Share how awesome the crack me was or where you struggle to finish it! Stay polite and do not spoil the solution/flag!
8:13 PM 04/26/2021
You are required to attack the target using brute force.Therefore,hook the function responsible for password checking then do brute force using it. Useful info: The password consists of 4 letters , all of which are of small letters and no special letters or numbers. Rules: 1. No patching 2. No reversing the check password function Extra: Innovative solutions are welcome and I hope, i would see some..
4epuxa on 9:07 AM 05/01/2021: pass: 'w70*
4epuxa on 9:20 AM 05/01/2021: pass: bibr
TrojanPoem on 10:19 AM 05/01/2021: Nice , you found a collision in the hash function :D
pranav on 10:53 AM 05/01/2021: That was kinda easier than 2.5, made a Keygen that lists out all possible keys.. Here is all valid keys for the crackme https://drive.google.com/file/d/1dNCX4QpN8b0B-NC1sqy5zD0Q0n5mY89a/view?usp=sharing
TrojanPoem on 11:27 AM 05/01/2021: I agree with you. However, I listed it for a higher difficulty as I was looking for mainly brute force solutions that utilize the crackme itself using e.g. a DLL or creating remote thread in the process to accomplish that.
kondeti on 4:01 PM 05/01/2021: Nice one, written a small keygen for it in python. https://pastebin.com/aQU2gQc4
TrojanPoem on 4:17 PM 05/01/2021: Awesome kondeti! I am a fan of Python
TuanNguyen on 12:42 PM 08/03/2021: Nice one.
as06zx on 10:19 AM 02/03/2022: 416 == str + str + str + str
featherfx on 2:28 AM 03/22/2023: the jumbled letters is a caeser cipher shifted 3 places to the left if you want to reveal the messages without cracking. If you want to crack it any 4 combo of characters that adds up to 19F in hex from their ascii will do it.