Steps | 622 | ||
1. | Pair the digits, from right to left - in ascending order; and then, from left to right in descending order. | In this case, the pairs are {2},{62},{6} Pairs are obtained by including one digit at a time from right to left (ascending), and then again excluding one digit at a time from left to right (descending) |
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2. | Find duplexes for each of the pairs. | In this case, the Duplexes are 4, 24, 36 (Calculating a Duplex is explained below) |
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3. | Sum the duplexes by setting-up one zero at a time, by simple carry over method. | In this case, 4 + 240 + 3600 = 3844, which is the answer! |
For n-digit number, the Duplex is: 1. For even-number of digits: Sum of twice the digits, that are equidistant from the ends, multiplied. 2. For odd-number of digits: Sum of twice the digits, that are equidistant from the ends, multiplied - summed with central-digit multiplied with itself, that is squared. Then, for a single-digit number (m), Duplex is m2 For a double-digit number (mn), Duplex is 2(m × n) For a 3-digit number (mno), Duplex is 2(m × n) + o2 For a 4-digit number (mnop), Duplex is 2(m × p) + 2(n × o) For a 5-digit number (mnopq), Duplex is 2(m × q) + 2(n × p) + o2 And so on. As examples, Duplex of 3 is 32 = 9 Duplex of 6 is 62 = 36 Duplex of 23 is 2(2 × 3) = 2 × 6 = 12 Duplex of 64 is 2(6 × 4) = 2 × 24 = 48 Duplex of 128 is 2(1 × 8) + 22 = 16 + 4 = 20 Duplex of 305 is 2(3 × 5) + 02 = 30 + 0 = 30 Duplex of 4231 is 2(4 × 1) + 2(2 × 3) = 8 + 12 = 20 Duplex of 7346 is 2(7 × 6) + 2(3 × 4) = 84 + 24 = 108 Note: That the Duplex of two numbers can be the same (like 128 and 4231 above), especially for similar numbers (like 23 and 32). |
Note that the pairs are {4}, {34}, {234}, {23}, {2} Duplex of 4 is 16 Duplex of 34 is 24, set-up one zero to 240 Duplex of 234 is 16 + 9 = 25, set-up two zeroes to 2500 Duplex of 23 is 12, set-up three zeroes to 12,000 Duplex of 2 is 4, set-up four zeroes to 4,0000 Sum of Duplexes = 16 240 2500 12000 + 40000 -------- 54756 Thus, 2342 = 54,756 |
Note that the pairs are {6}, {26}, {426}, {1426}, {142}, {14}, {1} Duplex of 6 is 36 Duplex of 26 is 24, set-up one zero to 240 Duplex of 426 is 48 + 4 = 52, set-up two zeroes to 5200 Duplex of 1426 is 12 + 16 = 28, set-up three zeroes to 28,000 Duplex of 142 is 4 + 16 = 20, set-up four zeroes to 20,0000 Duplex of 14 is 8, set-up five zeroes to 8,00000 Duplex of 1 is 1, set-up six zeroes to 1,000000 Sum of Duplexes = 36 240 5200 28000 200000 800000 + 1000000 ---------- 2033476 Thus, 14262 = 2,033,476 Note that, the professional approach is to write the answer from right to left, without adding the Zeroes - summing the Duplexes on the go: Step 1: 36; 6 (carry 3) Step 2: 3 + 24 = 27; 7 (carry 2) Step 3: 2 + 52 = 54; 4 (carry 5) Step 4: 5 + 28 = 33; 3 (carry 3) Step 5: 3 + 20 = 23; 3 (carry 2) Step 6: 2 + 8 = 10; 0 (carry 1) Step 7: 1 + 1 = 2 Hence, 2033476 |
Assuming a two-digit number, mn, then N = 10m + n As in example above, 62 = 60 + 2 Now, N2 = (10m + n)2 = 100m2 + 2(10mn) + n2 = 100(m2) + 10(2mn) + n2 This is exactly what this Upasūtra makes us do: 622 = (100 × 36) + (10 × 24) + 4 = 3844 Assuming a 3-digit number, mno, then N = 100m + 10n + o As in example above, 234 = 200 + 30 + 4 Now, N2 = (100m + 10n + o)2 = ((100m + 10n) + o)2 = (100m + 10n)2 + 2((100m + 10n) × o) + o2 = [10000m2 + 2(100m × 10n) + 100n2] + [2(100mo) + 2(10no)] + o2 = [10000(m2) + 1000(2mn) + 100(n2)] + [100(2mo) + 10(2no)] + o2 = 10000(m2) + 1000(2mn) + 100((2mo) + n2) + 10(2no) + o2 This is exactly what this Upasūtra makes us do: 2342 = (10000 × 4) + (1000 × 12) + (100 × 25) + (10 × 24) + 16 = 54756 Note that, as the number of digits increase - so does the number of steps that this Upasūtra makes us perform - in the same pattern as above. |
Steps | 143 | ||
1. | Find the ratio of the two digits. | In this case, the Ratio is 1:4 | |
2. | Pair 4 terms, the first being the cube of 1st digit, and the remaining terms as Geometric Progression as the Ratio. | In this case, {1},{4},{16},{64} As above, 13 = 1, and 1 × 4 = 4; 4 × 4 = 16; 16 × 4 = 64 |
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3. | Pair 4 more terms, with 2nd and 3rd terms being the double of the original values. The others being set-up to zero. | In this case, it is {0},{8},{32},{0} | |
