Brute zip with generating pass lists 5 bytes str + int caused for truncated

Get archive.zip and history and password.txt

History is like .bash_history :

2229  nano password.txt
2231  zip -e archive.zip maleo.log
2232  cp password.txt backuppassword.txt
2234  truncate -s -5 password.txt
2235  cat password.txt
2236  cat backuppassword.txt
2237  rm -rf backuppassword.txt
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and password is

703435356B756E
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'truncate -s 5' means remove 5 bytes. before it was i got tricked. because i think 5 bytes mean 5 chars. so make pass lists with 5 chars. but i'm wrong. because if i put 4 chars the size show me its 5 bytes. example:

so truncate -s -5 means delete 4 chars. ( 5 byte = 4 chars )

so this is my solver:

import string, itertools

text = string.ascii_letters.upper()
number = string.digits
original = "703435356B756E"


def generate_len(length: int) -> list:
    characters = number+text
    return [''.join(p) for p in itertools.product(characters, repeat=length)] 



if __name__ == "__main__":
    pass_lists = generate_len(4)
    for password in pass_lists:
        print(original+password)
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python3 gen.py > pass_4_char_and_digits.pass
zip2john archive.zip > hash_archive
john hash_archive --wordlist=pass_4_char_and_digits.pass
john --show archive_hash
archive.zip/maleo.log:703435356B756EAE3F:maleo.log:archive.zip::archive.zip

1 password hash cracked, 0 left
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analyse and grab the hidden text in .log file ( description results is more then 10 coins )

import re, base64

def main():
    pattern = r"send (.*?) coin\.\.\.\.\n.*give '(.*?)' for reward"
    with open('maleo.log', 'r') as f:
        matches = re.findall(pattern, f.read())
        output = ""
        for coins, reward in matches:
            if int(coins) >= 10:
                output += reward
                print(reward, end="")

        
        print(base64.b64decode(output))

if __name__ == "__main__":
    main()
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TechnoFairCTF{L0g_aja_b4ng_c333k}