Vedic Mathematics | Yavadunam Tavadunikritya Varganca Yojayet

YĀVADŨNAM TĀVADŨNĪKŖTYA
VARGAΡCA YOJAYET

The meaning of the Sutra is 'what ever the deficiency subtract that deficit from the number and write along side the square of that deficit'.

This Sutra can be applicable to obtain squares of numbers close to bases of powers of 10.

Method-1 : Numbers near and less than the bases of powers of 10.

     Eg 1:  9² Here base is 10.

        The answer is separated in to two parts by a’/’

        Note that deficit is 10 - 9 = 1

        Multiply the deficit by itself or square it

        1² = 1. As the deficiency is 1, subtract it from the number i.e., 9–1 = 8.

        Now put 8 on the left and 1 on the right side of the vertical line or slash i.e., 8/1.

        Hence 81 is answer.

Eg. 2:  96² Here base is 100.

        Since deficit is 100-96=4 and square of it is 16 and the deficiency subtracted from the number 96 gives 96-4 = 92, we get the answer 92 / 16  Thus 96² = 9216.

Eg. 3:  994² Base is 1000

        Deficit is 1000 - 994 = 6. Square of it is 36.

        Deficiency subtracted from 994 gives 994 - 6 = 988

        Answer is 988 / 036     [since base is 1000]

Eg. 4:   9988²   Base is 10,000.

        Deficit = 10000 - 9988 = 12.

        Square of deficit = 12² = 144.

        Deficiency subtracted from number = 9988 - 12 = 9976.

        Answer is 9976 / 0144     [since base is 10,000].

Eg. 5:   88²   Base is 100.

        Deficit = 100 - 88 = 12.

        Square of deficit = 12² = 144.

        Deficiency subtracted from number = 88 - 12 = 76.

        Now answer is 76 / ₁44 = 7744    [since base is 100]

Algebraic proof:

The numbers near and less than the bases of power of 10 can be treated as (x-y), where x is the base and y, the deficit.

        Thus
            (1) 9 = (10 -1) (2) 96 = ( 100-4) (3) 994 = (1000-6)

            (4) 9988 = (10000-12 ) (v) 88 = (100-12)

            ( x – y )² = x² – 2xy + y²
                         = x ( x – 2y ) + y²
                         = x ( x – y – y ) + y²
                         = Base ( number – deficiency ) + ( deficit )²
        Thus

            985² = ( 1000 – 15 )²
                   = 1000 ( 985 – 15 ) + (15)²
                   = 1000 ( 970 ) + 225
                   = 970000 + 225
                   = 970225.

or we can take the identity a² - b² = (a + b) ( a - b) and proceed as

                  a² - b² = (a + b) ( a - b).

        gives          a² = (a + b) ( a - b) + b²

    Thus for a = 985 and b = 15;

        a²= (a + b) ( a - b) + b²

    985² = ( 985 + 15 ) ( 985 - 15 ) + (15)²

           = 1000 ( 970 ) + 225

           = 970225.

Method. 2 : Numbers near and greater than the bases of powers of 10.

      Eg.(1):  13² .

        Instead of subtracting the deficiency from the number we add and proceed as in Method-1.

        for 13² , base is 10, surplus is 3.

        Surplus added to the number = 13 + 3 = 16.

        Square of surplus = 3² = 9

        Answer is 16 / 9 = 169.

      Eg.(2):   112²

           Base = 100, Surplus = 12,

        Square of surplus = 12² = 144

        add surplus to number = 112 + 12 = 124.

        Answer is 124 / ₁44 = 12544

    Or think of identity a² = (a + b) (a – b) + b²   for a = 112, b = 12:

                112² = (112 + 12) (112 – 12) + 12²
                        = 124 (100) + 144
                        = 12400 + 144
                        = 12544.

            (x + y)² = x² + 2xy + y²
                         = x ( x + 2y ) + y²
                         = x ( x + y + y ) + y²

        = Base ( Number + surplus ) + ( surplus )²

       gives
            112² = 100 ( 112 + 12 ) + 12²
                    = 100 ( 124 ) + 144
                    = 12400 + 144
                    = 12544.

