downloadbrowseDCrack's Crackme v1.0 by DCrack/FOFF

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This is my first crackme :)
No packing, no anti-olly tricks, no obfuscation, no junk code... as straightforward as it gets.
All I need is a valid serial.
I think I left a small window open so this can be crackable.
Feel free to write any comments or suggestions.

Difficulty: 3 - Getting harder
Platform: Windows
Language: Borland Delphi

Published: 18. Feb, 2008
Downloads: 845

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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 =).
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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