VEDIC MATHEMATICS TRICK

 Ekadhikena Purvena The Sutra (formula) Ekādhikena Pūrvena means: “By one more than the previous one”. i) Squares of numbers ending in 5 : Now we relate the sutra to the ‘squaring of numbers ending in 5’. Consider the example 25². Here the number is 25. We have to find out the square of the number. For the number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more than the previous one', that is, 2+1=3. The Sutra, in this context, gives the procedure'to multiply the previous digit 2 by one more than itself, that is, by 3'. It becomes the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S (right hand side) of the result is 5², that is, 25. Thus 25² = 2 X 3 / 25 = 625. In the same way, 35²= 3 X (3+1) /25 = 3 X 4/ 25 = 1225; 65²= 6 X 7 / 25 = 4225; 105²= 10 X 11/25 = 11025; 135²= 13 X 14/25 = 18225; Algebraic proof: a) Consider (ax + b)² = a². x²+ 2abx + b². This identity for x = 10 and b = 5 becomes (10a + 5) ² = a² . 10² + 2. 10a . 5 + 5² = a² . 10²+ a.10²+ 5² = (a²+ a ) . 10² +5² = a (a + 1) . 10² + 25. Clearly 10a + 5 represents two-digit numbers 15, 25, 35, -------,95 for the values a = 1, 2, 3, -- -----,9 respectively. In such a case the number (10a + 5)² is of the form whose L.H.S is a (a + 1) and R.H.S is 25, that is, a (a + 1) / 25. Thus any such two digit number gives the result in the same fashion. Example: 45 = (40 + 5)², It is of the form (ax+b)² for a = 4, x=10 and b = 5. giving the answer a (a+1) / 25 that is, 4 (4+1) / 25 = 4 X 5 / 25 = 2025. b) Any three digit number is of the form ax²+bx+c for x =10, a ≠ 0, a, b, c Є W. Now (ax²+bx+ c) ² = a² x⁴ + b²x² + c² + 2abx³ + 2bcx + 2cax² = a² x⁴+2ab.x³+(b²+ 2ca)x²+2bc . x+ c². This identity for x = 10, c = 5 becomes (a . 10² + b .10 + 5) ² = a².10⁴+ 2.a.b.10³ + (b²+ 2.5.a)10²+ 2.b.5.10 + 5² = a².10⁴+ 2.a.b.10³ + (b² + 10 a)10² + b.10²+ 5² = a².10⁴+ 2ab.10³+ b².10²+ a . 10³ + b 10²+ 5² = a².10⁴ + (2ab + a).10³ + (b²+ b)10² +5² = [ a².10² +2ab.10 + a.10 + b² + b] 10²+ 5² = (10a + b) ( 10a+b+1).10² + 25 = P (P+1) 10²+ 25, where P = 10a+b. Hence any three digit number whose last digit is 5 gives the same result as in (a) for P=10a + b, the ‘previous’ of 5. Example : 165²= (1 . 10² + 6 . 10 + 5) ². It is of the form (ax²+bx+c)² for a = 1, b = 6, c = 5 and x = 10. It gives the answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’. The answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.