matha namo namaha


TERMS AND OPERATIONS 
a) Ekadhika means ‘one more’
    e.g:  Ekadhika of 0 is 1
                 Ekadhika of 1 is 2
                 -----------------
                 Ekadhika of 8 is 9
                 -------------------
                 Ekadhika of 23 is 24
                 ---------------------
                 Ekadhika of 364 is 365------
b) Ekanyuna means ‘one less’
    e.g:   Ekanyuna of  1 2 3 ..... 8 ..... 14 .....69 ......
                   is                0 1 2 ..... 7 ......13 .... 68 ......
c) Purak means ‘ complement’
    e.g:   purak of   1 2 3 ..... 8., 9 from 10
                   is            9 8 7 ..... 2 1
d) Rekhank means ‘a digit with a bar on its top’. In other words it is a negative number.

                                     _
         e.g:  A bar on 7 is 7. It is called rekhank 7 or bar 7. We treat purak as a Rekhank.
               _             _
        e.g: 7 is 3 and 3 is 7
    At some instances we write negative numbers also with a bar on the top of the numbers as
                                             _
                -4 can be shown as 4.
                                               __
                -21 can be shown as 21.
e) Addition and subtraction using Rekhank.
    Adding a bar-digit i.e. Rekhank to a digit means the digit is subtracted.
                        _              _              _
          e.g:  3 + 1 = 2, 5 + 2 = 3, 4 + 4 = 0
    Subtracting a bar - digit i.e. Rekhank to a digit means the digit is added.
                      _             _              _
          e.g: 4 - 1 = 5, 6 - 2 = 8, 3 - 3 = 6
    1. Some more examples
     e.g:  3 + 4 = 7
                                    _    _    _
                (-2) + (-5) = 2 + 5 = 7 or -7
f) Multiplication and Division using rekhank.
1. Product of two positive digits or two negative digits ( Rekhanks )
                               _    _
        e.g: 2 X 4 = 8; 4 X 3 = 12 i.e. always positive
2. Product of one positive digit and one Rekhank
                     _    _          _          __
        e.g: 3 x 2 = 6 or -6; 5 X 3 = 15 or -15 i.e. always Rekhank or negative.
3. Division of one positive by another or division of one Rekhank by another Rekhank.
                               _    _
        e.g: 8 ÷ 2 = 4, 6 ÷ 3 = 2 i.e. always positive
4. Division of a positive by a Rekhank or vice versa.
                __          _        _    _
        e.g:  10 ÷ 5 = 2, 6 ÷ 2 = 3 i.e. always negative or Rekhank.
g) Beejank: The Sum of the digits of a number is called Beejank. If the addition is a two digit number, Then these two digits are also to be added up to get a single digit.
  e.g: Beejank of 27 is 2 + 7 = 9.
         Beejank of 348 is 3 + 4 + 8 = 15
         Further 1 + 5 = 6. i.e. 6 is Beejank.
         Beejank of 1567 --->1 + 5 + 6 + 7 -->19 -->1 + 9 -->1
         i.e. Beejank of 1567 is 1.
ii) Easy way of finding Beejank:
Beejank is unaffected if 9 is added to or subtracted from the number. This nature of 9 helps in finding Beejank very quickly, by cancelling 9 or the digits adding to 9 from the number.
eg 1: Find the Beejank of 632174.
        As above we have to follow
        632174 -->6 + 3 + 2 + 1 + 7 + 4 -->23 -->2 + 3 -->5
But a quick look gives 6 & 3 ; 2 & 7 are to be ignored because 6+3=9,2+7=9. Hence remaining 1 + 4  -->5 is the beejank of 632174.
eg 2:
Beejank of 1256847 -->1+2+5+6+8+4+7 -->33 -->3+3 -->6.
But we can cancel 1& 8, 2& 7, 5 & 4 because in each case the sum is 9. Hence remaining 6 is the Beejank.
h) Check by Beejank method:
The Vedic sutra - Gunita Samuccayah gives ‘the whole product is same’. We apply this sutra in this context as follows. It means that the operations carried out with the numbers have same effect when the same operations are carried out with their Beejanks.
Observe the following examples.
i) 42 + 39
Beejanks of 42 and 39 are respectively 4 + 2 = 6 and 3 + 9 = 12 and 1+2=3
Now 6 + 3 = 9 is the Beejank of the sum of the two numbers
Further 42 + 39 = 81. Its Beejank is 8+ 1 = 9.
we have checked the correctness.
ii)   64 + 125.

