happytown's CrackMe.HappyTown.VC.0040
| Download CrackMe.HappyTown.VC.0040.zip, 265 kb (password: crackmes.de) Browse contents of CrackMe.HappyTown.VC.0040.zip Math knowledge is a must!
Difficulty: 3 - Getting harder | RatingWaiting for at least 3 votes View profile of happytown » |
Solutions
Solution by redoC, published 02. oct, 2012; download (327 kb), password: crackmes.de or browse.
redoC has not rated this crackme yet.
Discussion and comments
| redoC 11. Sep 2012 | Final condition is Hash(Name) = pow (Serial, E) mod N E = 12754A064C91DD0A8E26385EC9335A268192B730DE8541535695C9EC68ADD24F22C5DDC3CD9D44EC38FA2F708640CB7189069E956FE84F10301128AEA613F70D N = 6F907AAA920DAF37AD19DD6974540903FBC772FE38F314F4B058076B097911FEA8E7BE75254BDB6536F96C1A2F5BDB8C69EF81C61E369837F3B9CBC188BDCFB9 Can anyone generate private exponent from this 256-bit public key? RSA-Tool always crashes. |
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| tomkol 11. Sep 2012 | It may take a while to factor this modulus N. It's 512bit and not 256bit. You may try 'msieve' to factor it and then any big calc to calculate private D. |
| tamaroth Moderator 14. Sep 2012 | You do not need to factorise anything. With 512 bits we have few options: * Weger attack - factors of N are close to each other meaning that |P-Q| is reasonably small * Wiener attack - special attack for small private key * low public key attack (e = 3, m^e < n) * cycling attack - you encrypt one message over and over until you get original, not really practical. * we're not actually dealing with decryption Note the last part, what if you already have private key but need to get exponent? |
| redoC 14. Sep 2012 | tamaroth, you dont help me so much, numbers are above, can you give some usefull informations? |
| tomkol 15. Sep 2012 | I don't want to spoil fun to you redoC. If you want to quick factor this challenge use Black-Eye RAT (RSA Attacking Tool). Just google for it and use tamaroth hints. Using RAT you can solve this in seconds. |
| redoC 16. Sep 2012 | Yes, private key is 0x10001. But second part is bit more difficult. |
| tamaroth Moderator 17. Sep 2012 | So, you have: e,d,n RSA m^d mod n = c c^e mod n = m Can you see it now? |
| redoC 18. Sep 2012 | I mean second part of crackme. There's another new algorithm for construction Magic string. |
| HMX0101 20. Sep 2012 | Algo used for second part its RC4, magic string is used as key to decipher goodboy msg, I guess actually you know this :P |
| redoC 21. Sep 2012 | Yes, it's RC4. Solution soon. |
| SFeS 17. Oct 2012 | for fun: http://rghost.ru/users/FeS1/releases/solution-happytownvc0040 (pure Python implementation: the hash algorithm from crackmes by happytown; RSA – Wiener attack; keygen) |
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