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उपसूत्र ८. अन्त्ययोर्दशकेऽपि

(Upasūtra 8. antyayordaśake'pi) - Last totaling 10.
 

The Upasūtra: antyayordaśake'pi (Last totaling 10) is used for multiplication (x × y) of numbers, which may be broken into two parts:
1. The 1st part should be the same.
2. The 2nd part should be such that their sum equals 10 (or 10n)
 
Due to its limited domain of operability, this Sūtra is not amongst the most popular of Vedic Mathematical principles, but practitioners of Vedic Mathematics find it a handy technique.
 
As an illustration, let us use this Upasūtra for:
18 × 12
(Note that, for both the numbers, the 1st part are their first digit, and the second part 8 + 2 = 10)
  Steps 14 × 12
1. Multiply both the second parts. In this case,
8 × 2 = 16
2. If the number of digits of the above result is less than (Number of digits of the second part + 1), set-up Zeroes in front. In this case, the 2nd part is of 1 digit and the result is 16, which is a 2-digit number - no Zeroes are set-up.
3. Multiply the 1st part with its ekādhika (1st part + 1). In this case,
The 1st part is 1, and 1 × (1+1) = 1 × 2 = 2
4. Join both the results to get the answer. In this case,
Joining 2 and 16, we get 216, which is the answer!
Let us take another example, for something like:
101 × 109
  Note, that these numbers have the same 1st part, which is 10. And, sum of 2nd parts, 1 + 9 = 10
 
Product of 2nd parts: 1 × 9 = 9
But it is less than (Number of digits of the 2nd part + 1), so we make it 09
 
Product of 1st part, with its ekādhika: 10 × (10 + 1) = 10 × 11 = 110
 
Joining both the results, we get 11009, which is the answer!
Again, for something like:
9988 × 9912
  Note, that these numbers have the same 1st part, which is 99. And, sum of 2nd parts, 88 + 12 = 100
 
Product of 2nd parts: 88 × 12 = 1056
The number of digits is greater than (Number of digits of the 2nd part + 1), so no Zeroes are set-up.
 
Product of 1st part, with its ekādhika: 99 × (99 + 1) = 99 × 100 = 9900
 
Joining both the results, we get 99001056, which is the answer!
This Upasūtra can be used for any (n-digit) number, but gets cumbersome for bigger numbers. Let us take an example to demonstrate the same, for something like:
8388 × 8312
  Note, that these numbers have the same 1st part, which is 83. And, sum of 2nd parts, 88 + 12 = 100
 
Product of 2nd parts: 88 × 12 = 1056
The number of digits is greater than (Number of digits of the 2nd part + 1), so no Zeroes are set-up.
 
Product of 1st part, with its ekādhika: 83 × (83 + 1) = 83 × 84 = 6972
 
Joining both the results, we get 69721056, which is the answer!
As above, 83 × 84 would probably mean using another Vedic Mathematical principle - which is the reason that this Upasūtra is discussed for 2-digit and 3-digit numbers.
 
This Upasūtra, however, finds its optimal usage in squaring numbers ending with 5 (or for that matter, number ending with 50, 500, ...), but is popularly discussed with Sūtra: ekādhikena pūrveṇa. Let us take another example, for something like:
352
  The second part is: 5 × 5 = 25
Product of 1st part, with its ekādhika: 3 × (3+1) = 3 × 4 = 12
 
Joining both the results, we get 1225, which is the answer!
So, for a practitioner of Vedic Mathematics, for something like:
99952
  The second part is: 5 × 5 = 25
Product of 1st part, with its ekādhika: 999 × (999+1) = 999 × 1000 = 999000
 
Joining both the results, we get 99900025
 
Thus, 99952 = 99,900,025
And, for something like:
100052
  The second part is: 5 × 5 = 25
Product of 1st part, with its ekādhika: 1000 × (1000+1) = 1000 × 1001 = 1001000
 
Joining both the results, we get 100100025
 
Thus, 100052 = 100,100,025
Finally, for something like:
100502
  The second part is: 50 × 50 = 2500
Product of 1st part, with its ekādhika: 100 × (100+1) = 100 × 101 = 10100
 
Joining both the results, we get 101002500
 
Thus, 100502 = 101,002,500
But, why does it work? For this Upasūtra (antyayordaśake'pi), let us consider the following:
  Assuming two numbers (x + a) and (x + b), where a + b = 10
Then, (x + a) × (x + b)
= x2 + ax + bx + ab
= x2 + x(a + b) + ab
But, a + b = 10, we get:
= x2 + 10x + ab
= x (x + 10) + ab
 
This is exactly what this Upasūtra makes us do.
 
Note that, if a + b = 10n, it may be similarly proven. The Upasūtra makes us adjust Zeroes, which takes care of any 10n. And, since the results are 'joined', (x + 10n) is always adjusted for (x + 100), which is (x + 1).
 
Needless to state, that under the given conditions, if the last digit is 5 (for example, 105) with the same 1st part - we will end up with squaring the number. And the function of this Upasūtra, will be the same as Sūtra: ekādhikena pūrveṇa, of which it is a corollary.
 
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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