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), set-up Zeroes in front. | In this case, the 2nd part is of 1 digit and the result is 16, which is a 2-digit number - no Zeroes are set-up. |
|
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 set-up. 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 set-up. 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, 99952 = 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, 100052 = 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, 100502 = 101,002,500 |
Assuming two numbers (x + a) and (x + b), where a + b = 10 Then, (x + a) × (x + b) = x2 + ax + bx + ab = x2 + x(a + b) + ab But, a + b = 10, we get: = x2 + 10x + ab = x (x + 10) + ab This is exactly what this Upasūtra makes us do. Note that, if a + b = 10n, it may be similarly proven. The Upasūtra makes us adjust Zeroes, which takes care of any 10n. And, since the results are 'joined', (x + 10n) is always adjusted for (x + 100), which is (x + 1). |
« Upasūtra 7 | Upasūtra 9 » |