Instantly Factorize Any Product Of Large Twin prime number.

Instantly Factorize Any Product Of Large Twin prime number.

Postby Guest » Sun Apr 11, 2021 12:28 am

Research paper at http://viXra.org/abs/2103.0181

Simple Method -
72 is the constant integer used in the process to find repeated addition in the
series.
First Step –
Repeated Addition Series.
Following the steps ask your colleague to add 72 and 36 as show below.
72 * 1 = 72 + 36 = 108
72 * 2 = 144 + 36 = 180
72 * 3 = 216 + 36 = 252
72 * 4 = 288 + 36 = 324 ......... 'Last Sum Of Series'
Counting can be done as many times like 72 *5 , 72 * 6, 72 * 7 ........ and
one time adding 36 for each series.
Series can go up to infinity.
Last sum of series is 324.
Second Step -
Finding ‘r ‘ Total Sum Of Series -
Ask your colleague to add all the sums together with number 35 to get total
sum of series ' r ' as shown below.
108+ 180 + 252 + 324 + 35 = 899 ...... 35 is the constant to be
added at last in total sum of series each time you calculate this series.
Here we get r = 899
Now get this two information from your colleague .
1) Last sum of series ie 324.
2) Total sum of series ie 899 .
You should know this to calculate the formula.
Therefore,
Ask your colleague to show the last sum of series i.e 324 and the total sum of
series i.e 899.
Note - Total sum of series is also a product of some two twin prime numbers
or prime number and composite number or may be of two composite
numbers.
So 899 is the product of p*q = 899 which we don’t know yet and we are going
to factorize it to know p & q using formula explained below.
Third step -
Now, ask your colleague that, 'can they immediately guess what is the
multiple factors of given total sum of series is, without factorizing ?'
Answer for your colleague must be 'no', since no one can easily guess or
reverse the p * q = n if the ''n ' is any large integer.
But wait, using my new researched method you can factor in few minutes, no
matter what large integer 'n' is.
So without showing your colleague, calculate the process explained below.
Calculation Process – Finding ‘s’.
There are two method to find ' s ’.
1) First method -
Notice the above 'repeated 72 series', those bold highlighted integers 72 * 1,
72 * 2, 72 * 3, 72 * 4 ........
Series of Integers in line i.e 1,2, 3, 4.....
Find the last integer i.e 4
Substitute 4 with 0 of 0.83 (constant).
We get 4.83
Therefore, s = 4.83.
Each time you calculate to find ' s’ always find the last integer in the line as
explained above.
2) Second Method -
You know that last sum of series is 324. ( you got the information from
your colleague)
Taking 324 / 72 = 4.5
Get the left hand side integer before the decimal point i.e 4.
Substitute 4 with 0 of 0.83 (constant).
As per second method, we get s = 4.83
Next,
Apply the ‘r’ and ‘s’ in the below formula.
r / s = m
m/ 6 = n
Where,
‘r’ is the total sum of series.
‘s’ in this case is the substitution of 4 with 0 of 0.83 constant to get as 4.83.
6 is the constant in the formula.
We got r = 899, s = 4.83
Finding ' m ' -
r / s = m
899 / 4.83 = 186.12836....
Notice integer on the left hand side before the decimal point i.e 186
So, consider only those integers as ' m ' and ignore integers on the right
hand side of decimal point.
Therefore, m = 186
Finding ' n ' -
m / 6 = n
186 / 6 = 31
n = 31 ....... is the answer.
Check it dividing 899 by 31.
899/ 31 = 29
So the factors of 899 is 31, 29.

Immediately show and surprise your colleague with the answer i.e 31 , 29.
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