AdventOfCTF 2024 Day 6: Epochrypt

Day 6: Epochrypt

It’s time to test out Tibel Elf’s new encryption method. He says once you encrypt it, you can’t unencrypt it. Sureeee…

The following function is used to perform the encryption of a string:

def epochrypt(enc):
    bits = bytes([(b + 3) % 256 for b in enc])
    based = b64.b64encode(bits)
    epc = str(int(time.time())).encode()
    final = xor(based, epc)
    print(final.hex())

Based on this, it seems relatively simple to decrypt a string. We just need to get an encrypted string, determine the (rough estimate) epoch timestamp of when it was encrypted, and perform the reverse of the encryption: XOR → Base64 decode → Shift the bytes

I first connected to the remote server to get an encrypted flag: 6b59695a5359570367607f5e6670515a7a5c7d0357707a0760637d0356065c7677650809

Then, I wrote a Python script that reverses the epochrypt functionality and bruteforces the past five minutes of epoch timestamps. This will yield a significant amount of false positives, but luckily the program has a flag checker built-in, so we can validate without needing to waste submissions on the actual CTF platform.

import time
import base64 as b64
from pwn import xor
 
def decrypt(enc, epoch):
    try:
        final = xor(enc, epoch)
        based = b64.b64decode(final)
        bits = bytes([(b - 3) % 256 for b in based])
        print(bits.decode())
    except Exception as e:
        pass
 
encrypted = bytes.fromhex("6b59695a5359570367607f5e6670515a7a5c7d0357707a0760637d0356065c7677650809")
 
# Try -5 minutes
for i in range(5 * 60):
    epoch = str(int(time.time()) - i)
    decrypt(encrypted, epoch.encode())
 
# csd{d3F0_M4d3_8y_4N_3lf}