Steps | 100 - 73 | ||
1. | If the number to be subtracted (subtrahend) has lesser number of digits, than the Zeroes of 10n (minuend) - prefix as many Zeroes in front. | In this case, 100 has 2 Zeroes, and 73 has 2 digits, so no Zeroes need to be prefixed. |
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2. | Subtract all the digits from 9, respectively - except for the last digit. | In this case, 9 - 7 = 2 Note that the last digit, 3, is the digit in the Unit's place. |
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3. | Subtract the last digit from 10. | In this case, the sets are 10 - 3 = 7 |
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4. | Join the numbers to get the answer. | In this case, Joining 2 and 7, we get 27, which is the answer! |
The last digit subtracted from 10 is 10 - 9 = 1 And, the other digits, each subtracted from 9 is: 9 - 2 = 7 9 - 4 = 5 Hence, 75 Joining both the results, we get: 751 Thus, 1000 - 249 = 751 |
Note that 1000000 has 7 Zeroes, and 32 has two digits. So, we consider the minuend as 0000032 Now, subtracting all (except the last digit) from 9, we get: 999996 And the last digit, subtrated from 10, gives: 8 Thus, 1000000 - 32 = 9999968 |
Assuming any number, xa, where a is the last digit. Then, N = 10x + a This also holds for number with more digits. Also, 10n = 999999 ... (n-1 times)..0 + 10 For example, 100 = 90 + 10 1000 = 990 + 10 100000 = 99990 + 10 So, 10n - N = 10n - (10x - a) = 999999 ... (n-1 times)..0 + 10 - (10x - a) = (999999 ... (n-1 times)..0 - 10x) + (10-a) = 10((999999 ... (n-1 times) - x) + (10-a) This is exactly what this Sūtra makes us do. |
Steps | 99 × 98 | ||
1. | Find a Base, that is near both the Operands - that is a power of 10. | In this case, Base = 100 | |
2. | Write the differences from the Base on the right hand side of the respective numbers. | In this case, the differences are 99 = 100 - 1, and 98 = 100 - 2 × 99 -1 × 98 -2 ----------- |
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3. | Derive the result in two parts. The RHS being the product of the differences, and the LHS being (Any of the Original Numbers) added with the other number's difference. | In this case, × 99 -1 × 98 -2 ----------- × [97 , 2] Note: -1 × -2 = 2, and 99 - 2 = 97 |
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4. | Convert the number to Base10, by multiplying the LHS with Base, added with the RHS. | In this case,
97 × 100 = 9700 And, 9700 + 2 = 9702, which is the answer! |
Note that, both are close to 100, which is the Base. × 97 -3 × 104 +4 ------------ × [101 ,-12] Note: -3 × 4 = -12, and 97 + 4 = 101 [101,-12] in Base10 = (101 × 100) - 12 = 10100 - 12 = 10088, which is the answer! |
Note that, both are close to 10,000, which is the Base. × 9989 -11 × 9998 -2 -------------- × [9987 ,22] Note: -11 × -2 = 22, and 9989 - 2 = 9987 [9987,22] in Base10 = (9987 × 10,000) + 22 = 99870000 + 22 = 99870022 Thus, 9989 × 9998 = 99,870,022 |
Note that, both are close to 1,000,000, which is the Base. × 999991 -9 × 1000003 +3 ----------------- × [999994 ,-27] Note: -9 × 3 = -27, and 999991 + 3 = 999994 [999994,-27] in Base10 = (999994 × 1,000,000) - 27 = 999994000000 - 27 = 999993999973 Thus, 999991 × 1000003 = 999,993,999,973 Note that the 999994000000 - 27, in the final stages - can be very easily done by using the same Sūtra (for subtracting from 10n): 999994000000 - 27 = 999993000000 + 1000000 - 000027 = 999993000000 + 999973 = 999993999973 |
Assuming two numbers, x and y, close to a common Base, 10n Then, x = Base + a, and y = Base + b Now, x × y = (Base + a)(Base + b) = (Base)2 + a × Base + b × Base + ab = (Base)2 + Base (a + b) + ab But, Base = x - a = y - b So, substituting Base = (x - a) = Base (x - a) + Base (a + b) +ab = Base (x - a + a + b) + ab = Base (x - b) + ab = 10n(x - b) + ab This is exactly what this Sūtra makes us do. |
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