           64 6 + 4 10 1 + 0 1

           125 1 + 2 + 5 8
    Sum of these Beejanks 8 + 1 = 9
    Note that

            64 + 125 = 189 1 + 8 + 9 18 1 + 8 9
we have checked the correctness.
iii)  134 - 49

           134 1 + 3 + 4 8

            49 4+9 13 1 + 3 4
Difference of Beejanks 8 -4 4, note that 134 – 49 = 85
Beejanks of 85 is 8 + 5 85 8 + 5 13 1 + 3 4 verified.
iv)  376 - 284

        376 7 ( 6 + 3 9)

        284 2 + 8 + 4 14 1 + 4 5
Difference of Beejanks = 7 – 5 = 2
   376 – 284 = 92
Beejank of  92 9 + 2 11 1 + 1 2
Hence verified.
v)  24 X 16 = 384
Multiplication of Beejanks of
    24 and 16 is 6 X 7 = 42 4 + 2 6
Beejank of 384 3 + 8 + 4 15 1 + 5 6
Hence verified.
vi)  237 X 18 = 4266
Beejank of 237 2 + 3 + 7 12 1 + 2 3
Beejank of 18 1 + 8 9
Product of the Beejanks = 3 X 9 27 2 + 7 9
Beejank of 4266 4 + 2 + 6 + 6 18 1 + 8 9
Hence verified.
vii)  242 = 576
Beejank of 24 2 + 4 6
square of it 62 36 9
Beejank of result = 576 5 + 7 + 6 18 1 + 8 9
Hence verified.
viii)  3562 = 126736
Beejank of 356 3 + 5 + 6 5
Square of it = 52 = 25 2 + 5 7
Beejank of result 126736 1 + 2 + 6 + 7 + 3 + 6 1 + 6 7
( 7 + 2 = 9; 6 + 3 = 9) hence verified.
ix) Beejank in Division:
Let P, D, Q and R be respectively the dividend, the divisor, the quotient and the remainder.
Further the relationship between them is P = ( Q X D ) + R
eg 1:  187 ÷ 5
we know that 187 = ( 37 X 5 ) + 2 now the Beejank check.
We know that 187 = (37 X 5) +2 now the Beejank check.
187 1 + 8 + 7 7( 1 + 8 = 9)
(37 X 5) + 2 Beejank [(3 + 7) X 5] + 2
            5 + 2 7
Hence verified.
eg 2:  7986 ÷ 143
    7896 = (143 X 55) + 121
Beejank of 7986 7 + 9 + 8 + 6 21
( 9 is omitted) 2 + 1 3
Beejank of 143 X 55 (1 + 4 + 3) (5 + 5)
8 X 10 80 (8 + 0) 8
Beejank of (143 X 55) + 121 8 + (1 + 2 + 1)
8 + 4 12 1 + 2 3
hence verified.
Check the following results by Beejank method

        1.  67 + 34 + 82 = 183          2.  4381 - 3216 = 1165

        3.  632 = 3969                        4.  (1234)2 = 1522756

        5.  (86X17) + 34 = 1496      6.  2556 ÷ 127 gives Q =20, R = 16
i) Vinculum : The numbers which by presentation contains both positive and negative digits are called vinculum numbers.

ii) Conversion of general numbers into vinculum numbers.
We obtain them by converting the digits which are 5 and above 5 or less than 5 without changing the value of that number.
Consider a number say 8. (Note it is greater than 5). Use it complement (purak - rekhank) from 10. It is 2 in this case and add 1 to the left (i.e. tens place) of 8.
                             _
    Thus  8 = 08 = 12.

