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EKANYŨŅENA PŨRVENA
The Sutra Ekanyunena purvena comes as a Sub-sutra to Nikhilam which gives
the meaning 'One less than the previous' or 'One less than the one before'. 1) The use of this sutra in case of multiplication by 9,99,999.. is as follows . Method : a) The left hand side digit (digits) is ( are) obtained by applying the ekanyunena purvena i.e. by deduction 1 from the left side digit (digits) . e.g. ( i ) 7 x 9; 7 – 1 = 6 ( L.H.S. digit ) b) The right hand side digit is the complement or difference between the multiplier and the left hand side digit (digits) . i.e. 7 X 9 R.H.S is 9 - 6 = 3. c) The two numbers give the answer; i.e. 7 X 9 = 63. Example 1: 8 x 9 Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit ) Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit ) Step ( c ) gives the answer 72 Example 2: 15 x 99 Step ( a ) : 15 – 1 = 14 Step ( b ) : 99 – 14 = 85 ( or 100 – 15 ) Step ( c ) : 15 x 99 = 1485 Example 3: 24 x 99 Answer :  Example 4: 356 x 999 Answer :  Example 5: 878 x 9999 Answer :  Note the process : The multiplicand has to be reduced by 1 to obtain the LHS and the rightside is mechanically obtained by the subtraction of the L.H.S from the multiplier which is practically a direct application of Nikhilam Sutra. Now by Nikhilam 24 – 1 = 23 L.H.S. x 99 – 23 = 76 R.H.S. (100–24) _____________________________ 23 / 76 = 2376 Reconsider the Example 4: 356 – 1 = 355 L.H.S. x 999 – 355 = 644 R.H.S. ________________________ 355 / 644 = 355644 and in Example 5: 878 x 9999 we write 0878 – 1 = 877 L.H.S. x 9999 – 877 = 9122 R.H.S. __________________________ 877 / 9122 = 8779122 Algebraic proof : As any two digit number is of the form ( 10x + y ), we proceed ( 10x + y ) x 99 = ( 10x + y ) x ( 100 – 1 ) = 10x . 102 – 10x + 102 .y – y = x . 103 + y . 102 – ( 10x + y ) = x . 103 + ( y – 1 ) . 102 + [ 102 – ( 10x + y )] Thus the answer is a four digit number whose 1000th place is x, 100th place is ( y - 1 ) and the two digit number which makes up the 10th and unit place is the number obtained by subtracting the multiplicand from 100.(or apply nikhilam). Thus in 37 X 99. The 1000th place is x i.e. 3 100th place is ( y - 1 ) i.e. (7 - 1 ) = 6 Number in the last two places 100-37=63. Hence answer is 3663.
Apply Ekanyunena purvena to find out
the products
1. 64 x 99 2. 723 x 999 3. 3251 x 9999 4. 43 x 999 5. 256 x 9999 6. 1857 x 99999 i) When the multiplicand and multiplier both have the same number of digits ii) When the multiplier has more number of digits than the multiplicand. In both the cases the same rule applies. But what happens when the multiplier has lesser digits? i.e. for problems like 42 X 9, 124 X 9, 26325 X 99 etc., For this let us have a re-look in to the process for proper understanding. Multiplication table of 9. a b 2 x 9 = 1 8 3 x 9 = 2 7 4 x 9 = 3 6 - - - - - - - - - - 8 x 9 = 7 2 9 x 9 = 8 1 10 x 9 = 9 0 Observe the left hand side of the answer is always one less than the multiplicand (here multiplier is 9) as read from Column (a) and the right hand side of the answer is the complement of the left hand side digit from 9 as read from Column (b) Multiplication table when both multiplicand and multiplier are of 2 digits. a b 11 x 99 = 10 89 = (11–1) / 99 – (11–1) = 1089 12 x 99 = 11 88 = (12–1) / 99 – (12–1) = 1188 13 x 99 = 12 87 = (13–1) / 99 – (13–1) = 