20171103, 07:02  #1 
Nov 2017
2^{4} Posts 
Help with discrete logarithm
Hi all!
I try to solve reversing chanllenge. I understand algo: 2^x mod 1000000000000000000000000000000000000000000 = 7642587499893475369658291795841 So I have to find x. I read some papers about discrete logarithm but can't found any suitable solution. Could you help me? 
20171103, 07:17  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41·229 Posts 
Can you find any x>0 that 2^{x} is odd? No, all of these will be even.
Can you find any even numbers A>0 and B>0 such that A mod B will be odd? of course not Now look back at your problem... what do you see? 
20171103, 07:36  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5×1,223 Posts 
I guess it is meant to be "2^x  1 mod ..."

20171103, 08:02  #4 
Nov 2017
10_{16} Posts 

20171103, 08:22  #5 
Nov 2017
2^{4} Posts 
That's correct:
2^x mod 1000000000000000000000000000000000000000000 = 15285174999786950739316583591682 
20171103, 08:27  #6 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5×1,223 Posts 

20171103, 08:46  #7 
Nov 2017
2^{4} Posts 

20171103, 09:52  #8  
Dec 2012
The Netherlands
3·19·29 Posts 
Quote:
http://www.mersenneforum.org/showthread.php?t=21692 

20171103, 10:05  #9 
"Robert Gerbicz"
Oct 2005
Hungary
10110110001_{2} Posts 

20171103, 10:23  #10  
Nov 2017
2^{4} Posts 
Quote:
1. Factorization modulus  1 (10000000000000000000000000000000000000000001) = [ <3, 3>, <7, 2>, <11, 1>, <13, 1>, <37, 1>, <43, 1>, <127, 1>, <239, 1>, <1933, 1>, <2689, 1>, <4649, 1>, <459691, 1>, <909091, 1>, <10838689, 1> ] 2. For each factor: 2 = 15285174999786950739316583591682 ^ p mod 9 2 = 15285174999786950739316583591682 ^ q mod 49 2 = 15285174999786950739316583591682 ^ z mod 11 ... 3. And use Chinese Remainder Theorem But I stuck at 2 step. Can't find (PohligHellman) p, q ... 

20171103, 10:40  #11 
Dec 2012
The Netherlands
3165_{8} Posts 
To use the Chinese Remainder Theorem here, you work in the integers modulo \(2^{42}\) and modulo \(5^{42}\) instead of the integers modulo \(10^{42}\).
But, as R. Gerbicz pointed out, there is still a mistake in the number on the right (or no solution). 
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