    The number 1 contains both positive and negative digits
                   _           _
    i.e. 1 and 2 . Here 2 is in unit place hence it is -2 and value of 1 at tens place is 10.
              _
    Thus 12 = 10 - 2 = 8

Conveniently we can think and write in the following way
 General Number
 Conversion
 Vinculum number
   _
 6
 10 - 4
 14
    _
 97
 100 - 3
 103
    __
 289
 300 - 11
  311
etc.,,
The sutras ‘Nikhilam Navatascharamam Dasatah’ and ‘Ekadhikena purvena’ are useful for conversion.
eg 1:  289, Edadhika of 2 is 3
                                                      _
          Nikhilam from 9 : 8 - 9 = -1 or 1
                                                           _
          Charmam from 10 :9 -10 = -1 or  1
                                                __
          i.e. 289 in vinculum form 311
eg 2:  47768

        ‘Ekadhika’ of 4 is 5
                                             ___
        ‘Nikhilam’ from 9 (of 776) 223
                                             _
        ‘Charmam from 10 (of 8)  2
                                          ____
        Vinculum of 47168 is 5 2232
eg 3:  11276.
Here digits 11 are smaller. We need not convert. Now apply for 276 the two sutras Ekadhika of 2 is 3
                                           __
     ‘Nikhilam Navata’ for 76 is 24
                           __
        11276 = 11324
                 __               
      i.e. 11324 = 11300 - 24 = 11276.
The conversion can also be done by the sutra sankalana vyavakalanabhyam as follows.
eg 4:  315.
sankalanam (addition) = 315+315 = 630.
                                                                     _
     Vyvakalanam (subtraction) = 630 - 315 = 325

     Working steps :            _
                            0 - 5 = 5

                            3 - 1 = 2

                            6 - 3 = 3
Let’s apply this sutra in the already taken example 47768.
Samkalanam = 47768 + 47768 = 95536
Vyavakalanam = 95536 - 47768.

            
Consider the convertion by sankalanavyavakalanabhyam and check it by Ekadhika and Nikhilam.
eg 5: 12637
1. Sankalana ....... gives,  12637 + 12637 = 25274
                                                                                                        _ _
        25274 – 12637 = (2 – 1) / (5 – 2) / (2 – 6) / (7 – 3) / (4 – 7) = 13443
2. Ekadhika and Nikhilam gives the following.
As in the number 1 2 6 3 7, the smaller and bigger digits (i.e. less than 5 and; 5, greater than 5) are mixed up, we split up in to groups and conversion is made up as given below.
Split 1 2 6 and 3 7
                                                   _                 _
     Now the sutra gives 1 2 6 as 134 and 37 as 43
                               _  _
     Thus  12637 = 13443
                                                                                                        _
    Now for the number 315 we have already obtained vinculum as 325 by "sankalana ... " Now by ‘Ekadhika and Nikhilam ...’ we also get the same answer.
315 Since digits of 31 are less than 5,
We apply the sutras on 15 only as
Ekadhika of 1 is 2 and Charman of 5 is 5 .
Consider another number which comes under the split process.
eg 6:   24173
As both bigger and smaller numbers are mixed up we split the number 24173 as 24 and 173 and write their vinculums by Ekadhika and Nikhilam sutras as
                  _                  __
        24 = 36 and 173 = 227
                               _  __
        Thus 24173 = 36227
Convert the following numbers into viniculum number by
    i. Ekadhika and Nikhilam sutras  ii. Sankalana vyavakalana sutra. Observe whether in any case they give the same answer or not.

        1.  64                      2.  289             3.  791

        4.  2879                  5.  19182         6.  823672

        7. 123456799         8.  65384738
ii) Conversion of vinculum number into general numbers.
The process of conversion is exactly reverse to the already done. Rekhanks are converted by Nikhilam where as other digits by ‘Ekanyunena’ sutra. thus:
     _
i) 12 = (1 – 1) / (10 – 2) Ekanyunena 1 – 1
                                                 _
        = 08 = 8               Nikhilam. 2 = 10 – 2
     __
ii) 326 = (3 – 1) / (9 – 2) / (10 – 6)

          = 274
      _  _
iii) 3344 = (3 – 1) / (10 – 3) / (4 – 1) / (10 – 4)

            = 2736         (note the split)
        __ __
iv) 20340121 = 2/(0–1)/(9–3)/(10–4)/(0–1)/(9–1)/(10–2)/1
                        _   _
                   = 21661881
                        _          _
                   = 21 / 6 / 61 / 881. once again split

                   = (2 – 1) / (10 –1) / 6 / (6 –1) / (10 –1) / 881

                   = 19659881
         ___
v)  303212 = 3 / 0321 / 2

                = 3 / (0-1) / (9-3) / (9-2) / (10-1) / 2
                        _
                   3 / 1 / 6792

