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YĀVADŨNAM TĀVADŨNĪKŖTYA
The meaning of the Sutra is 'what ever the deficiency subtract that deficit
from the number and write along side the square of that deficit'. VARGAÑCA YOJAYET This Sutra can be applicable to obtain squares of numbers close to bases of powers of 10. Method-1 : Numbers near and less than the bases of powers of 10. Eg 1: 92 Here base is 10. The answer is separated in to two parts by a’/’ Note that deficit is 10 - 9 = 1 Multiply the deficit by itself or square it 12 = 1. As the deficiency is 1, subtract it from the number i.e., 9–1 = 8. Now put 8 on the left and 1 on the right side of the vertical line or slash i.e., 8/1. Hence 81 is answer. Eg. 2: 962 Here base is 100. Since deficit is 100-96=4 and square of it is 16 and the deficiency subtracted from the number 96 gives 96-4 = 92, we get the answer 92 / 16 Thus 962 = 9216. Eg. 3: 9942 Base is 1000 Deficit is 1000 - 994 = 6. Square of it is 36. Deficiency subtracted from 994 gives 994 - 6 = 988 Answer is 988 / 036 [since base is 1000] Eg. 4: 99882 Base is 10,000. Deficit = 10000 - 9988 = 12. Square of deficit = 122 = 144. Deficiency subtracted from number = 9988 - 12 = 9976. Answer is 9976 / 0144 [since base is 10,000]. Eg. 5: 882 Base is 100. Deficit = 100 - 88 = 12. Square of deficit = 122 = 144. Deficiency subtracted from number = 88 - 12 = 76. Now answer is 76 / 144 = 7744 [since base is 100] Algebraic proof: The numbers near and less than the bases of power of 10 can be treated as (x-y), where x is the base and y, the deficit. Thus (1) 9 = (10 -1) (2) 96 = ( 100-4) (3) 994 = (1000-6) (4) 9988 = (10000-12 ) (v) 88 = (100-12) ( x – y )2 = x2 – 2xy + y2 = x ( x – 2y ) + y2 = x ( x – y – y ) + y2 = Base ( number – deficiency ) + ( deficit )2 Thus 9852 = ( 1000 – 15 )2 = 1000 ( 985 – 15 ) + (15)2 = 1000 ( 970 ) + 225 = 970000 + 225 = 970225. or we can take the identity a2 - b2 = (a + b) ( a - b) and proceed as a2 - b2 = (a + b) ( a - b). gives a2 = (a + b) ( a - b) + b2 Thus for a = 985 and b = 15; a2= (a + b) ( a - b) + b2 9852 = ( 985 + 15 ) ( 985 - 15 ) + (15)2 = 1000 ( 970 ) + 225 = 970225. Method. 2 : Numbers near and greater than the bases of powers of 10. Eg.(1): 132 . Instead of subtracting the deficiency from the number we add and proceed as in Method-1. for 132 , base is 10, surplus is 3. Surplus added to the number = 13 + 3 = 16. Square of surplus = 32 = 9 Answer is 16 / 9 = 169. Eg.(2): 1122 Base = 100, Surplus = 12, Square of surplus = 122 = 144 add surplus to number = 112 + 12 = 124. Answer is 124 / 144 = 12544 Or think of identity a2 = (a + b) (a – b) + b2 for a = 112, b = 12: 1122 = (112 + 12) (112 – 12) + 122 = 124 (100) + 144 = 12400 + 144 = 12544. (x + y)2 = x2 + 2xy + y2 = x ( x + 2y ) + y2 = x ( x + y + y ) + y2 = Base ( Number + surplus ) + ( surplus )2 gives 1122 = 100 ( 112 + 12 ) + 122 = 100 ( 124 ) + 144 = 12400 + 144 = 12544. Eg. 3: 100252 = ( 10025 + 25 ) / 252 = 10050 / 0625 [ since base is 10,000 ] = 100500625. Method - 3: This is applicable to numbers which are near to multiples of 10, 100, 1000 .... etc. For this we combine the upa-Sutra 'anurupyena' and 'yavadunam tavadunikritya varganca yojayet' together. Example 1: 3882 Nearest base = 400. We treat 400 as 4 x 100. As the number is less than the base we proceed as follows Number 388, deficit = 400 - 388 = 12 Since it is less than base, deduct the deficit i.e. 388 - 12 = 376. multiply this result by 4 since base is 4 X 100 = 400. 376 x 4 = 1504 Square of deficit = 122 = 144. Hence answer is 1504 / 144 = 150544 [since we have taken multiples of 100]. Example 2: 4852 Nearest base = 500.  Example 3: 672 Nearest base = 70  Here surplus = 16 and 400 = 4 x 100  Surplus = 12  1. 72 2. 982 3. 9872 4. 142 5. 1162 6. 10122 7. 192 8. 4752 9. 7962 10. 1082 11. 99882 12. 60142.
Cubing of Numbers:
Example : Find the cube of the number 106. We proceed as follows:
i) For 106, Base is 100. The surplus is 6.
Now proceeding from right to left and adjusting the
carried over, we get the answerHere we add double of the surplus i.e. 106+12 = 118. (Recall in squaring, we directly add the surplus) This makes the left-hand -most part of the answer. i.e. answer proceeds like 118 / - - - - - ii) Put down the new surplus i.e. 118-100=18 multiplied by the initial surplus i.e. 6=108. Since base is 100, we write 108 in carried over form 108 i.e. . As this is middle portion of the answer, the answer proceeds like 118 / 108 /.... iii) Write down the cube of initial surplus i.e. 63 = 216 as the last portion i.e. right hand side last portion of the answer. Since base is 100, write 216 as 216 as 2 is to be carried over. Answer is 118 / 108 / 216 119 / 10 / 16 = 1191016. Eg.(1): 1023 = (102 + 4) / 6 X 2 / 23 = 106 = 12 = 08 = 1061208. Observe initial surplus = 2, next surplus =6 and base = 100. Eg.(2): 943 Observe that the nearest base = 100. Here it is deficit contrary to the above examples.
i) Deficit = -6. Twice of it -6 X 2 = -12
add it to the number = 94 -12 =82. ii) New deficit is -18. Product of new deficit x initial deficit = -18 x -6 = 108 iii) deficit3 = (-6)3 = -216.
__
Since 100 is base 1 and -2 are the carried over. Adjusting the carried over
in order, we get the answerHence the answer is 82 / 108 / -216 ( 82 + 1 ) / ( 08 – 03 ) / ( 100 – 16 ) = 83 / = 05 / = 84 = 830584 __ 16 becomes 84 after taking1 from middle most portion i.e. 100. (100-16=84). _ Now 08 - 01 = 07 remains in the middle portion, and 2 or 2 carried to it makes the middle as 07 - 02 = 05. Thus we get the above result. Eg.(3): 9983 Base = 1000; initial deficit = - 2. 9983 = (998 – 2 x 2) / (- 6 x – 2) / (- 2)3 = 994 / = 012 / = -008 = 994 / 011 / 1000 - 008 = 994 / 011 / 992 = 994011992. 1. 105 2. 114 3. 1003 4. 10007 5. 92 6. 96 7. 993 8. 9991 9. 1000008 10. 999992. |
