2ourc3's Baby Keygen 3

Author:
2ourc3

Language:
C/C++

Upload:
2022-12-20 10:51

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Platform:
Windows

Difficulty:
1.4

Quality:
5.2

Arch:
x86-64

Downloads:
24

Size:
76.86 KB

Writeups:
3

Comments:
10

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/ )

Comments (10)
Writeups (3)

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maxodrom1 on 2022-12-22 13:39: 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 2022-12-26 18:45: [Click to reveal]

0xr0sc0 on 2023-01-08 15:36: 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 2023-02-27 17:09: 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 2023-03-11 13:21: key length must be 11, 1st char 'A', 4th and 8th char '-'

Ivy04 on 2023-03-23 15:11: First ever Crack Me done!!! Thanks for this.

konfety on 2023-07-04 22:30: my first crackMe :D https://ibb.co/T44bh6s

iwn on 2024-10-23 10:25: 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 2025-01-10 15:43: 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 2025-08-10 10:59: Really easy keygen, my first cracked keygen, Thanks, key - AFF-FFF-FFF, F - random symbol, pass - nanyKeygen

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Solution by interrrp on 2023-03-11 13:32:
nice :D this is the first crackme i've ever actually tried to solve instead of patching it, nice job 2ourc3

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Solution by zeykey on 2024-02-14 19:38:
solution tells you how to make key.

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Solution by nilhiu on 2024-08-04 13:13:

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crackmes.one

crackmes.one

2ourc3's Baby Keygen 3