ikad's Employee
| Download Employee.zip, 34 kb (password: crackmes.de) Browse contents of Employee.zip Crack it!
Difficulty: 1 - Very easy, for newbies | RatingWaiting for at least 3 votes View profile of ikad » |
Solutions
Solution by gibleMac, published 25. aug, 2016; download (634 b), password: crackmes.de or browse.
gibleMac has not rated this crackme yet.
Solution by UltraM4N, published 25. aug, 2016; download (35 kb), password: crackmes.de or browse.
UltraM4N has rated this crackme as quite nice.
Discussion and comments
| zazoutaz 16. Jul, 09:41 | EAX must != 0 at 1679. |
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| asor 22. Jul, 22:54 | 004016D6: NOP 004016D7: NOP :) |
| sadfud 03. Aug, 18:00 | too easy. thanks |
| arlkers 24. Aug, 09:05 | Someone writes a generator, F1 (code) =f2 (pass) |
| IWannaBeTheBest 26. Aug, 23:58 | Is it possible to solve this without using NOP's? |
| Kirdneh 27. Aug, 14:10 | @IWannaBeTheBest 004016D6: JMP SHORT 004016D8 works as well. its like always jump to access granted |
| IWannaBeTheBest 27. Aug, 16:40 | @Kirdneh Oh I see. Thanks. |
| 10SecSleeper 10. Sep, 13:34 | Searched for string references. Found what jmp to access granted. Reversed from JE to JNE. |
| vmlinuz11221 15. Sep, 21:45 | 0x40166e: NOP 0x40166f: NOP 0x401670: NOP 0x401671: NOP |
| dahek 11. Oct, 11:12 | 004016d6 je->jg |
| allocer 23. Nov, 22:27 | Is it possible to solve it by find out how correct passcode is generated for given number? |
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