| I'm at the beginning of my reverse engineering studies, and this was a real challenge for me. When I received the message that the debugger wasn't allowed, I initially tried to find a way to solve it without using one. However, I couldn't find any alternative, since the main part of the code is loaded into memory during execution.
Eventually, I discovered the password because it appeared in a register during analysis. I then looked for the logic behind the password and found that it was: imul eax, eax, 539.
After some further analysis, I also discovered the backdoor through a register. I believe the backdoor is hardcoded, as I couldn’t find any logic for its generation.
Thank you very much for the challenge. |
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| No clue if debugging was off-limits, but I noped the debugger check and flipped a few conditionals till it dropped “awesome123” |
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