# Matemática Vedica

Páginas: 2 (308 palabras) Publicado: 12 de octubre de 2012
Matemáticas védicas
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 same44 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 x9) = (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 arecurring 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 ==> 421multiplying 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 ==> .0next 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 6divided 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
andso 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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