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Powerful digit “9”

The result of multiplication by multiplying a number with another number and result of multiplication by multiplying the reverse of numbers, then subtracting the results of multiplication and then by adding all digits of the subtracted result number we always get 9.

For example:

              By reversing:         54 × 23 = 1242

                                                54 × 32 = 1728

                                                45 × 23 = 1035

                                                45 × 32 = 1440

            By subtracting:        1728 – 1242 = 486 = 4 + 8 + 6 = 18 = 9

                                                1728 – 1035 = 693 = 6 + 9 + 3 = 18 =9

                                                1440 – 1035 = 405 = 4 + 0 + 5 = 9

*

* Method and formula for solving the number having a power by using points.

* (design of points on triangle) -                  on points                   sum of points                                                                                                                                         

                                     1³      = 1

                                     2³      = 1 + 7 =8 or 1 + 7 + 2 × 1 × 0 = 8

 3³      = 1 + 7 + 13 + 3 × 2 × 1 = 21 + 6 = 27

 4³      = 1 + 7 + 13 + 19 + 4 × 3 × 2

          = 40 + 24 = 64

 5³      = 1 + 7 + 13 + 19 + 25 + 5 × 4 × 3

          = 65 + 60 = 125

Note: By using this method we not only got the method to solve number having a power, but also come to know that counting is started from zero (0) not from the one (1). 

"Square Root Table"

N₂ ² = 0+(20n₁ -19) n+ (2n-2) n

     2    

N₂ ² = 100+(20n₁ -19) n+ (2n-2) n

     2    

N₂ ² = 400+(20n₁ -19) n+ (2n-2) n

     2    

N₂ ² = 900+(20n₁ -19) n+ (2n-2) n

     2    

For Example

(1) 22² = 400 +(22 * 3 -19) 2+ (2*2-2)/2)2

            = 400 +(60 -19)2 +((4-2)/2)2

            =400 +(41*2)+(2/2)2

            =400 +82 +2

            =400+84

            =484

(2) 11² = 100 +(20 * 2 -19) 1+ (2*1-2)/2)2

            = 100 +40 -19 +0

            =100 +21

            =121

Note : n1=2 and n2=11

(3) 25² = 400 +(20 * 3 -19) 5+ ((2*5-2)/2)5

            = 400 +(60 -19 )5+20

            =400 +41*5+20

            =420+205

            =625

(4) 9²    = 0 +(20 n 1 -19) n+ ((2 n-2)/2)n

            = 0+(20 * 10 -19)9+ ((2* 9 -2)/2)9

            =0+(200-19)9+(16/2)9

            =0+(

(5) 11² = 100 +(20 * 2 -19) 1+ (2*1-2)/2)2

            = 100 +40 -19 +0

            =100 +21

            =121