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Patchme.py

03/15/2022

By: Jonawhogood

Tags: rev PicoCTF-2022

Problem Description:

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


When you open the python file, you see this:

### THIS FUNCTION WILL NOT HELP YOU FIND THE FLAG --LT ########################
def str_xor(secret, key):
    #extend key to secret length
    new_key = key
    i = 0
    while len(new_key) < len(secret):
        new_key = new_key + key[i]
        i = (i + 1) % len(key)        
    return "".join([chr(ord(secret_c) ^ ord(new_key_c)) for (secret_c,new_key_c) in zip(secret,new_key)])
###############################################################################


flag_enc = open('flag.txt.enc', 'rb').read()



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")



level_1_pw_check()

You notice that the program checks if the password the user entered is “ak98-=90adfjhgj321sleuth9000”, so then you just input that the next time you run the program with python3.

Flag: picoCTF{p47ch1ng_l1f3_h4ck_f01eabfa}