ECM Distributed Computing Project and Mr. Stolyarov Discover Factor for 279-Digit Number (“9” Surrounded by 139 Instances of “7” Per Side)

ECM Distributed Computing Project and Mr. Stolyarov Discover Factor for 279-Digit Number (“9” Surrounded by 139 Instances of “7” Per Side)

The New Renaissance Hat
G. Stolyarov II
November 21, 2012
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I am pleased to report that a large prime factor for (7·10279+18·10139-7)/9 (visualized as a “9” surrounded by 139 instances of “7” on each side – for a total of  279 digits) was discovered on November 19, 2012, through my participation in the ECM distributed computing project (organized via Yoyo@home). This is the fourth discovery made on my computer via the ECM project (see posts about previous discoveries here and here).

The prime factor is a 53-digit number: 42684752427275029312252733896207947190538122452468697. I am credited with the discovery here and here.

I continue to be impressed by the potential of individual hyper-empowerment through distributed computing, and I encourage my readers to also donate their idle computer time to projects that attract their interest.

One thought on “ECM Distributed Computing Project and Mr. Stolyarov Discover Factor for 279-Digit Number (“9” Surrounded by 139 Instances of “7” Per Side)

  1. This is the best idea about integer factorization, written here is to let more people know and participate.
    A New Way of the integer factorization
    1+2+3+4+……+k=Ny,(k<N/2),"k" and "y" are unknown integer,"N" is known Large integer.
    True gold fears fire, you can test 1+2+3+…+k=Ny(k<N/2).
    How do I know "k" and "y"?
    "P" is a factor of "N",GCD(k,N)=P.

    Two Special Presentation:
    N=5287
    1+2+3+…k=Ny
    Using the dichotomy
    1+2+3+…k=Nrm
    "r" are parameter(1;1.25;1.5;1.75;2;2.25;2.5;2.75;3;3.25;3.5;3.75)
    "m" is Square
    (K^2+k)/(2*4)=5287*1.75 k=271.5629(Error)
    (K^2+k)/(2*16)=5287*1.75 k=543.6252(Error)
    (K^2+k)/(2*64)=5287*1.75 k=1087.7500(Error)
    (K^2+k)/(2*256)=5287*1.75 k=2176(OK)
    K=2176,y=448
    GCD(2176,5287)=17
    5287=17*311

    N=13717421
    1+2+3+…+k=13717421y
    K=4689099,y=801450
    GCD(4689099,13717421)=3803
    13717421=3803*3607

    The idea may be a more simple way faster than Fermat's factorization method(x^2-N=y^2)!
    True gold fears fire, you can test 1+2+3+…+k=Ny(k<N/2).
    More details of the process in my G+ and BLOG.

    My G+ :https://plus.google.com/u/0/108286853661218386235/posts
    My BLOG:http://hi.baidu.com/s_wanfu/item/00cd4d3c5a2fd089f5e4ad0a
    Email:wanfu.sun@gmail.com

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