(A classic “crack‑me” style reverse‑engineering challenge) 1. Overview | Item | Description | |------|-------------| | Challenge name | Adeko 9 Crack 56 | | Category | Reverse Engineering / Binary Cracking | | Platform | Windows 10 (x86‑64) – compiled with Visual Studio 2019 | | File size | ≈ 82 KB (PE32+ executable) | | Protection | No packer, but includes basic anti‑debug tricks and a custom serial‑check routine | | Goal | Produce a valid serial key that makes the program display “Correct!” (or the equivalent success message). | 2. Setup # Create a clean analysis environment mkdir adeko9-crack56 && cd adeko9-crack56 cp /path/to/Adeko9Crack56.exe . Tools used
# 1. Undo the final XOR (none in this binary) – not needed # 2. Reverse CRC over 9 bytes # We can use a known library that provides reverse CRC; however for clarity # we implement a straightforward brute‑force over the 9‑byte space using # the linearity property. # Here we employ the `crcmod` module which can compute CRC with an # *initial* value; we simply walk backwards using the known table. Adeko 9 Crack 56
# Instead of a complicated generic reverse, we exploit the fact that # CRC‑32 with polynomial 0xEDB88320 is reversible byte‑by‑byte. # The following tiny routine does it: def reverse_crc_bytes(target, nbytes): crc = target out = [] for _ in range(nbytes): # The low byte of the CRC is the byte that was processed last, # after the forward step it becomes (crc ^ byte) & 0xFF. # So to reverse, we take the low byte as the original data byte. b = crc & 0xFF out.append(b) crc = (crc ^ TABLE[b]) >> 8 return list(reversed(out)) Setup # Create a clean analysis environment mkdir
If we denote the post‑transform byte as b_i = t(i) , the CRC algorithm is applied to the sequence b_0 … b_8 . Reverse CRC over 9 bytes # We can
#!/usr/bin/env python3 import binascii import struct
t(i) = ROL8( c_i XOR 0x5A, 3 ) ROL8 rotates an 8‑bit value left by 3 bits.
def crc32_step_rev(crc, b): """Reverse one CRC‑32 step (process byte b at the *end* of the stream).""" # The forward step is: crc = (crc >> 8) ^ TABLE[(crc ^ b) & 0xFF] # Reversing: idx = (crc ^ b) & 0xFF prev_crc = (crc ^ TABLE[idx]) << 8 prev_crc |= idx return prev_crc & 0xFFFFFFFF