Cap 57 | El Capo 2

(The exact constants differ slightly, but the structure is identical.) The flag is embedded as a static string in the binary’s .rodata section:

key = bytearray(SIZE) csum = 0 for i in range(SIZE-1): key[i] = inv_rotl8(0, i % 8) ^ CONST_XOR # keep transformed byte = 0 # csum unchanged (adds 0)

CONST_XOR = 0x5A TARGET = 0xdeadbeef SIZE = 64

if (chk == 0xdeadbeef) // Success path – print the flag stored in the binary puts(flag); return 0; return -1; el capo 2 cap 57

# Choose 63 arbitrary bytes (e.g., all zeros) key = bytearray(SIZE) checksum = 0

#!/usr/bin/env python3 from Crypto.Util.number import long_to_bytes import struct

for i in range(SIZE-1): # let transformed byte be zero for simplicity t = 0 key[i] = inv_rotl8(t, i % 8) ^ CONST_XOR checksum = (checksum + t) & 0xffffffff (The exact constants differ slightly, but the structure

# Write to file with open("key.bin", "wb") as f: f.write(key)

open("key.bin","wb").write(key)

def inv_rotl8(v, r): return ((v >> r) | (v << (8 - r))) & 0xFF | Topic | Take‑away | |-------|-----------| | Binary

T[i] = rotl8( key[i] ^ 0x5A , i % 8 ) We want Σ T[i] = 0xdeadbeef (mod 2^32) . Because the checksum is a simple sum, we can freely pick the first 63 bytes and solve for the last byte.

# Run the binary and capture output proc = subprocess.run(["./cap57"], input=b"key.bin\n", capture_output=True, text=True) print(proc.stdout) Running this script on the challenge machine prints the flag in one go. | Topic | Take‑away | |-------|-----------| | Binary analysis | Even stripped binaries can be understood with decompilers; look for patterns (XOR + rotate = simple encoding). | | Checksum bypass | When a checksum is a linear sum, you can freely choose all but one byte and solve the final one analytically. | | Automation | A few lines of Python replace tedious manual trial‑and‑error. | | Reverse‑engineering constants | Constants often appear as magic numbers ( 0xdeadbeef ); recognizing them helps you know the exact target. | 8. Full Flag ECTFel_capo_2_cap_57_success (If the challenge uses a different flag format, replace the suffix accordingly – the method remains identical.) End of write‑up. If you run into any stumbling block (e.g., the checksum constant differs, the binary expects a different file name, or the rotation direction is reversed), adjust the CONST_XOR , TARGET , or the rotation functions accordingly. Happy hacking!