RSS
 

Posts Tagged ‘AES256.pbkdf2’

Hackvent 2019: Day 21

21 Dec 2019
CTF: Hackvent 2019
Link to challenge: https://academy.hacking-lab.com
Date Completed: 21 December 2019

Challenge

HV19.21 Happy Christmas 256

Solution

We review the clues the elves gave us and first start by trying to find Santa password that was leaked 10 years ago. We are looking for data breaches in 2009 so we look at a list of data breaches. We find that the rockyou breach was the biggest breach that year and that is password dumps are readily available (with usernames stripped out) so we download this dump. We also know that Santa’s password is of length sqrt(256) = 16. There are roughly 118k passwords in the dump that meet the length requirement. Another clue tells us that the AES256 key can be derived with pbkdf2_hmac with the salt TwoHundredFiftySix and with 256*256*256 iterations. We also know that Santa’s password is of length sqrt(256) = 16. We attempt to bruteforce the AES key but realise very quickly this is very slow and would take a long time. To be clear, it is quite feasible to crack the AES key this way over the course of say 48 hours but we want a faster solution. We also tried to guess that the password would contain the text 256 and used this to limit our bruteforce space for AES key cracking. This unfortunately failed but was a cool idea! (If only we used the word santa instead).

Instead, we look at Santa’s private keys and notice how the corresponding private key is his password with SHA256 encryption. This is much, much faster to bruteforce as we can generate potential private keys and attempt to sign and verify a test message. If our verification is successful, we will have our original password which we can then use to decrypt the AES cipertext directly.

Putting all of this together we write our Python script:

Running the above gives us Santa’s password and our flag after 30 seconds:

Flag:  HV19{sry_n0_crypt0mat_th1s_year}

 
No Comments

Posted in Hackvent 2019