Забытый пароль 2

Platform: duckerz | Category: Reversing | Type: Challenge | Difficulty: Easy | OS: NA | Author: XESXOR | Date: 2026-01-09 | Status: Solved Techniques: xor_decryption

Summary

Task: Python script with XOR-encrypted flag using key "0x2a". Solution: Applied reverse XOR operation with the same key to decrypt the flag.

Забытый пароль 2 — duckerz

Description

The task provided a Python script rev2.py which contained an encrypted flag and a simple XOR encryption logic using the key "0x2a".

Analysis

When analyzing the provided script rev2.py, the following was discovered:

1. The flag variable contains an encrypted byte sequence: b"t-q*u*h\x1a@L\x00SGH\x05o;US\x13\x01%o\x0f\x03\x15X'jQB\x05".
2. The key used is key = "0x2a".
3. The encryption algorithm is character-by-character XOR between the message and a cyclically repeating key:

    python read-only • copy allowed Copy
   for i in range(0, len(message)):
       list_output.append(chr(ord(message[i])^ord(key[i % len(key)])))

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Since the XOR operation is reversible, to decrypt the flag it is sufficient to apply the same operation with the same key to the encrypted bytes.

Solution

A script solve_rev2.py was written:

 python read-only • copy allowed Copy
#!/usr/bin/env python3

flag_enc = b"t-q*u*h\x1a@L\x00SGH`\x05o;`US\x13\x01%o\x0f\x03\x15X'jQB\x05"
key = "0x2a"
flag = ""
for i in range(len(flag_enc)):
    flag += chr(flag_enc[i] ^ ord(key[i % len(key)]))
print(flag)

Running the script gives us the flag. ‍​‌‌​​‌​​​‌‌​​‌​​​​‌‌​‌​​​​‌‌‌​​‌​​‌‌​​​‌​​‌‌‌​​​​​‌‌​‌‌​​​‌‌​​​​‍