Steps  42 × 47  
1.  Find a Base, that is near both the Operands  that is a multiple of power of 10.  In this case, Base = 50 = 5 × 10  
2.  Write the differences from the Base on the right hand side of the respective numbers.  In this case, the differences are 42 = 50  8, and 47 = 50  3 × 42 8 × 47 3  

3.  Derive the result in two parts. The RHS being the product of the diffences, and the LHS being (Any of the Original Numbers) added with the other number's difference.  In this case, × 42 8 × 47 3  × [39 , 24] Note: 8 × 3 = 24, and 42  3 = 39 

4.  Convert the number to Base10, using the same proportion as Base10 is to the considered Base.  In this case, Base10:Base50 = 1:5 [39,24] in Base50 = [39 × 5, 24] in Base10 = [195,24] = 1950 + 24 = 1974, which is the answer! 
Assuming Base50, × 43 7 × 52 +2  × [45 ,14] Note: 7 × 2 = 14, and 43 + 2 = 45 [45,14] in Base10 = [45 × 5,14] = [225,14] = 2250  14 = 2236, which is the answer! 
Assuming Base500, × 469 31 × 487 13  × [456 ,403] Note: 31 × 13 = 403, and 469  13 = 456 [456,403] in Base10 = [456 × 50,403] = [22800,403] = 228000 + 403 = 228403 Thus, 469 × 487 = 228,403 
Assuming Base3000, × 2988 12 × 3014 +14  × [3002 ,168] Note: 12 × 14 = 168, and 2988 + 14 = 3002 [3002,168] in Base10 = (3002 × 3000)  168 = 9006000  168 = 9005832 Thus, 2988 × 3014 = 9,005,832 
Assuming Base50000, × 49999 1 × 50003 +3  × [50002 ,3] Note: 1 × 3 = 3, and 50003  1 = 50002 [50002,3] in Base10 = (50002 × 50000)  3 = 250010,0000  3 = 250009,9997 Thus, 49999 × 50003 = 2,500,099,997 
Assuming two numbers, x and y Assuming any Base, x = Base + a, and y = Base + b Then, 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 This is exactly what this Upasūtra makes us do. Note that, if the Base is 10  no further conversion is required. For any other Base, it is converted to Base10. 
« Sūtra 16  Upasūtra 2 » 