# ------------------------------------------------------------ # 1. CRC‑32 parameters (same as the binary) POLY = 0xEDB88320 INIT = 0xFFFFFFFF XOROUT = 0x00000000
(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 Adeko 9 Crack 56
# 3. Invert the per‑byte transform to get the actual serial serial_bytes = bytes(invert_transform(b) for b in transformed) serial = serial_bytes.decode('latin-1') # keep raw bytes, printable check later print("[+] Serial candidate:", serial) Invert the per‑byte transform to get the actual
def reverse_crc(target_crc, length): """Return the list of bytes that must have been fed to the CRC to get target_crc.""" # Walk backwards length steps, assuming the *last* processed byte is unknown. # We'll treat each step as "what byte could we have processed last?" # Because CRC is linear, we can just brute‑force each step (256 possibilities) # and keep the one that leads to a feasible state. With 9 steps it is trivial. bytes_rev = [] crc = target_crc for _ in range(length): # Find a byte b such that there exists a previous CRC value. # Because the CRC algorithm is bijective for a fixed length, any byte works; # we simply pick the one that yields a CRC that is a multiple of 2**8. # The easiest way: try all 256 possibilities and keep the first that makes # the high‑byte of the previous CRC zero (which will be the case for the # correct sequence). for b in range(256): # Reverse the step prev = ((crc ^ TABLE[(crc ^ b) & 0xFF]) << 8) | ((crc ^ b) & 0xFF) prev &= 0xFFFFFFFF # After reversing one byte, the CRC must be divisible by 2**8 for the # next reverse step (since we are moving leftwards). This property holds # for the true sequence. if (prev & 0xFF) == 0: bytes_rev.append(b) crc = prev >> 8 break else: raise RuntimeError("No suitable byte found – something went wrong") return list(reversed(bytes_rev)) any byte works