     Eg. 3:   10025²

                = ( 10025 + 25 ) / 25²

                = 10050 / 0625     [ since base is 10,000 ]

                = 100500625.

Method - 3: This is applicable to numbers which are near to multiples of 10, 100, 1000 .... etc. For this we combine the upa-Sutra 'anurupyena' and 'yavadunam tavadunikritya varganca yojayet' together.

Example 1:   388²  Nearest base = 400.

        We treat 400 as 4 x 100. As the number is less than the base we proceed as follows

        Number 388, deficit = 400 - 388 = 12

        Since it is less than base, deduct the deficit

            i.e. 388 - 12 = 376.

        multiply this result by 4 since base is 4 X 100 = 400.

                    376 x 4 = 1504

        Square of deficit = 12² = 144.

        Hence answer is 1504 / ₁44 = 150544    [since we have taken multiples of 100].

Example 2:  485² Nearest base = 500.

        Treat 500 as 5 x 100 and proceed
            
                    

Example 3:   67² Nearest base = 70

               
     Example 4:   416² Nearest ( lower ) base = 400

            Here surplus = 16 and 400 = 4 x 100

                                   

Example 5:   5012² Nearest lower base is 5000 = 5 x 1000

        Surplus = 12

           

Apply yavadunam to find the following squares.

            1.  7²              2.  98²            3.  987²           4.  14²         

            5.  116²         6.  1012²        7.  19²             8.  475²

            9.  796²       10.  108²        11.  9988²      12.  6014².

So far we have observed the application of yavadunam in finding the squares of number. Now with a slight modification yavadunam can also be applied for finding the cubes of numbers.

i)   For 106, Base is 100. The surplus is 6.

    Here we add double of the surplus i.e. 106+12 = 118.

    (Recall in squaring, we directly add the surplus)

    This makes the left-hand -most part of the answer.

            i.e. answer proceeds like 118 / - - - - -

ii)  Put down the new surplus i.e. 118-100=18 multiplied by the initial surplus

            i.e. 6=108.

  Since base is 100, we write 108 in carried over form 108 i.e. .

  As this is middle portion of the answer, the answer proceeds like 118 / ₁08 /....

iii)  Write down the cube of initial surplus i.e. 6³ = 216 as the last portion

          i.e. right hand side last portion of the answer.

     Since base is 100, write 216 as ₂16 as 2 is to be carried over.

     Answer is 118 / ₁08 / ₂16

    Now proceeding from right to left and adjusting the carried over, we get the answer

                    119 / 10 / 16 = 1191016.

Eg.(1):   102³ = (102 + 4) / 6 X 2 / 2³

                          =   106      =    12  =  08

                          = 1061208.

        Observe initial surplus = 2, next surplus =6 and base = 100.

Eg.(2):   94³

        Observe that the nearest base = 100. Here it is deficit contrary to the above examples.

i) Deficit = -6. Twice of it -6 X 2 = -12
                    add it to the number = 94 -12 =82.

ii) New deficit is -18.
        Product of new deficit x initial deficit  = -18 x -6 = 108

iii) deficit³ = (-6)³ = -216.

Since 100 is base 1 and -2 are the carried over. Adjusting the carried over in order, we get the answer

            ( 82 + 1 ) / ( 08 – 03 ) / ( 100 – 16 )

                = 83   /    = 05      /     = 84          = 830584
      __
     16 becomes 84 after taking1 from middle most portion i.e. 100. (100-16=84).
                                                                                     _
        Now 08 - 01 = 07 remains in the middle portion, and 2 or 2 carried to it makes the middle as 07 - 02 = 05. Thus we get the above result.

Eg.(3):
              998³  Base = 1000; initial deficit = - 2.

              998³ = (998 – 2 x 2)  /  (- 6 x – 2)  /  (- 2)³

                      =    994         /     = 012      /  = -008

                      =    994  /  011  /  1000 - 008

                     =     994  /  011  /  992

                     =     994011992.

Find the cubes of the following numbers using yavadunam sutra.

        1.  105     2.  114         3.  1003         4.  10007         5.  92

        6.  96       7.  993         8.  9991         9.  1000008     10.  999992.

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