2ourc3's Baby Keygen 2

Author:
2ourc3

Language:
C/C++

Upload:
2022-12-17 12:32

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Platform:
Windows

Difficulty:
1.6

Quality:
4.2

Arch:
x86-64

Downloads:
15

Writeups:
3

Comments:
9

Description

This challenge is quite easy, find the valid key to get the pass, and write a small, simple, keygen to generate other valid key. Made by 2ourc3 ( https://www.bushido-sec.com/ )

Comments (9)
Writeups (3)

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maxodrom1 on 2022-12-22 13:16: Quite easy. The valid key would be "t123-" , and the only thing that matters is the "t" and the "-" can be until 9 characters total, for example "t123-4567" or "t435-ftgr" or "tgth-" etc...

Suasaki on 2022-12-31 06:58: Password is BabyKeygen10....and key can be anything if you are doing patching

2ourc3 on 2023-01-01 13:01: @Suasaki: Patching isn't part of the solution except if especially mentioned. On the other hand, the goal of this chall is to find the criteria of a valid key in order to produce other. Thanks for your feedback, enjoy the other chall coming :)

CyGhost on 2023-01-19 19:03: If you are not foolish enough you can type any key that first character is "t" and fifth "-" rest you can just type random

NisheshTyagi on 2023-06-12 12:31: keygen in python:- import random import string #created by NisheshTyagi def generateKeys(): letters = string.ascii_lowercase keys = [] while len(keys) != 5: key = 't' for i in range(8): if len(key) == 4: key += '-' else: key += letters[random.randint(0,25)] keys.append(key) return keys

0x0Frenchie on 2024-05-18 10:20: very good crackme brother

iwn on 2024-10-23 09:46: SPOILER else if ( user_input[0] == 116 && user_input[4] == 45 ) figured they were ascii characters. found it and the first character had to be a 't' and the fifth a hyphen.

iwn on 2024-10-23 10:04: keygen import string, random chars, key = string.ascii_letters + string.digits, "" for i in range(random.randint(4, 9)): key += random.choice(chars) key = 't' + key[1:4] + '-' + key[5:] print(key)

jhosuakz on 2025-01-10 15:22: import random import string character = string.ascii_letters PASSWORD = [''] * 9 PASSWORD[0] = chr(116) PASSWORD[4] = chr(45) for i in range(len(PASSWORD)): if PASSWORD[i] == '': PASSWORD[i] = random.choice(character) print(''.join(pw for pw in PASSWORD))

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Solution by Xartrick on 2022-12-28 05:48:

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Solution by kontrol on 2023-04-30 22:09:

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Solution by jhosuakz on 2025-01-10 15:21:
Thanks

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crackmes.one

2ourc3's Baby Keygen 2