Author:
2ourc3

Language:
C/C++

Upload:
10:51 AM 12/20/2022

Download

Platform
Windows

Difficulty:
1.5

Quality:
5.0

Arch:
x86-64

Description

Hi everyone! This challenge is quite easy, find the valid keys to get the pass. The goal is to write a little keygen that can generate valid key for the challenge, try to not bypass this logic, even if it is more simple. But aim to understand the condition required to obtain a valid key Have fun! Made by 2ourc3 ( https://www.bushido-sec.com/ )

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maxodrom1 on 1:39 PM 12/22/2022: a valid key would be for example : "A33-333-333" or "Adf-rfr-sds" , the only thing that matters is that it have to have the format "xxx-xxx-xxx" and the first have to be "A" so "Axx-xxx-xxx" would be valid also. And for some reason you have to introduce it 3 times before you get the password

jimsktsmith on 6:45 PM 12/26/2022: The key is: nanyJeygen the key format is: A00X00000X the first character always 'A' the third and seven character in string is 'X'

0xr0sc0 on 3:36 PM 01/08/2023: Real keygen in asm (not packed you can debug): https://www65.zippyshare.com/v/4MwossPS/file.html sha hash and Virustotal link added in zip

Phil on 5:09 PM 02/27/2023: Don't know why one of the comments say the 3. and 7. must be an X. It adds 3 and 7 to the counter and checking for an "-". First one must be an A, 4. an "-" and 8. an "-" and ist has the lenght of 11.

interrrp on 1:21 PM 03/11/2023: key length must be 11, 1st char 'A', 4th and 8th char '-'

Ivy04 on 3:11 PM 03/23/2023: First ever Crack Me done!!! Thanks for this.

konfety on 10:30 PM 07/04/2023: my first crackMe :D https://ibb.co/T44bh6s

iwn on 10:25 AM 10/23/2024: take a look at this compiled code to find out the valid format: if ( (unsigned int)strlen(localkey) == 11 ) { if ( *localkey == 65 && localkey[3] == 45 && localkey[7] == 45 ) {

jhosuakz on 3:43 PM 01/10/2025: import random import string character = string.ascii_letters PASSWORD = [''] * 11 PASSWORD[0] = chr(65) PASSWORD[3] = chr(45) PASSWORD[7] = chr(45) for i in range(len(PASSWORD)): if PASSWORD[i] == '': PASSWORD[i] = random.choice(character) print(''.join(pw for pw in PASSWORD))

TeRrArIsT on 10:59 AM 08/10/2025: Really easy keygen, my first cracked keygen, Thanks, key - AFF-FFF-FFF, F - random symbol, pass - nanyKeygen

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