# haxXor> Points: 281
## Description> you either know it or not take this and get your flag>>`5552415c2b3525105a4657071b3e0b5f494b034515`## SolutionI did a XOR brute force first in [CyberChef](https://gchq.github.io/CyberChef/) but nothing useful came out. 
So I tried a partial plain text attack.
I knew the start of the flag was **darkCTF{** (partial plain text) so I xored it with the given hexvalue to get the `key`.
Once I got the key I padded it to match the length of the cipher. Finally I xored the padded key and cipher to get the flag.
```py#!/bin/env python3
hex_string = "5552415c2b3525105a4657071b3e0b5f494b034515"flag = b'darkCTF{'
key = []f = []
#finding the secret keyb = bytes.fromhex(hex_string)	for c, m in zip(b, flag):	key.append(chr(c ^ m))print("Secret key : " + ''.join(key))
#using the secret key to decrypt secret = b'1337hack1337hack1337h'
for c, m in zip(b, secret):	f.append(chr(c ^ m))final_flag = ''.join(f)print("Final flag is : " + final_flag)```
## Flag> darkCTF{kud0s_h4xx0r}