Crack 56: Adeko 9

# ------------------------------------------------------------ # 2. Reverse the custom transform def invert_transform(b): """Given transformed byte b = ROL8(c ^ 0x5A, 3), recover original c.""" # Inverse of ROL8 by 3 is ROR8 by 3 r = ((b >> 3) | (b << 5)) & 0xFF c = r ^ 0x5A return c

# ------------------------------------------------------------ if __name__ == "__main__": TARGET = 0x56C9A4F2 Adeko 9 Crack 56

(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 Compute a 32‑bit “hash” of the transformed buffer

# Pre‑compute forward CRC table (standard) def crc32_table(): tbl = [] for i in range(256): c = i for _ in range(8): c = (c >> 1) ^ POLY if (c & 1) else c >> 1 tbl.append(c & 0xFFFFFFFF) return tbl ++i) (c &gt

// 2. Compute a 32‑bit “hash” of the transformed buffer uint32_t h = 0xFFFFFFFF; for (int i = 0; i < 9; ++i) h ^= buf[i]; for (int j = 0; j < 8; ++j) if (h & 1) h = (h >> 1) ^ 0xEDB88320; // CRC‑32 (polynomial 0xEDB88320) else h >>= 1;

The program uses the insecure gets_s but limits to 63 characters – no overflow. The real work is in check_serial . 3.3. The serial‑checking routine In Ghidra the function is named check_serial (address 0x140001560 ). Its decompiled pseudo‑code (after some renaming) looks like this:

// 1. Transform each character: xor with 0x5A, then rotate left 3 bits for (int i = 0; i < 9; ++i) (c >> 5); // rol 3