                   (3 –1) / (10 –1) / 6792

                = 296792.

iii) Single to many conversions.
It is interesting to observe that the conversions can be shown in many ways.
eg 1:  86 can be expressed in following many ways
                             __
        86 = 90 - 4 =94
                                  __
            = 100 – 14 = 114
                                      ___
            = 1000 – 914 = 1914
                      _      __      ___      ____
    Thus 86 = 94 = 114 = 1914 = 19914 = ………….
eg 2 :
                  _                  _
                07 = -10 +3 = 13
               __                       _
               36 = -100 + 64 = 164
              ___                         _
              978 = -1000 + 22 = 1022. etc.,
* Convert by Vedic process the following numbers into vinculum numbers.

        1)  274         2)  4898         3)  60725         4)  876129.
* Convert the following vinculum numbers into general form of numbers ( normalized form)

        1) 283         2)  3619          3) 27216

        4) 364718                           5) 60391874
(iv) Vedic check for conversion:
The vedic sutra "Gunita Samuctayah" can be applied for verification of the conversion by Beejank method.
Consider a number and find its Beejank. Find the vinculum number by the conversion and find its Beejank. If both are same the conversion is correct.
eg.
                       _
            196 = 216 . Now Beejank of 196 1 + 6 7
Beejank of 216 2 + ( -1 ) + 6 7.    Thus verified.
But there are instances at which, if beejank of vinculum number is rekhank i.e. negative. Then it is to be converted to +ve number by adding 9 to Rekhank ( already we have practised) and hence 9 is taken as zero, or vice versa in finding Beejank.
eg:
             __
            213 = 200 - 13 = 187
                                       _
            Now Beejank of 213 = 2 + ( -1 ) + (-3 ) = -2

            Beejank of 187 = 1 + 8 + 7 = 16 1 + 6 = 7
The variation in answers can be easily understood as
                            _    _
                            2 = 2 + 9 - 2 + 9 = 7 Hence verified.
Use Vedic check method of the verification of the following result.
                       _                                    _  _
        1.  24 = 36                2.  2736 = 3344.
               __                             _  _
        3.  326 = 274            4.  23213 = 17187
Addition and subtraction using vinculum numbers.
eg 1:  Add 7 and 6 i.e., 7+6.
i) Change the numbers as vinculum numbers as per rules already discussed.
                           _               _
                { 7 = 13 and 6 = 14 }
ii) Carry out the addition column by column in the normal process, moving from top to bottom or vice versa.

               
iii) add the digits of the next higher level i.e.,, 1 + 1 = 2
                          _
                        13
                          _
                        14
                       ____
                          _
                        27
iv) the obtained answer is to be normalized as per rules already explained. rules already explained.
                _
        i.e., 27 = (2 - 1) (10- 7) = 13 Thus we get 7 + 6 = 13.
eg 2 : Add 973 and 866.
                                  _                               _
                   973 = 1 0 3 3                      1 0 3 3
                                _ _ _                        _ _ _
                   866 = 1 1 3 4                      1 1 3 4
                                                             ______
                                                                _ _ _
                                                             2 1 6 1
                    ___
            But 2161 = 2000 - 161 = 1839.
Thus 973+866 by vinculum method gives 1839 which is correct.
Observe that in this representation the need to carry over from the previous digit to the next higher level is almost not required.
eg 3 : Subtract 1828 from 4247.

            i.e.,,  4247
                   -1828
                   ______
                                                            ____
    Step (i) : write –1828 in Bar form i.e.,, 1828
                                                        ____
           (ii) : Now we can add 4247 and 1828 i.e.,,

                        4247
                        ____
                      +1828
                     _______
                          _  _
                        3621
                      _    _        _              _    _        _
        since 7 + 8 = 1, 4 + 2 = 2, 2 + 8 = 6, 4 + 1 = 3
                                               _  _
         (iii) Changing the answer 3621 into normal form using Nikhilam, we get
                  _  _      _     _
                3621 = 36 / 21 split

                       = (3 –1) / (10 – 6) / (2 – 1) / (10 – 1) = 2419

                       4247 – 1828 = 2419
Find the following results using Vedic methods and check them

        1)  284 + 257             2) 5224 + 6127

        3)  582 - 464              4) 3804 - 2612

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