|sas0|'s w32crackme1
| Download w32crackme.zip, 3 kb (password: crackmes.de) Browse contents of w32crackme.zip Helo...my first win32 crackme...goal is to find out serial that will correctly decode "good msg" and show it in Status
Difficulty: 3 - Getting harder | RatingWaiting for at least 3 votes View profile of |sas0| » |
Solutions
Solution by ty123, published 01. jan, 2007; download (100 kb), password: crackmes.de or browse.
ty123 has not rated this crackme yet.
Discussion and comments
| geeko 22. Sep 2006 | u are out of your mind! the algo is too long and complicated. It's not worth to reverse. It does to checks: 1 for first 4 chars and then for last 4 chars. Self generating code! ugly. It took me a day just to step into the algo till the end. IS not very very hard, but is very long.I gave up. |
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| |sas0| Author 22. Sep 2006 | It's not that hard as it look like.Algo is rather long,but if you analize the code you'll see that there are about 5 or 6 checks...so If you want to crack it just study evry check for itself :)...yea well...I include smc just to make a bit harder to analize :) |
| l0calh0st 22. Sep 2006 | Good work m8....Analysing now :) |
| geeko 25. Sep 2006 | u said to find the serial, not to patch, right? |
| |sas0| Author 25. Sep 2006 | If you patch the code where it checks if the entered password is correct you won't do any good becouse "good message" is encrypted with "password" and with bytes that indicates that password is correct...decryption algo isn't hard...you can ofcourse manualy enter some good text in crackme and then display it but I think that is not the goal of the crackme :) |
| ty123 15. Dec 2006 | @|sas0| plz confirm the following DW (BCED5428h) is correct. Thanks! 004012F1 > 35 2854EDBC XOR EAX,BCED5428 I would rate this crackme as 4/10. You will know the reason from my solution. |
| |sas0| Author 15. Dec 2006 | Yep ty123 0xBCED5428 is right dword so the eax == 0 You're on the right track :) |
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