## SolutionThis challenge can be solved by brute force, or more precisely. The brute force approach is to try every printable character at each location of the flag until the number of characters correct in the response from the server increases. This is implemented in [one_byte.py](https://gitlab.com/malcrypt/ctf-archive/-/blob/master/2021/tenable/misc/one_byte/one_byte.py). The more precise method is to recognize that the flag is being XORed by a 4-byte key, and since we know 5 bytes of the plaintext, we can recover the key, then decrypt each value we receive from the server to determine the flag.
## Flag**flag{f0ll0w_th3_whit3_r@bb1t}**