int main(int argc, char **argv) char input[64]; puts("Enter your serial: "); gets_s(input, sizeof(input)); if (check_serial(input) == 0) puts("Invalid serial! Try again."); return 1; puts("Correct! Welcome, Adeko."); return 0;
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;
If we denote the post‑transform byte as b_i = t(i) , the CRC algorithm is applied to the sequence b_0 … b_8 . Adeko 9 Crack 56
# 4. Verify with the original CRC routine (optional) def crc32
(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 int main(int argc, char **argv) char input[64]; puts("Enter
// 1. Transform each character: xor with 0x5A, then rotate left 3 bits for (int i = 0; i < 9; ++i) (c >> 5); // rol 3
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)) 3) | (b <
# ------------------------------------------------------------ # 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
t(i) = ROL8( c_i XOR 0x5A, 3 ) ROL8 rotates an 8‑bit value left by 3 bits.
# ------------------------------------------------------------ # 1. CRC‑32 parameters (same as the binary) POLY = 0xEDB88320 INIT = 0xFFFFFFFF XOROUT = 0x00000000