# सूत्र २. निखिलं नवतश्चरमं दशत:

## (Sūtra 2. nikhilaṃ navataścaramaṃ daśataḥ) - All from 9, and the last from 10. The Sūtra: nikhilaṃ navataścaramaṃ daśataḥ (All from 9, and the last from 10) is used to subtract a number from 10n, and multiply two numbers close to 10n. This is a popular Sūtra, and practitioners of Vedic Mathematics find it very effective for number close to 10n, even over the famed Sūtra: ūrdhva tiryagbhyāṃ.

Subtraction of number from 10n:
As an illustration, let us use this Sūtra for:
100 - 73
 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. 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. 3. Subtract the last digit from 10. In this case, the sets are 10 - 3 = 7 4. Join the numbers to get the answer. In this case, Joining 2 and 7, we get 27, which is the answer!
So, for a practitioner of Vedic Mathematics, for something like:
1000 - 249
 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
Again, for something like:
1000000 - 32
 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
But, why does it work? For this Sūtra (nikhilaṃ navataścaramaṃ daśataḥ), let us consider the following:
 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.

Multiplication of numbers close to 10n:
Note that, this method is called the 'Nikhilam Method' of multiplication, and uses the same technique as the Upasūtra: ānurūpyeṇa (discussed here »).

As an illustration, let us use this Sūtra for:
99 × 98
 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 ----------- 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 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!
Let us take another example, for something like:
97 × 104
 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!
So, for a practitioner of Vedic Mathematics, for something like:
9989 × 9998
 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
And, for something like:
999991 × 1000003
 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
But, why does it work? For this Sūtra (nikhilaṃ navataścaramaṃ daśataḥ), let us consider the following:
 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.

The Sūtra: nikhilaṃ navataścaramaṃ daśataḥ is also discussed for division of a number - by a number, lesser than 10n. At Upavidhi, this technique is discussed with the Sūtra: parāvartya yojayet, which is a generic method.

Lastly, please remember that, as in any other form of mathematics, the mastery of Vedic Mathematics require practice and the judgement of applying the optimal method for a given scenario - a guideline of which, is presented in Applications »

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