20200802, 06:38  #12 
Jun 2003
2^{2}·5·239 Posts 

20200802, 06:40  #13 
Jun 2003
2^{2}×5×239 Posts 
Ah! You're thinking 99% as an exact number, whereas I'm pretty sure OP intended to give the general flavor of the structure of the number. Meaning, b^n part is the dominant term.

20200802, 07:25  #14 
Jun 2003
3^{2}·5^{2}·7 Posts 

20200802, 16:47  #15  
Jun 2003
2^{2}·5·239 Posts 
Quote:
Quote:
When N is about 1e5 bits, the first n that satisfies m*n > b is n_min~=4500, and b_max ~=4.5 million When N is about 1e6 bits, it is n_min ~= 39600 and b_max ~=39.6 million. For the above numbers, it should be feasible to brute force all prime b's < b_max. Basic outline of the algorithm would be to do c = N % (b^k), (prime b from 2 to b_max) where b^k is about 256 bits (>> our target c with is < 64 bits). If c < 2^64, we may have a potential solution, so trial factor Nc upto b_max and see if we get a complete factorization. 

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