Steps  14 × 12  
1.  Multiply both the second parts.  In this case, 8 × 2 = 16 

2.  If the number of digits of the above result is less than (Number of digits of the second part + 1), setup Zeroes in front.  In this case, the 2nd part is of 1 digit and the result is 16, which is a 2digit number  no Zeroes are setup. 

3.  Multiply the 1st part with its ekādhika (1st part + 1).  In this case, The 1st part is 1, and 1 × (1+1) = 1 × 2 = 2 

4.  Join both the results to get the answer.  In this case, Joining 2 and 16, we get 216, which is the answer! 
Note, that these numbers have the same 1st part, which is 10. And, sum of 2nd parts, 1 + 9 = 10 Product of 2nd parts: 1 × 9 = 9 But it is less than (Number of digits of the 2nd part + 1), so we make it 09 Product of 1st part, with its ekādhika: 10 × (10 + 1) = 10 × 11 = 110 Joining both the results, we get 11009, which is the answer! 
Note, that these numbers have the same 1st part, which is 99. And, sum of 2nd parts, 88 + 12 = 100 Product of 2nd parts: 88 × 12 = 1056 The number of digits is greater than (Number of digits of the 2nd part + 1), so no Zeroes are setup. Product of 1st part, with its ekādhika: 99 × (99 + 1) = 99 × 100 = 9900 Joining both the results, we get 99001056, which is the answer! 
Note, that these numbers have the same 1st part, which is 83. And, sum of 2nd parts, 88 + 12 = 100 Product of 2nd parts: 88 × 12 = 1056 The number of digits is greater than (Number of digits of the 2nd part + 1), so no Zeroes are setup. Product of 1st part, with its ekādhika: 83 × (83 + 1) = 83 × 84 = 6972 Joining both the results, we get 69721056, which is the answer! 
The second part is: 5 × 5 = 25 Product of 1st part, with its ekādhika: 3 × (3+1) = 3 × 4 = 12 Joining both the results, we get 1225, which is the answer! 
The second part is: 5 × 5 = 25 Product of 1st part, with its ekādhika: 999 × (999+1) = 999 × 1000 = 999000 Joining both the results, we get 99900025 Thus, 9995^{2} = 99,900,025 
The second part is: 5 × 5 = 25 Product of 1st part, with its ekādhika: 1000 × (1000+1) = 1000 × 1001 = 1001000 Joining both the results, we get 100100025 Thus, 10005^{2} = 100,100,025 
The second part is: 50 × 50 = 2500 Product of 1st part, with its ekādhika: 100 × (100+1) = 100 × 101 = 10100 Joining both the results, we get 101002500 Thus, 10050^{2} = 101,002,500 
Assuming two numbers (x + a) and (x + b), where a + b = 10 Then, (x + a) × (x + b) = x^{2} + ax + bx + ab = x^{2} + x(a + b) + ab But, a + b = 10, we get: = x^{2} + 10x + ab = x (x + 10) + ab This is exactly what this Upasūtra makes us do. Note that, if a + b = 10^{n}, it may be similarly proven. The Upasūtra makes us adjust Zeroes, which takes care of any 10^{n}. And, since the results are 'joined', (x + 10^{n}) is always adjusted for (x + 10^{0}), which is (x + 1). 
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