README.md

patchme.py

Author

LT 'syreal' Jones

Description

Can you get the flag? Run this Python program in the same directory as this encrypted flag.

Hints

(None)

Approach

If we look at the code, we'll see it's supposed to run level_1_pw_check and if we look at the function, we see this:

def level_1_pw_check():
    user_pw = input("Please enter correct password for flag: ")
    if( user_pw == "ak98" + \
                    "-=90" + \
                    "adfjhgj321" + \
                    "sleuth9000"):
        print("Welcome back... your flag, user:")
        decryption = str_xor(flag_enc.decode(), "utilitarian")
        print(decryption)
        return
    print("That password is incorrect")

It checks to see if the user inputted "ak98" + "-=90" + "adfjhgj321" + "sleuth9000" which means the password is "ak98-=90adfjhgj321sleuth9000".

$ python3 patchme.flag.py
Please enter correct password for flag: ak98-=90adfjhgj321sleuth9000
Welcome back... your flag, user:
picoCTF{p47ch1ng_l1f3_h4ck_4d5af99c}

Note to run Python, you will use either python or python3, not necessarily python3

Flag

picoCTF{p47ch1ng_l1f3_h4ck_4d5af99c}