1287 ------------------------------------------------- 18 x 99 = 17 82 ---------------------------- 19 x 99 = 18 81 20 x 99 = 19 80 = (20–1) / 99 – (20–1) = 1980 The rule mentioned in the case of above table also holds good here Further we can state that the rule applies to all cases, where the multiplicand and the multiplier have the same number of digits. Consider the following Tables. (i) a b 11 x 9 = 9 9 12 x 9 = 10 8 13 x 9 = 11 7 ---------------------- 18 x 9 = 16 2 19 x 9 = 17 1 20 x 9 = 18 0 (ii) 21 x 9 = 18 9 22 x 9 = 19 8 23 x 9 = 20 7 ----------------------- 28 x 9 = 25 2 29 x 9 = 26 1 30 x 9 = 27 0 (iii) 35 x 9 = 31 5 46 x 9 = 41 4 53 x 9 = 47 7 67 x 9 = 60 3 -------------------------so on. From the above tables the following points can be observed: 1) Table (i) has the multiplicands with 1 as first digit except the last one. Here L.H.S of products are uniformly 2 less than the multiplicands. So also with 20 x 9 2) Table (ii) has the same pattern. Here L.H.S of products are uniformly 3 less than the multiplicands. 3) Table (iii) is of mixed example and yet the same result i.e. if 3 is first digit of the multiplicand then L.H.S of product is 4 less than the multiplicand; if 4 is first digit of the multiplicand then, L.H.S of the product is 5 less than the multiplicand and so on. 4) The right hand side of the product in all the tables and cases is obtained by subtracting the R.H.S. part of the multiplicand by Nikhilam. Keeping these points in view we solve the problems: Example1 : 42 X 9 i) Divide the multiplicand (42) of by a Vertical line or by the Sign : into a right hand portion consisting of as many digits as the multiplier. i.e. 42 has to be written as 4/2 or 4:2 ii) Subtract from the multiplicand one more than the whole excess portion on the left. i.e. left portion of multiplicand is 4. one more than it 4 + 1 = 5. We have to subtract this from multiplicand i.e. write it as 4 : 2 :-5 --------------- 3 : 7 This gives the L.H.S part of the product. This step can be interpreted as "take the ekanyunena and sub tract from the previous" i.e. the excess portion on the left. iii) Subtract the R.H.S. part of the multiplicand by nikhilam process. i.e. R.H.S of multiplicand is 2 its nikhilam is 8 It gives the R.H.S of the product i.e. answer is 3 : 7 : 8 = 378. Thus 42 X 9 can be represented as 4 : 2 :-5 : 8 ------------------ 3 : 7 : 8 = 378. Example 2 : 124 X 9 Here Multiplier has one digit only . We write 12 : 4 Now step (ii), 12 + 1 = 13 i.e. 12 : 4 -1 : 3 ------------ Step ( iii ) R.H.S. of multiplicand is 4. Its Nikhilam is 6 124 x 9 is 12 : 4 -1 : 3 : 6 ----------------- 11 : 1 : 6 = 1116 The process can also be represented as 124 x 9 = [ 124 – ( 12 + 1 ) ] : ( 10 – 4 ) = ( 124 – 13 ) : 6 = 1116 Example 3: 15639 x 99 Since the multiplier has 2 digits, the answer is [15639 – (156 + 1)] : (100 – 39) = (15639 – 157) : 61 = 1548261 1. 58 x 9 2. 62 x 9 3. 427 x 99 4. 832 x 9 5. 24821 x 999 6. 111011 x 99 Thus we have a glimpse of majority of the Sutras. At some places some Sutras are mentioned as Sub-Sutras. Any how we now proceed into the use of Sub-Sutras. As already mentioned the book on Vedic Mathematics enlisted 13 Upa-Sutras. But some approaches in the Vedic Mathematics book prompted some serious research workers in this field to mention some other Upa-Sutras. We can observe those approaches and developments also. |
