Santhacklaus 2019: Jacques ! Au secours !

Published December 23, 2019 • 3 minutes read

Before we start! This post is authored by @pwnh4. I'm really proud to have him for this guest post as he started computer security only a few months ago with my 2019 Piscine but already performs very well in CTF. Go follow him!

Jacques ! Au secours! is a cryptography challenge of the Santhacklaus 2019 CTF.


Let's start by downloading and extracting the provided .zip file:

➜  chall_files tree
├── vacation pictures
│   ├── DCIM-0533.jpg.hacked
│   ├── DCIM-0534.jpg.hacked
│   ├── DCIM-0535.jpg.hacked
│   ├── DCIM-0536.jpg.hacked
│   └── READ_THIS.txt
└── virus.cpython-37.pyc

The .pyc file is some Python bytecode. It is not human readable so let's use some online Python decompiler to retrive the actual code source of what looks to be a Python ransomware.

The Ransomware

from Crypto.Cipher import AES
from Crypto.Random import get_random_bytes
import hashlib, os, getpass, requests
TARGET_DIR = 'C:\\Users'
C2_URL = ''
TARGETS = ['Scott Farquhar', 'Lei Jun', 'Reid Hoffman', 'Zhou Qunfei', 'Jeff Bezos', 'Shiv Nadar', 'Simon Xie', 'Ma Huateng', 'Ralph Dommermuth', 'Barry Lam', 'Nathan Blecharczyk', 'Judy Faulkner', 'William Ding', 'Scott Cook', 'Gordon Moore', 'Marc Benioff', 'Michael Dell', 'Yusaku Maezawa', 'Yuri Milner', 'Bobby Murphy', 'Larry Page', 'Henry Samueli', 'Jack Ma', 'Jen-Hsun Huang', 'Jay Y. Lee', 'Joseph Tsai', 'Dietmar Hopp', 'Henry Nicholas, III.', 'Dustin Moskovitz', 'Mike Cannon-Brookes', 'Robert Miller', 'Bill Gates', 'Garrett Camp', 'Lin Xiucheng', 'Gil Shwed', 'Sergey Brin', 'Rishi Shah', 'Denise Coates', 'Zhang Fan', 'Michael Moritz', 'Robin Li', 'Andreas von Bechtolsheim', 'Brian Acton', 'Sean Parker', 'John Doerr', 'David Cheriton', 'Brian Chesky', 'Wang Laisheng', 'Jan Koum', 'Jack Sheerack', 'Terry Gou', 'Adam Neumann', 'James Goodnight', 'Larry Ellison', 'Wang Laichun', 'Masayoshi Son', 'Min Kao', 'Hiroshi Mikitani', 'Lee Kun-Hee', 'David Sun', 'Mark Scheinberg', 'Yeung Kin-man', 'John Tu', 'Teddy Sagi', 'Frank Wang', 'Robert Pera', 'Eric Schmidt', 'Wang Xing', 'Evan Spiegel', 'Travis Kalanick', 'Steve Ballmer', 'Mark Zuckerberg', 'Jason Chang', 'Lam Wai Ying', 'Romesh T. Wadhwani', 'Liu Qiangdong', 'Jim Breyer', 'Zhang Zhidong', 'Pierre Omidyar', 'Elon Musk', 'David Filo', 'Joe Gebbia', 'Jiang Bin', 'Pan Zhengmin', 'Douglas Leone', 'Hasso Plattner', 'Paul Allen', 'Meg Whitman', 'Azim Premji', 'Fu Liquan', 'Jeff Rothschild', 'John Sall', 'Kim Jung-Ju', 'David Duffield', 'Gabe Newell', 'Scott Lin', 'Eduardo Saverin', 'Jeffrey Skoll', 'Thomas Siebel', 'Kwon Hyuk-Bin']

def get_username():
    return getpass.getuser().encode()

# ....

def lock_files():
    username = get_username()
    if username in TARGETS:
        for directory, _, filenames in os.walk(TARGET_DIR):
            for filename in filenames:
                if filename.endswith('.hacked'):
                fullpath = os.path.join(directory, filename)
                print('Encrypting', fullpath)

        with open(os.path.join(TARGET_DIR, 'READ_THIS.txt'), 'wb') as (fo):
            fo.write('We have hacked all your files. Buy 1 BTC and contact us at\n')

if __name__ == '__main__':

The way the ransomware works is pretty simple. It has a list of usernames defined in TARGET. If the victim is one of them, it encrypts his files and asks for a ransom.

