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12:32 PM 12/17/2022
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/ )
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maxodrom1 on 1:16 PM 12/22/2022: 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 6:58 AM 12/31/2022: Password is BabyKeygen10....and key can be anything if you are doing patching
2ourc3 on 1:01 PM 01/01/2023: @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 7:03 PM 01/19/2023: 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 12:31 PM 06/12/2023: 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 10:20 AM 05/18/2024: very good crackme brother
iwn on 9:46 AM 10/23/2024: 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 10:04 AM 10/23/2024: 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 3:22 PM 01/10/2025: 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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