Steps | 14 × 12 | ||
1. | Arrange the numbers vertically. For non-equidigit numbers, prefix as many Zeroes in front. | In this case, × 14 × 12 ----- |
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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) |
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3. | Find the sum of the cross-products - of the digits equidistant from the edges, to represent each set. | In this case, (1 × 1), (1 × 2) + (4 × 1), (4 × 2) |
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4. | Re-format the number in Base10 (Normalized) form, if required. | In this case, [1,6,8] = 168, which is the answer! |
For n-digit set, the Cross-Product is: 1. For even-number of digits: Sum of the cross-wise product of the digits, that are equidistant from the ends. 2. For odd-number of digits: Sum of the cross-wise product of the digits, that are equidistant from the ends - summed with central-digits multiplied with each other. Then, for a single-digit set, m1 m2 Cross-Product is is m1 × m2 For a double-digit set, m1n1 m2n2 Cross-Product is (m1 × n2)+(n1 × m2) For a 3-digit set, m1n1o1 m2n2o2 Cross-Product is (m1 × o2)+(o1 × m2)+(n1 × n2) For a 4-digit set, m1n1o1p1 m2n2o2p2 Cross-Product is (m1 × p2)+(p1 × m2)+(n1 × o2)+(o1 × n2) For a 5-digit set, m1n1o1p1q1 m2n2o2p2q2 Cross-Product is (m1 × q2)+(q1 × m2)+(n1 × p2)+(p1 × n2)+(o1 × o2) 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 right-most 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 516 Again, for the set: 196 671, we get (1 × 1) + (6 × 6) + (9 × 7) = 1 + 36 + 63 = 100 Considering the carry-over 5, this becomes 100 + 5 = 105 So, our answer becomes 10516 For the next set: 7196 4671, we get (7 × 1)+(6 × 4)+(1 × 7)+(9 × 6) = 7 + 24 + 7 + 54 = 92 Considering the carry-over 10, this becomes 91 + 10 = 102 So, our answer becomes 102516 Now, for the set: 719 467, we get (7 × 7)+(9 × 4)+(1 × 6) = 49 + 36 + 6 = 71 Considering the carry-over 10, this becomes 71 + 10 = 81 So, our answer becomes 812516 For the next set: 71 46, we get (7 × 6)+(1 × 4) = 42 + 4 = 48 Considering the carry-over 8, this becomes 48 + 8 = 56 So, our answer becomes 5612516 Finally, for the set: 7 4, we get 7 × 4 = 28 Considering the carry-over 5, this becomes 28 + 5 = 33 Which, gives us: 33612516 Thus, 7196 × 4671 = 33,612,516 |
Assuming 2 two-digit numbers, mn and op, Let N1 = 10m + n And, N2 = 10o + p N1 × N2 = (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 three-digit numbers, mno and pqr, Let N1 = 100m + 10n + o And, N2 = 100p + 10q + r N1 × N2 = (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. |
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