more than the previous one"
1. Square of numbers ending in 5
65 x 65 = (6
x (6+1) ) 25 = (6x7)
25 = 4225
45 x 45 = (4
x (4+1) ) 25 = (4x5)
25 = 2025
105 x 105 = (10
x (10+1) 25 = (10
x 11) 25 = 11025
2. When sum of the last digits is the base(10) and
previous parts are the same
44 x 46 = (4
x (4+1)) (4 x 6) = (4
x 5) (4 x 6) = 2024
37 x 33 = (3
x (3+1)) (7 x 3) = (3
x 4) (7 x 3) = 1221
11 x 19 = (1
x (1+1)) (1 x 9) = (1
x 2) (1 x 9) = 209
3. 1 divided by 19, 29, 39,..............
Consider 1/19 since 19 is not divisible by 2 or 5 it is a
purely a recurring decimal
take last digit 1
multiply this with 1+1
(one more) i.e 2 (this is the key digit) ==>21
multiply 2 by 2 ==> 421
multiplying 4 by 2 ==> 8421
multiply 8 by 2 ==> 68421
carry 1
multiply 6 by 2 =12 +
carry 1= 13 ==> 368421
carry 1
continuing (till 18
digits =denominator-numerator)
the result is 0.052631578947368421
4.
1/19 using divisions
divide 1 by 2, answer
is 0 with remainder 1 ==> .0
next 10 divided by 2 is
5 ==> .05
next 5 divided by 2 is
2 remainder 1 ==> 0.052
next 12 (remainder 2)
divided by 2 is 6 ==> 0.0526
next 6 divided by 2 is
3 ==> 0.05263
next 3 divided by 2 is
1 remainder 1 ==> 0.052631
next 11 divided by 2 is
5 remainder 1 ==> 0.0526315
and so on...
5. 1/7 = 7/49
previous digit is 4 so multiply by 4+1 i.e. by 5
7->
57 -> 857 ->
42857 -> 0.142857 (stop
after 7-1= 6 digits)
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