Let's focus on the encryption process:

def xorbytes(a, b):
    assert len(a) == len(b)
    res = ''
    for c, d in zip(a, b):
        res += bytes([c ^ d])

    return res

def lock_file(path):
    # 1
    username = get_username()
    hsh ='md5')
    key = hsh.digest()
    cip =, 1)

    # 2
    iv = get_random_bytes(16)

    params = (('target', username), ('path', path), ('iv', iv))
    requests.get(C2_URL, params=params)

    with open(path, 'rb') as (fi):
        with open(path + '.hacked', 'wb') as (fo):
            block =
            while 1:
                if block:
                    while 1:
                        if len(block) < 16:
                            block += bytes([0])
                    cipherblock = cip.encrypt(xorbytes(block, iv))
                    iv = cipherblock
                    block =


We can observe multiple interesting things:

  1. The key used for encryption is the MD5 hash value of the victim username. We can also see the declaration of an AES cipher in EBC mode (, 1))
  2. We can see that a variable IV is defined. An Initialization vector is used in AES CBC mode which means that this function is implementing its own AES CBC encryption. The IV is set to a random 16 bytes value.
  3. There is a request to what should be a C&C (useless for our case)
  4. The encryption algorithm: for each 16 bytes block of the file to encrypt, we xor it with the IV and we encrypt the result with the AES ECB cipher defined in step 1. Then, the result is written in the .hack file and the IV is set to this current encrypted block.

This algorithm is beautifully represented in this Wikipedia schema:

Wikipedia CBC

The problem

We can uncipher every block in the .hack file except the first one, as it is the IV that was initialized to a random 16 bytes value.

But luckily we can bypass this! We know that we want to retrieve a JPG files. Thus, we could try to guess the first plaintext block based on the standard header of the JPG format. Let's try with:


An other problem is that we should try all targeted usernames as key. But let's make some guess and consider that the key is Jack Sheerack (because of the challenge name). We will check the unciphered files generated with this key first.

Ok, so now that we have solutions for our issues, let's decode the hacked files!

The script

for target in TARGETS:
    key = hashlib.md5(target.encode()).digest()
    f = sys.argv[1]

    # used to store all 16 bytes block
    blocks = []
    with open(f, "rb") as f:
        while not len(blocks) or len(blocks[-1]):
    del blocks[-1]

    cip =, 1)
    with open("/tmp/out-"+target+".jpg", "wb") as fo:
        # write 16 bytes of jpg header
        # decode each ith block using the i-1th block as initialization vector
        for i in range(1, len(blocks)):
            fo.write(xorbytes(cip.decrypt(blocks[i]), blocks[i-1]))

We try to open the file associated with the Jack Sheerack key but it is invalid. Just like all the other files...

At this time I was quite lost and tried to figure out why my generated JPG files were invalid.

Black Magic

After some research, I found the ImageMagick tool, which is great to help me debug my generated files.

With the identify command, you can detect all the corruption sources.

➜  identify -regard-warnings -verbose /tmp/out-Jack\ Sheerack.jpg

  filename: /tmp/out-Jack Sheerack.jpg
  verbose: true
Tainted: False
Filesize: 174736B
Number pixels: 990000
Pixels per second: 62.6515MB
User time: 0.010u
Elapsed time: 0:01.015
Version: ImageMagick 6.9.10-67 Q16 x86_64 2019-10-04
identify: Corrupt JPEG data: 18 extraneous bytes before marker 0xfe `/tmp/out-Jack Sheerack.jpg' @ warning/jpeg.c/JPEGWarningHandler/389.

Thats a good thing! Our JPG is corrupted. How is that a good thing? Well now we know that it is almost valid ! If I try the same thing on a generated file with the key Jack Ma, identify simply doesn't work:

identify: Unsupported JPEG process: SOF type 0xc3 `/tmp/out-Jack Ma.jpg' @ error/jpeg.c/JPEGErrorHandler/338.

At this time I discovered another tool part of the ImageMagick suite: mogrify, which helps correct the corrupted bytes in my generated JPG. I simply have to indicate the error I want it to correct:

➜  mogrify -set comment 'Extraneous bytes removed' /tmp/out-Jack\ Sheerack.jpg

mogrify: Corrupt JPEG data: 18 extraneous bytes before marker 0xfe `/tmp/out-Jack Sheerack.jpg' @ warning/jpeg.c/JPEGWarningHandler/389.

I can now open my file and see a wonderful photo of Jacques Chirac!


The picture with the flag was DCIM-0534.jpg.hacked.

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