README.md

patchme.py

Write-up author: jon-brandy

DESCRIPTION:

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

HINT:

• NONE

STEPS:

1. Download all the files given.
2. Open the python source code and change the == characters to != at line 19.

### 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()

3. Now, run the source code in the same directory as the encrypted flag text file.
4. When the program asks for input, type aaaa then press enter.

5. Because we change the "if statements" algorithm by giving a flag if the input does not match what it should be, it is certain that the program will give an output that is a decrypted flag.

6. Finally, we got the flag!

---

FLAG

picoCTF{p47ch1ng_l1f3_h4ck_21d62e33}