**Yaavadunam Tavadunikrtya Sutra**

__Definition__: Whatever the deficiency subtract that deficit from the number and write along side the square of that deficit

We have placed '0' before 36 as the final output should be a six-digit number . The above method is similar for numbers like 96

^{ 2}, 9988^{ 2 }, etc...In case of Surplus , the above sutra sounds true for any numbers close to the powers of 10

Carrying over is done for each value of hundredth place value except for '0'. The above method is similar for numbers like 106

^{2}, 908^{2 }, etc...

__Algebric Proof :__Let 'x' be the base and 'y' be the deficit .The numbers near or less than power of "10" can be treated as ( x - y ).

That is , 9 = ( 10 - 1)

994 = ( 1000 - 6 )

96 = ( 100 - 4 ) , etc...

( x - y )

^{ 2 }= x

^{2}- 2 x y + y

^{2}

= x ( x - 2y) + y

^{2}

= x [ ( x - y ) - y ] + y

^{2}

=

**base [ (number) - deficiency ] + deficit**

^{2}^{Thus ,}

985

^{2}

which can be written as ( 1000 - 15 )

^{2 }

(or) = 1000 [ ( 985 - 15 )] + 15

^{2}

= 970000 + 225

= 970225

**Therefore , 985**

^{2 }= 970225

**Read Also :**

**Simple and Effective Trick for BASE 50**

I've used this method. But now i find multiplying 3 digits using another one easier than the above said. This method reduces time only if the number has more 9s.

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