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

// 1. Transform each character: xor with 0x5A, then rotate left 3 bits for (int i = 0; i < 9; ++i) (c >> 5); // rol 3 Adeko 9 Crack 56

int __cdecl check_serial(const char *s) uint8_t buf[9]; // 9‑byte “key” derived from input size_t len = strlen(s); if (len != 9) // must be exactly 9 characters return 0; Setup # Create a clean analysis environment mkdir

# ------------------------------------------------------------ # 1. CRC‑32 parameters (same as the binary) POLY = 0xEDB88320 INIT = 0xFFFFFFFF XOROUT = 0x00000000 CRC‑32 parameters (same as the binary) POLY =

# ------------------------------------------------------------ if __name__ == "__main__": TARGET = 0x56C9A4F2

# ------------------------------------------------------------ # 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

Find an input string s (9 bytes) such that CRC32( b_0 … b_8 ) == 0x56C9A4F2 . 4.2. CRC‑32 is linear over GF(2) CRC‑32 with a fixed polynomial is a linear operation: