20200802, 06:38  #12 
Jun 2003
4,969 Posts 

20200802, 06:40  #13 
Jun 2003
4,969 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
11000101011_{2} Posts 

20200802, 16:47  #15  
Jun 2003
4,969 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. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
More Power!!!!  petrw1  Teams  10  20191015 17:36 
TDP as power used?  CRGreathouse  Hardware  9  20160206 18:46 
Power 2  JohnFullspeed  Miscellaneous Math  45  20110710 20:13 
testing, if a number is a power  bitblit  Math  8  20090424 02:48 
IBM Power 6  Unregistered  Information & Answers  7  20080830 14:36 