Challenge 16: convertme.py

Description

Run the Python script and convert the given number from decimal to binary to get the flag. Download Python script.

Tags

Beginner picoMini 2022 General Skills base python

Points

100

Solution

If we run the script, we get the following output:

$ python3 convertme.py

If 48 is in decimal base, what is it in binary base?
Answer: 

We need to convert the decimal number 48 to binary. We can do this in another terminal by using the bin() function in python. We can run the following command:

$ python3 -c "print(bin(48))"
0b110000

We can then enter the answer into the script:

$ python3 convertme.py

If 48 is in decimal base, what is it in binary base?
Answer: 110000

That is correct! Here's your flag: picoCTF{4ll_y0ur_b4535_762f748e}

Or we can take a look at the source code of the script:

import random



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 = chr(0x15) + chr(0x07) + chr(0x08) + chr(0x06) + chr(0x27) + chr(0x21) + chr(0x23) + chr(0x15) + chr(0x5f) + chr(0x05) + chr(0x08) + chr(0x2a) + chr(0x1c) + chr(0x5e) + chr(0x1e) + chr(0x1b) + chr(0x3b) + chr(0x17) + chr(0x51) + chr(0x5b) + chr(0x58) + chr(0x5c) + chr(0x3b) + chr(0x42) + chr(0x53) + chr(0x5c) + chr(0x0d) + chr(0x5e) + chr(0x50) + chr(0x4d) + chr(0x00) + chr(0x13)


num = random.choice(range(10,101))

print('If ' + str(num) + ' is in decimal base, what is it in binary base?')

ans = input('Answer: ')

try:
  ans_num = int(ans, base=2)
  
  if ans_num == num:
    flag = str_xor(flag_enc, 'enkidu')
    print('That is correct! Here\'s your flag: ' + flag)
  else:
    print(str(ans_num) + ' and ' + str(num) + ' are not equal.')
  
except ValueError:
  print('That isn\'t a binary number. Binary numbers contain only 1\'s and 0\'s')

as we can see, we can just hack our way through the script and get the flag:

import random



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 = chr(0x15) + chr(0x07) + chr(0x08) + chr(0x06) + chr(0x27) + chr(0x21) + chr(0x23) + chr(0x15) + chr(0x5f) + chr(0x05) + chr(0x08) + chr(0x2a) + chr(0x1c) + chr(0x5e) + chr(0x1e) + chr(0x1b) + chr(0x3b) + chr(0x17) + chr(0x51) + chr(0x5b) + chr(0x58) + chr(0x5c) + chr(0x3b) + chr(0x42) + chr(0x53) + chr(0x5c) + chr(0x0d) + chr(0x5e) + chr(0x50) + chr(0x4d) + chr(0x00) + chr(0x13)

flag = str_xor(flag_enc, 'enkidu')
print(flag)

Remember my friend, that you can always look at the source code of a script to get the flag. It’s not always the best way to solve a challenge, but it’s a good way to get the flag if you’re stuck.

And Remember, we are HACKERS, not script kiddies. We don’t need to follow the rules. 😉

Flag

picoCTF{4ll_y0ur_b4535_762f748e}