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Thread: Trick On Instantly Factoring Large Products Of Two Multiplied Twin Prime Numbers.

  1. #1 Trick On Instantly Factoring Large Products Of Two Multiplied Twin Prime Numbers. 
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    Note -
    1) Get your colleague, friends in the calculation process to surprise them with this new method.
    2) Use only simple calculators( not having auto square root finding).

    3) One can take this new method as an alternate to square root method but with fast result.

    4) This new method is all about the magic number i.e the 'Last Sum Of Series' in this case is 10, as explained in the below method. This magic number instantly revails the two factors of any small or large product of multiplied twin prime numbers.

    5) If we simply multiply p×q =n and get n product to factor, then this method won't work. Therefore, repeated 72 series is required to calculate to get Last Sum Of Series' therby we can calculate formula to find factors p×q, as shown below in the method.

    Method -
    First Step –
    Repeated 72 Series.


    You Calculate (72 × 1 + 36) + (72 × 2 + 36) + (72 × 3 + 36) +...... + (72 × n + 36) to get whole sum.


    72 * 1 = 72 + 36 = 108
    72 * 2 = 72 + 36 = 180
    72 * 3 = 72 + 36 = 252
    72 * 4 = 72 + 36 = 324
    72 * 5 = 72 + 36= 396
    72 * 6 = 72 + 36 = 468
    72 * 7 = 72 + 36 = 540
    72 * 8 = 72 + 36 = 612
    72 * 9 = 72 + 36 = 684
    72 * 10 = 72 + 36 = 756
    Note the last sum of series i.e. 72 * 10 = 72 + 36 = 756

    (Here below i have shown e.g. up till 10 steps. You can go for as many further steps to get bigger digits of sums)

    Second Step -
    Finding’ r’,

    Take 'Total Sum Of Series'.
    Add 35 once to get 'r' as shown below.

    108+ 180 + 252 + 324 + 396 + 468 + 540 + 612 + 684 + 756 + 35 =
    4355 ......35 is the constant to be added at last in total sum of series.

    Here we get r = 4355

    4355 is also a product of some p×q =4355 which both you and your colleague dont no the two factors.
    For sure your colleague would take sqrt finding aproach.

    So, ask your colleague to find factor of 4355 using simple calcuator in 1 min.
    ( It would be hard and take time for them to calculate using any other method to factor with simple calculator.)

    But you can factor
    r = 4355 in just few seconds. No matter how big is 'r'

    So, without showing them calcuate the below method and find two factors immediately.

    Third Step -
    Finding ‘s’.

    You know the 'last sum of series' i.e
    72 * 10 = 72 + 36 = 756

    Notice the multiplied term with 72 i.e 10
    Substitute 10 with 0 of 0.83 (constant).
    We get 10.83

    Therefore, s = 10.83.
    (Each time you calculate to find ' s ’ always find the last integer and substitute it with 0.83, as explained above).

    We got,
    r = 4355
    s = 10.83

    Next,
    Apply the ‘r’ and ‘s’ in the below formula.
    r / s = m
    m / 6 = n


    Finding 'm'.

    r / s = m
    4355 / 10.83 = 402.12373..... (402 is 'm')
    Get those integers left hand side before the decimal point
    i.e 402
    Therefore, m = 402

    Finding ' n '.

    m / 6 = n
    402 / 6 = 67
    n = 67 ....... is the answer.
    Check it dividing 4355 by 67.
    4355/ 67 = 65
    So the factors of 4355 is 65, 67.

    Immediately surprise your colleague with the answer.

    This way you can count Repeated 72 Series and increase the steps to get bigger digits of 'r'
    and follow the same method to instantly find factors and surprise your colleague.
    Even if the 'r' = n (which is n= p×q) is 100 + digits the time to find factors is less than 2 minutes.

    Research paper at,
    vixra


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