Steps  14 × 12  
1.  Arrange the numbers vertically. For nonequidigit numbers, prefix as many Zeroes in front.  In this case, × 14 × 12  

2.  Consider sets of vertical columns, preferably from left to right, in ascending order, and then in descending order.  In this case, the sets are 1 14 4 1 , 12 , 2 Sets are obtained by including one vertical column at a time from left to right (ascending), and then again excluding one vertical column at a time from right to left (descending) 

3.  Find the sum of the crossproducts  of the digits equidistant from the edges, to represent each set.  In this case, (1 × 1), (1 × 2) + (4 × 1), (4 × 2) 

4.  Reformat the number in Base10 (Normalized) form, if required.  In this case, [1,6,8] = 168, which is the answer! 
For ndigit set, the CrossProduct is: 1. For evennumber of digits: Sum of the crosswise product of the digits, that are equidistant from the ends. 2. For oddnumber of digits: Sum of the crosswise product of the digits, that are equidistant from the ends  summed with centraldigits multiplied with each other. Then, for a singledigit set, m_{1} m_{2} CrossProduct is is m_{1} × m_{2} For a doubledigit set, m_{1}n_{1} m_{2}n_{2} CrossProduct is (m_{1} × n_{2})+(n_{1} × m_{2}) For a 3digit set, m_{1}n_{1}o_{1} m_{2}n_{2}o_{2} CrossProduct is (m_{1} × o_{2})+(o_{1} × m_{2})+(n_{1} × n_{2}) For a 4digit set, m_{1}n_{1}o_{1}p_{1} m_{2}n_{2}o_{2}p_{2} CrossProduct is (m_{1} × p_{2})+(p_{1} × m_{2})+(n_{1} × o_{2})+(o_{1} × n_{2}) For a 5digit set, m_{1}n_{1}o_{1}p_{1}q_{1} m_{2}n_{2}o_{2}p_{2}q_{2} CrossProduct is (m_{1} × q_{2})+(q_{1} × m_{2})+(n_{1} × p_{2})+(p_{1} × n_{2})+(o_{1} × o_{2}) And so on. 
× 873 × 234  Clearly, the sets are: 8 87 873 73 3 2 , 23 , 234 , 34 , 4 Which means, [8 × 2, (8 × 3)+(7 × 2), (8 × 4)+(3 × 2)+(7 × 3), (7 × 4)+(3 × 3), (3 × 4)] = [16, 24+14, 32+21+6, 28+9, 12] = [16, 38, 59, 37, 12] By śūddhikaran, we get: = [20,4,2,8,2] = 204282 Thus, 873 × 234 = 204,282 Note: For starters, the śūddhikaran in the final stages may get confusing. This has been discussed here » 
× 7196 × 4671  Clearly, the sets are: 7 71 719 7196 196 96 6 4 , 46 , 467 , 4671 , 671 , 71 , 1 Let us take a professional approach, from right to left. Considering the rightmost set, gives 6 × 1 = 6, as the last digit of the answer. Taking it ahead, for the set: 96 71, we get (9 × 1)+(6 × 7) = 51 So, our answer becomes ^{5}16 Again, for the set: 196 671, we get (1 × 1) + (6 × 6) + (9 × 7) = 1 + 36 + 63 = 100 Considering the carryover 5, this becomes 100 + 5 = 105 So, our answer becomes ^{10}516 For the next set: 7196 4671, we get (7 × 1)+(6 × 4)+(1 × 7)+(9 × 6) = 7 + 24 + 7 + 54 = 92 Considering the carryover 10, this becomes 91 + 10 = 102 So, our answer becomes ^{10}2516 Now, for the set: 719 467, we get (7 × 7)+(9 × 4)+(1 × 6) = 49 + 36 + 6 = 71 Considering the carryover 10, this becomes 71 + 10 = 81 So, our answer becomes ^{8}12516 For the next set: 71 46, we get (7 × 6)+(1 × 4) = 42 + 4 = 48 Considering the carryover 8, this becomes 48 + 8 = 56 So, our answer becomes ^{5}612516 Finally, for the set: 7 4, we get 7 × 4 = 28 Considering the carryover 5, this becomes 28 + 5 = 33 Which, gives us: 33612516 Thus, 7196 × 4671 = 33,612,516 
Assuming 2 twodigit numbers, mn and op, Let N_{1} = 10m + n And, N_{2} = 10o + p N_{1} × N_{2} = (10m + n)(10o + p) = 100om + 10mp + 10on + np = 100om + 10(mp + on) + np This is exactly what this Sūtra makes us do. Similarly, for 2 threedigit numbers, mno and pqr, Let N_{1} = 100m + 10n + o And, N_{2} = 100p + 10q + r N_{1} × N_{2} = (100m + 10n + o)(100p + 10q + r) = 10000mp + 1000mq + 100mr + 1000np + 100nq + 10nr + 100op + 10oq + or = 10000mp + 1000(mq + np) + 100(mr + nq + op) + 10 (nr + oq) + or This is exactly what this Sūtra makes us do. Note that, as the number of digits increase  so does the number of steps that this Sūtra makes us perform  in the same pattern as above. 
« Sūtra 2  Sūtra 4 » 