20100525, 12:12  #1 
Oct 2008
2^{4} Posts 
who can factor this 128 digits number?
my friend let me to factor it ,but it is difficult for me to factor it,
who can help me to factor it ? 67265468438158925156029310985265926133792339483088426060945720579389614045244727695101069671293650665794246539609007915477818493 
20100525, 12:14  #2  
Nov 2003
2^{6}·113 Posts 
Quote:
You might also reveal the reason for wanting to factor it. If you want others to help, you need to provide a reason for them to spend their time and resources. 

20100525, 12:19  #3 
May 2010
Prime hunting commission.
2^{4}·3·5·7 Posts 
"my friend let me to factor it ,but it is difficult for me to factor it,
who can help me to factor it ? 67265468438158925156029310985265926133792339483088426060945720579389614045244727695101069671293650665794246539609007915477818493" Hmm. At least it wasn't a result of jabbing the number pad repeatedly, as it isn't at all repetitive. The 
20100525, 12:23  #4 
Oct 2008
2^{4} Posts 
I know that msieve can factor it ,but my computer is too poor to factor it

20100525, 13:00  #5  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
Quote:
http://gilchrist.ca/jeff/factoring/n...ers_guide.html It might be worthwhile to first do some ECM, if you're not sure that the factors are all larger than about 50 digits. An easy way to do this is with aliqueit, which can also start factmsieve.py once it's done ECMing. Aliqueit is available here, and the tools you need along with it can be downloaded here. If you expect anyone to help you with the work of factoring this number, first tell us why we should care about this number. It may have some value of curiosity to you and your friend, but if that's all it is, you may have trouble getting help with the NFS sieving. Last fiddled with by MiniGeek on 20100525 at 13:07 

20100816, 09:29  #6 
Jul 2005
2×193 Posts 
OK, factored (I was bored and needed to soak test a new blade server for a week; if it wasn't behind a firewall I would have given it some Homogenous Cunningham numbers to do...)
p64 x p65 Looks RSAish, given the similarly sized factors and also as both P+1 and P1 factoring wouldn't have been worthwhile spending time on (the appropriate B2 would have been way too big, the p1 of the p65 is 2^5 * p28 * p35 for example). So, to get the factors we all need you to answer the questions asked above about why this particular number and why it is important that it is factored. Last fiddled with by Greenbank on 20100816 at 09:31 
20100816, 11:26  #7 
Tribal Bullet
Oct 2004
DA1_{16} Posts 
It's unusual to get a even split like that, so my guess is that it's an RSA key. The small size suggests it's a 'final exam' for somebody's crackme. For those who don't know, these are mini challenges that the reverse engineers give to each other to show off how much they know.

20100824, 08:57  #8 
Jul 2005
2×193 Posts 
Still waiting for a response to the questions in order to post the factors.
Don't anyone else waste time factoring this number as it has already been done. It raises an interesting theoretical question though, is there any way that I can prove that I've factored the number without revealing enough information to give away the factors? I'm guessing not. 
20100824, 09:00  #9 
Nov 2008
2·3^{3}·43 Posts 
aaa120 hasn't been on the forum since 24 June.

20100824, 10:02  #10  
Jun 2003
The Texas Hill Country
2·541 Posts 
Quote:
Assume that it is used to construct an RSA key. Since only you (and others who know the factors) can construct the private key, use that private key to "sign" a sample message. If you post the public key, the message, and the signature, then anyone can verify the signature and, by extension, infer that you know the factors. 

20100824, 10:17  #11  
Dec 2009
89 Posts 
Quote:
Last fiddled with by warut on 20100824 at 10:26 

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