Greetings from OTA,
So it was a short binary given, on analyzing we found it was using aes_cbc_256 
```cint  setup_crypter(__int64 a1){  __int64 v1; // ST08_8@1  __int64 v2; // rax@1  __int64 v3; // rax@1  __int64 v4; // rax@1  void *v5; // ST18_8@1  __int64 v6; // rax@1  v1 = a1;  v2 = HMAC_CTX_new();  a1 + 160 = v2;  v3 = EVP_sha256();  HMAC_Init_ex(*(_QWORD *)(v1 + 160), v1 + 64, 32LL, v3, 0LL);  v4 = EVP_CIPHER_CTX_new();  a1 + 168 = v4;  v5 = calloc(0x10, 0x1);  v6 = EVP_aes_256_cbc(16LL, 1LL);  ######### USING AES 256CBC  return EVP_EncryptInit_ex(*(_QWORD *)(v1 + 168), v6, 0LL, v1 + 32, v5);  ###### KEY i.e v5 IS NEVER INITIALIZED}```So lets try all 32 bit combination 
```pythonfrom Crypto.Cipher import AES
maybeKey = open('core', 'r').read().strip()flag = open("flag.txt.enc", "rb").read()
for i in range(0,len(maybeKey)-32):    key = maybeKey[i:i+32]    aes = AES.new(key, AES.MODE_CBC, IV="\xff"*16)    text = aes.decrypt(flag)    if "CTF" in text:        print (key, text)        break
```Reveals the flag `CTF{core_dump_your_secrets}`