DCrack's Crackme v1.0 by DCrack/FOFF
| Download Crackme_1.0_by_DCrack-FOFF.zip, 228 kb (password: crackmes.de) Browse contents of Crackme_1.0_by_DCrack-FOFF.zip This is my first crackme :)
Difficulty: 3 - Getting harder | RatingWaiting for at least 3 votes View profile of DCrack » |
Solutions
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Discussion and comments
| DigitalAcid 18. Feb 2008 | Is bruteforce needed or can it be solved by using lots of grey matter ? Ps: finding character 1, 5 and 10 was easy, 12 more to go =). |
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| Ox87k 18. Feb 2008 | I don't think bruteforce is needed, just a brain! I found a serial that don't show me the error msgbox but... congratz message? Where is it? >I think I left a small window open so this can be crackable. It means that if you try a good serial, no congratz message appears to us? |
| DCrack Author 18. Feb 2008 | When a valid key is used, the crackme should display the good boy message. You got it right DigitalAcid :) |
| Till.ch 18. Feb 2008 | @Ox87k I think he more meant with this, that the 3 easy chars can give us a little 'startpoint' on reversing the encrypted function. Like the PUSHAD and RETN on the end it decrypts. I have atm. spare time for this, but indeed interesting crackme :) |
| Ox87k 18. Feb 2008 | @Till.ch: I know this! :) |
| DCrack Author 18. Feb 2008 | Hey Till.ch that's exactly what it is :) No traditional cracking here, just making sense of the decrypted code or bruteforcing. You can see how this can easily be uncrackable may adding perhaps a better encryption, stronger key and no comparison at all (letting the SEH take care of everything); perhaps the last one isn't even necessary. Good work mate. If anyone wants the valid key just write me. I think I'll publish a tute on how this one was coded and how does it work. |
| DCrack Author 18. Feb 2008 | Ps: I don't think bruteforcing is viable (please correct me if I'm wrong). The key has a length of 15, we know 3 already, so that leaves us with 12 characters. Let's say we're not gonna take the whole ASCII codes, but up to 126 decimal (wich are the ones most commonly used, besides I know all the characters are in this range :)). So we have 12 spaces to brute-force that can be filled with, let's say 94 characters (starting at 32). Isn't this 94^12? which equals: 475920314814253376475136 possible combinations. You do the time math :) |
| apuromafo 19. Feb 2008 | good crackme but ~i wait the solution jiji |
| Absolom1 20. Feb 2008 | cracked!!!by ClS|AbsshA in Crackslatinos, i dont know if that guy has an account here... |
| DCrack Author 21. Feb 2008 | Congrats to whoever did it! Let´s wait for the solution :) |
| AbsshA 21. Feb 2008 | Yes :) i'll send it to moderators. I hope don't mind if the languaje is spanish.. |
| DCrack Author 21. Feb 2008 | I saw the solution, this crackme was destroyed by AbsshA :) Thanks for the tutorial bro and congrats! Now I´ll go and make something a little harder :) |
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