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उपसूत्र ६. यावदूनं तावदुनं

(Upasūtra 6. yāvadūnaṃ tāvadūnaṃ) - Lessen by the deficiency.

The Upasūtra: yāvadūnaṃ tāvadūnaṃ (Lessen by the deficiency) is used for cubing (x3) a number, that is close to a power of ten (10n). The technique followed by this Upasūtra uses Rekhanks & Vinculum Numbers (discussed here »).
 
As an illustration, let us use this Upasūtra for:
963
  Steps 963
1. Consider the nearest power of 10 as the Base In this case, the Base, closest to 96, is 100.
2. Find the deficiency, and represent as Rekhank for negative deficiency. In this case, the Deficiency =
100 - 96 = 4
3. Subtract the twice the deficiency from that number In this case,
96 - (2 × 4) = 88
4. Set-up twice of as many Zeroes, as the Base In this case, 100 has two Zeroes.
So, we get 88,0000
5. Multiply thrice the Deficiency, with the Deficiency - and set-up as many Zeroes as the Base. In this case, (3 × 4) × 4 = 12 × 4 = 48
And set-up with two Zeroes, we get 4800
6. Cube the Deficiency, and subtract from the sum previous numbers to get the answer. In this case, 43 = 64
And, 88,00,00 + 48,00 - 64 = 884736, which is the answer!
Let us take another example, for something like:
1033
  The Base is 100 and, the Deficiency is: 100 - 103 = -3 = 3
And, 33 = 27
 
Also, (3 × 3) × 3 = 27
 
Now, 103 - (2 × 3) = 103 + 6 = 109
Also, 109,00,00 + 2700 = 1092700
And, 1092700 - 27 = 1092700 + 27 = 1092727, which is the answer!
So, for a practitioner of Vedic Mathematics, for something like:
9913
  Clearly, the Deficiency is 9
And, 93 = 729
 
Also, (3 × 9) × 9 = 243
 
Now, 991 - (2 × 9) = 991 - 18 = 973
Also, 973,000,000 + 243,000 = 973,243,000
And, 973,243,000 - 729 = 973242271
 
Thus, 9913 = 973,242,271
 
Again, for something like:
100063
  Clearly, the Deficiency is 6
And, 63 = 216
 
Also, (3 × 6) × 6 = 108
 
Now, 10006 - (2 × 6) = 10006 + 12 = 10018
Also, 10018,0000,0000 + 108,0000 = 10018,0108,0000
And, 10018,0108,0000 - 216 = 10018,0108,0000 + 216 = 10018,0108,0216
 
Thus, 100063 = 1,001,801,080,216
Similarly, one take take any (n-digit) number and execute the steps above to obtain the desired calculated value.
 
But, why does it work? For the above Upasūtra (yāvadūnaṃ tāvadūnaṃ), let us consider the following:
  Assuming N is a number close (and less) to a power of 10, then N = a - b
'a' being the power of 10, and 'b' being the Deficiency
 
Now, N3 = (a - b)3
= a3 - 3a2b + 3ab2 - b3
= a3 - a2b - 2a2b + 3ab2 - b3
= a2(a - b - 2b) + a(3b × b) + b3
But, N = a - b. So, substituting a = N + b
N3 = a2(N + b - b - 2b) + a(3b × b) + b3
= a2(N - 2b) + a(3b × b) + b3
 
This is exactly what this Upasūtra makes us do.
 
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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