Impossible

This is a writeup of the misc challenge impossible from the DUCTF(https://play.duc.tf/) CTF

Level: beginner, Score: 100

Premise

Lol I thought it would be funny to commit a CTF iceberg taboo and make an unsolvable CTF challenge, then gaslight competitors in support tickets.

There is no solution. I don’t even know why I published files for this challenge anyway.

Challenge files:

impossible.zip containing a python webapp

The link: https://misc-impossible-2fb66cb8eaca.2023.ductf.dev

Observations

When looking at the unzipped files in impossible, we see a app folder containing the contents of the webapp.

Within this folder we have an additional interesting folder: utils.

In utils, we find crypto.py, which, when looked through contains what looks like an encrypted message. The file contains the following content:

from Crypto.Protocol.KDF import PBKDF2
from Crypto.Hash import SHA512
from Crypto.Random import get_random_bytes
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad, unpad
import binascii

# lol u thought it would be easy that i gave you these values in the sauce code
# i have changed them all on the challenge instance just to make sure it is impossible to solve
FLAG_ENC='5700668797b137bb540ff5c7febf0f48687911940afae0d5e98ca7c80c25ab475ccdc7f4cd4c04b2890d38f0048abd88bf9d10339b7d37962b1cc0668f9d945e192286d517add90fdd39c923424e282b5f34c1067f467bcfdca1f2e819e3aa667e4452c7da74880cba6d31f10b88eb72c7e6215eb445cc9c9e9d435bf2eae1e5b1d6593f43a26193d5cb8e8745e5a8d81f26efd4aa9f47e364f529d0a334613a'
KEY='8deee6793066fa6529a9af2ef31723ca6d78da5c175d78412294604a2a62af82'
IV='104e20c2abf12b342fe27064cead0c4f'

def gen_key(passphrase: str) -> bytes:
    salt = get_random_bytes(16)
    return PBKDF2(passphrase, salt, 32, count=1000000, hmac_hash_module=SHA512)

def encrypt_print(flag: str, key: bytes):
    iv = get_random_bytes(16)
    cipher = AES.new(key, AES.MODE_CBC, iv=iv)
    ct = cipher.encrypt(pad(flag.encode(), 32))
    print(f"FLAG_ENC='{ct.hex()}'")
    print(f"KEY='{key.hex()}'")
    print(f"IV='{iv.hex()}'")

def decrypt(key_hex: str) -> str:
    if key_hex != KEY:
        return "Lol we told you this is impossible to solve!"
    key = binascii.unhexlify(key_hex)
    flag_enc = binascii.unhexlify(FLAG_ENC)
    iv = binascii.unhexlify(IV)

    cipher = AES.new(key, AES.MODE_CBC, iv=iv)
    pt = unpad(cipher.decrypt(flag_enc), 32)

    try:
        return pt.decode()
    except:
        return "Cannot decode plaintext!"
    
if __name__ == "__main__":
    flag = input("flag: ")
    passphrase = input("passphrase: ")
    key = gen_key(passphrase)
    encrypt_print(flag, key)

As we can see, there is a commented message saying that we wont be able to decrypt the message. Doing this anyway with a tool like CyberChef gives us the following result.

So as we can see, we need to try something different to solve the challenge.

Solution

In the utils folder, we additionally see the folder __pycache__. In the __pycache__ folder, we find the file crypto.cpython-310.pyc which contains a cached version of the crypto.py script and thereby also the previously used FLAG_ENC, KEY and IV variables.

Opening the file, we find the old variables,

If we try to decrypt the message this time with our new variables, we get the following output:

Giving us our flag

DUCTF{o0p5_i_f0rG0t_aBoUt_pYc4Ch3!!1!}

Tools used:

• Cyberchef
• Any text editor