4. | Sum the terms, by simple carry over method. | In this case, {1 + 0},{4 + 8},{16 + 32},{64 + 0} = 2744, which is the answer! Note: [1,4,16,64] + [0,8,32, 0] -------------- Step 1: 64 + 0 = 64; 4 (carry 6) Step 2: 6 + 16 + 32 = 54; 4 (carry 5) Step 3: 5 + 4 + 8 = 17; 7 (carry 1) Step 4: 1 + 1 + 0 = 2 Hence, 2744 |
Clearly, the ratio is 1:8 And, 13 = 1 So, the first pair of terms are: {1},{8},{64},{512} And, the second pair of terms are: {0},{16},{128},{0} Sum of the pairs is: 5832 Note: [1, 8, 64,512] + [0,16,128, 0] ----------------- Step 1: 512 + 0 = 512; 2 (carry 51) Step 2: 51 + 64 + 128 = 243; 3 (carry 24) Step 3: 24 + 8 + 16 = 48; 8 (carry 4) Step 4: 4 + 1 + 0 = 5 Hence, 5832 Thus, 183 = 5,832 |
Clearly, the ratio is 2:8 = 1:4 And, 23 = 8 So, the first pair of terms are: {8},{32},{128},{512} And, the second pair of terms are: {0},{64},{256},{0} Sum of the pairs is: 21952 Note: [8,32,128,512] + [0,64,256, 0] ----------------- Step 1: 512 + 0 = 512; 2 (carry 51) Step 2: 51 + 128 + 256 = 435; 5 (carry 43) Step 3: 43 + 32 + 64 = 139; 9 (carry 13) Step 4: 13 + 8 + 0 = 21 Hence, 21952 Thus, 283 = 21,952 |
Clearly, the ratio is 7:7 = 1:1 And, 73 = 343 So, the first pair of terms are: {343},{343},{343},{343} And, the second pair of terms are: {0},{686},{686},{0} Sum of the pairs is: 456533 Note: [343,343,343,343] + [ 0,686,686, 0] -------------------- Step 1: 343 + 0 = 343; 3 (carry 34) Step 2: 34 + 343 + 686 = 1063; 3 (carry 106) Step 3: 106 + 343 + 686 = 1135; 5 (carry 113) Step 4: 113 + 343 + 0 = 456 Hence, 456533 Thus, 773 = 456,533 |
If we split the number into two, say: 1 and 21 Clearly, the ratio is 1:21 And, 13 = 1 So, the first pair of terms are: {1},{21},{441},{9261} And, the second pair of terms are: {0},{42},{882},{0} Sum of the pairs is: 1771561 Note: [1,21, 441,9261] + [0,42, 882, 0] ------------------- [1,63,1323,9261] Note again, that since the second part has 2 digits, the working Base is 100 So, [1,63,1323,9261] = 1,000000 + 63,0000 + 1323,00 + 9261 = 1771561 Thus, 1213 = 1,771,561 |
If we split the number into two, say: 2 and 78 Clearly, the ratio is 1:39 And, 23 = 8 So, the first pair of terms are: {8},{312},{12168},{474552} And, the second pair of terms are: {0},{624},{24336},{0} Sum of the pairs is: 21484952 Note: [8,312,12168,474552] + [0,624,24336, 0] ----------------------- [8,936,36504,474552] Note again, that since the second part has 2 digits, the working Base is 100 So, [8,936,36504,474552] = 8,000000 + 936,0000 + 36504,00 + 474552 = 21484952 Thus, 2783 = 21,484,952 |
If we split the number into two, say: 11 and 48 Clearly, the ratio is 11:48 = 1:(48/11) And, 113 = 1331 So, the first pair of terms are: {1331},{5808},{25344},{110592} And, the second pair of terms are: {0},{11616},{76032},{0} Sum of the pairs is: 1512953792 Note: [1331, 5808,25344,110592] + [ 0,11616,50688, 0] ---------------------------- [1331,17424,76032,110592] Note again, that since the second part has 2 digits, the working Base is 100 So, [1331,17424,76032,110592] = 1331,000000 + 17424,0000 + 76032,00 + 110592 = 1512953792 Thus, 11483 = 1,512,953,792 Again, if we split the number into two, say: 1 and 148 Clearly, the ratio is 1:148 And, 13 = 1 So, the first pair of terms are: {1},{148},{21904},{3241792} And, the second pair of terms are: {0},{296},{43808},{0} Sum of the pairs is: 1512953792 Note: [1,148,21904,3241792] + [0,296,43808, 0] ------------------------ [1,444,65712,3241792] Note again, that since the second part has 3 digits, the working Base is 1000 So, [1,444,65712,3241792] = 1,000000000 + 444,000000 + 65712,000 + 3241792 = 1512953792 Thus, 11483 = 1,512,953,792 |
Assuming a two-digit number, mn, then N = 10m + n As in examples, 14 = 10 + 4, 18 = 10 + 8, 28 = 20 + 8, 77 = 70 + 7 Now, N3 = (10m + n)3 = (10m + n)(10m + n)2 = (10m + n)(100m2 + 2(10mn) + n2) = [1000m3 + 2(100m2n) + 10mn2] + [100m2n + 20mn2 + n3] = 1000m3 + 3(100m2n) + 3(10mn2) + n3 This can be written as: = 1000m3 + 100m2n + 10mn2 + n3 + 100(2m2n) + 10(2mn2) This is exactly what this Upasūtra makes us do: 143 = 1000(1) + 100(4) + 10(16) + 64 + 100(8) + 10(32) The others, also being 2-digit numbers, are similar - as above. Note that, the ratios are important because that is how the power of n is being distributed. But more importantly, m3:m2n = m2n:n3 = m:n Similarly, numbers with more digits may be proven. |
« Upasūtra 15 |