N1ghtm4r3's KeygenMe4
| Download KeygenMe4.zip, 2 kb (password: crackmes.de) Browse contents of KeygenMe4.zip No patching!
Difficulty: 2 - Needs a little brain (or luck) | RatingVotes: 4 View profile of N1ghtm4r3 » |
Solutions
Solution by robbje, published 07. apr, 2011; download (133 kb), password: crackmes.de or browse.
robbje has rated this crackme as awesome.
Discussion and comments
| Beezdul 26. Mar 2011 | Ok can i have a little help? I am getting stuck at the part after the check for serial length and that the 7th letter is a dash. |
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| indole 27. Mar 2011 | It starts checking the first half of the key and gets each characters position inside a string in the program. It does some math on it to get a sum and then does a similiar thing on the second half of the key. I was able to solve this earlier but haven't written up a tutorial. I enjoyed it as my first crackme. Here are some keys that I generated: 2WAEON-DN2RC5 , EOHK2R-DFCMKE . |
| Fear2oo0 28. Mar 2011 | :), good crackme. |
| robbje 29. Mar 2011 | I liked it so much, that i decided to write a solution :) Enjoy |
| Borgiman 29. Mar 2011 | Yo, really nice KeyGenMe. I will also try to write a keygen. (hope i'll get it :)) |
| Borgiman 07. Apr 2011 | i think difficult 2 is still to different for me :P patching & cracking yes, but keygenning not yet :D nice tut robbje |
| N1ghtm4r3 Author 07. Apr 2011 | well done robbje :) |
| Borgiman 08. Apr 2011 | oh god, my english :D i mean difficulty 2 is still too difficult for me :P |
| Coccinell 19. Apr 2011 | i have found : code1 = ( (serial[0] * position(serial[0])) + (serial[1] * position(serial[1])) + (serial[2] * position(serial[2])) + (serial[3] * position(serial[3])) ) * 78 code2 - code1 - 1 = 6486568h code2 = 6486568h + 1 + code1 but i don't success to decode asm code for code2 ! i don't understand this : f7 * 24^5 + f8 * 24^4 + f9 * 24^3 + f10 * 24^2 + f11 * 24 + f12 why ^5, ^4, ^3, etc ? can you help me to understand that ? thanks |
| robbje 19. Apr 2011 | Check again what the asm code does when it calculates the checksum of key[7..12]. It takes the position of the key in the string and adds it to the accumulator, which will get multiplied by 18h (=24d). If you expand this sum, you will see that f7 gets multiplied by 24 5 times, f6 4 times, and so one. Therefore ^5, ^4, ^3... |
| Coccinell 20. Apr 2011 | i have found that : code2 = (((((position(serial[7]) * 24) + position(serial[8]) * 24) + position(serial[9]) * 24) + position(serial[10]) * 24) + position(serial[11]) * 24) + position(serial[12]) code2 = (position(serial[7]) * 24^5) + (position(serial[8]) * 24^4) + (position(serial[9]) * 24^3) + (position(serial[10]) * 24^2) + (position(serial[11]) * 24^1) + (position(serial[12]) * 24^0) i understand now, thanks robbje |
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