Steps  100  73  
1.  If the number to be subtracted (subtrahend) has lesser number of digits, than the Zeroes of 10^{n} (minuend)  prefix as many Zeroes in front.  In this case, 100 has 2 Zeroes, and 73 has 2 digits, so no Zeroes need to be prefixed. 

2.  Subtract all the digits from 9, respectively  except for the last digit.  In this case, 9  7 = 2 Note that the last digit, 3, is the digit in the Unit's place. 

3.  Subtract the last digit from 10.  In this case, the sets are 10  3 = 7 

4.  Join the numbers to get the answer.  In this case, Joining 2 and 7, we get 27, which is the answer! 
The last digit subtracted from 10 is 10  9 = 1 And, the other digits, each subtracted from 9 is: 9  2 = 7 9  4 = 5 Hence, 75 Joining both the results, we get: 751 Thus, 1000  249 = 751 
Note that 1000000 has 7 Zeroes, and 32 has two digits. So, we consider the minuend as 0000032 Now, subtracting all (except the last digit) from 9, we get: 999996 And the last digit, subtrated from 10, gives: 8 Thus, 1000000  32 = 9999968 
Assuming any number, xa, where a is the last digit. Then, N = 10x + a This also holds for number with more digits. Also, 10^{n} = 999999 ... (n1 times)..0 + 10 For example, 100 = 90 + 10 1000 = 990 + 10 100000 = 99990 + 10 So, 10^{n}  N = 10^{n}  (10x  a) = 999999 ... (n1 times)..0 + 10  (10x  a) = (999999 ... (n1 times)..0  10x) + (10a) = 10((999999 ... (n1 times)  x) + (10a) This is exactly what this Sūtra makes us do. 
Steps  99 × 98  
1.  Find a Base, that is near both the Operands  that is a power of 10.  In this case, Base = 100  
2.  Write the differences from the Base on the right hand side of the respective numbers.  In this case, the differences are 99 = 100  1, and 98 = 100  2 × 99 1 × 98 2  

3.  Derive the result in two parts. The RHS being the product of the differences, and the LHS being (Any of the Original Numbers) added with the other number's difference.  In this case, × 99 1 × 98 2  × [97 , 2] Note: 1 × 2 = 2, and 99  2 = 97 

4.  Convert the number to Base10, by multiplying the LHS with Base, added with the RHS.  In this case,
97 × 100 = 9700 And, 9700 + 2 = 9702, which is the answer! 
Note that, both are close to 100, which is the Base. × 97 3 × 104 +4  × [101 ,12] Note: 3 × 4 = 12, and 97 + 4 = 101 [101,12] in Base10 = (101 × 100)  12 = 10100  12 = 10088, which is the answer! 
Note that, both are close to 10,000, which is the Base. × 9989 11 × 9998 2  × [9987 ,22] Note: 11 × 2 = 22, and 9989  2 = 9987 [9987,22] in Base10 = (9987 × 10,000) + 22 = 99870000 + 22 = 99870022 Thus, 9989 × 9998 = 99,870,022 
Note that, both are close to 1,000,000, which is the Base. × 999991 9 × 1000003 +3  × [999994 ,27] Note: 9 × 3 = 27, and 999991 + 3 = 999994 [999994,27] in Base10 = (999994 × 1,000,000)  27 = 999994000000  27 = 999993999973 Thus, 999991 × 1000003 = 999,993,999,973 Note that the 999994000000  27, in the final stages  can be very easily done by using the same Sūtra (for subtracting from 10^{n}): 999994000000  27 = 999993000000 + 1000000  000027 = 999993000000 + 999973 = 999993999973 
Assuming two numbers, x and y, close to a common Base, 10^{n} Then, x = Base + a, and y = Base + b Now, x × y = (Base + a)(Base + b) = (Base)^{2} + a × Base + b × Base + ab = (Base)^{2} + Base (a + b) + ab But, Base = x  a = y  b So, substituting Base = (x  a) = Base (x  a) + Base (a + b) +ab = Base (x  a + a + b) + ab = Base (x  b) + ab = 10^{n}(x  b) + ab This is exactly what this Sūtra makes us do. 
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