Flare-on 9 - Challenge 9 and 11

Flare-on 9 - Challenge 9 and 11


Flare-on 9 ( Challenge 9 and 11 )


Flare-on 9 has officially been declared over!


This year of flare-on is the third try and the first that I have ever completed so far! Definitely did have my share of pain and joy during this time of challenges. For this post, I will share my writeup on challenge 9 and 11.

Challenge 9 - Encryptor

This challenge took me quite a bit of time to realise what algorithms were used. The two main algorithms that we have to familiarize ourselves with are RSA and ChaCha20. ChaCha20’s key was used to encrypt files in the specified directory for files extension .EncryptMe. However, it is a common practice as well to encrypt the key with asymmetric encryption. In this case it was using RSA to do so.

Here is the high level view of the main function with some reversing and renaming.


RSA initialization

The reason that there were much trouble in figuring out that this is RSA is due to the fact that c/c++ does not handle very huge numbers. Therefore, it makes sense to have functions that can deal with them. Since RSA is more secure with very very large random prime numbers, this code section would looked a little more cryptic than the formula we would see online.

Here is the renamed init_RSA_sub_4021d0 which is responsible for the initialization of RSA needed for asymmetric encryption.


In init_RSA_sub_4021d0, RSA key values including:

  • p
  • q
  • N -> p*q
  • phi(N) -> (p-1)*(q-1)
  • Decryption Key -> d
  • Exponent e -> 0x10001


Next, in the Encrypt function that I labeled in main function, we can see that chacha20 was used to encrypt the content of the sample buffer before the key is being encrypted with RSA.



To solve this, we have to understand how RSA encryption algorithm works. To encrypt :

# Encryption
cipher = (msg ^ exponent) % N

# Decryption
msg = (cipher ^ decrypt_key) % N

Additionally, in the ransom note given, we are given some hardcoded values, the N value, the encrypted content and the encrypted chacha20 key. However, during the encryption of the file content, the encryption was calculated mistakenly.

cipher = (msg ^ decrypt_key) % N

Instead of encryption, this has just turned into a “verification” where you encrypt the content with the decrypt key rather than the exponent to “prove” that you are the legitimate. But since this was done this way and that the exponent is hardcoded to 0x10001, we can decrypt with the following:

Chacha20_key = (enc_chacha20_key ^ exponent) % N

This works mainly because of the condition in RSA since the computation of exponent according to https://www.di-mgt.com.au/rsa_theory.html states that ` ed≡1(modϕ(n))` . This means that with the mistaken encryption

cipher = (msg ^ decrypt_key) % N

Therefore, to get back msg, we need to xor the cipher with the hardcoded exponent

Cipher ^ exponent = (msg ^ decrypt_key ^ exponent) % N
Cipher ^ exponent = (msg ^ 1(modϕ(n))) % N
(Cipher ^ exponent) % N = (msg ^ 1(modϕ(n))) % N
(Cipher ^ exponent) % N = msg

Now to obtain the chacha20 keystream, we can make use of the given values from the ransom note.


The keystream shown in the above image should be read as little endian.

Now that we get the decrypted keystream, we can start the actual decryption of the ` SuspiciousFile.txt.Encrypted` file. What I did to decrypt the encrypted file was to manually insert the bytes of the initial state of ChaCha20 cipher.


During the actual solve, I have done this dynamically by replacing the data after the expand 32-byte k string. However, we can solve it in a script as well since the decrypted state contains both the key and the nonce.


Running this should give us the flag :


Troubles Faced

One of the troubles I faced was confusing Salsa20 with ChaCha20 since this was the first time hearing about these. After googling the expand 32-byte k keyword, Salsa20 was the first algorithm I saw.


However, after some time, I noticed that during the hashing, the rotate values were not the same as the source implementation of salsa20. Therefore, it has to be ChaCha20 and they did match as seen in the following image.


As for RSA functions, it took some actual trial and errors to confirm them since I was only familiar with the simpler implementation of RSA and not considering generation of prime numbers and more.

At the end of the challenge, though the logic is pretty easy to understand, I think that the thing about this is to be able to have some intuition about relevant functions.

Challenge 11 - The challenge that shall not be named

This challenge took me a pretty long time due to quite a number of rabbit holes. Even so, I think that what I did was not the intended solution, that is by memory inspection to look for the flag which should also make sense once I explain things.

This challenge binary is actually a PE32+ that was compiled via PyInstaller. Upon extraction with pyinstxtractor, we can guess that 11.pyc is the main function for the executable. Using decompyle3, we can see that the decompiled shows that the main function was compiled with pyarmor in strong mode (the value 2 after the encrypted byte codes).


To run the main function, I have compiled python3.7.exe ( more about this later ) and placed it in the PYZ-00.pyz_extracted directory as the extracted data so that we can run python in it and that many of the modules are contained in the same directory. Running 11.pyc with python outside PYZ-00.pyz_extracted would give an error if run outside of this directory indicating that the module _crypt is not found.

If you want to see the solution straight without going through what I have failed, then feel free to jump straight to the solution segment.

The Different Attempts

At this point, I have tried quite a number of methods to attempt to get the bytecode.

Dumping Python serialized Code Object

First thing I have tried was to look at _PyEval_EvalFrameDefault function in python 3.7 source. This is because pyarmor would decrypt the bytecode so that this function could run it before encrypting it back. I have also attempted to dump the bytecodes that were my separately compiled python executable ( the reason why I compiled Python 3.7)


In this added code, PyMarsahl_WriteObjectToFile function is used to serialize python object. The ones that we are interested are the ones that are the PyCodeObject or Code Objects as seen in http://www.goldsborough.me/python/low-level/2016/10/04/00-31-30-disassembling_python_bytecode/. However, after extracting all the bytecodes from both going on in 11.pyc in x64Dbg and the stored code_object text file, we see that the pyarmor code object was returned and I was not able to step into it just like that.


Furthermore, I read that pyarmor has their own implementation of PyEval which suggest that this method would not work or is just not marshallable back to its original code.

API Monitor

It gave a really good overview (though not complete) about what the program was doing.


We first see that there is an attempt to resolve for www.evil.flare-on.com. After that, I have setup a server with resolvable IP address.

Since, I did not know what the server actually does, i thought we have to interact with the server somehow and to do that, I have tried to create a server that can send bytes back assuming it is some form of a server.


With the server setup as follows and editing the Windows Host file, we receive the base64 encrypted flag via post request.


However, api monitor did not have any other information about this. But it did give us some important information about the program.

Patching pytransform.pyd

I was thinking of patching pytransform.pyd mainly because this is mostly responsible for the obfuscation and restriction of pyarmor code. While I think this is possible, I did not manage to successfully do this before getting the flag. The restriction comes mainly from trying to edit pyarmor code.

Solution - challenge 11

Firstly, I have decompiled both 11.pyc and crypt.pyc (after noticing that it was pyarmor’d) into 11.py and crypt.py

Next, I configured the environment so that it would run with the python.exe from the pyz_extracted folder. I have also used the following configuration in Visual Studio Code to allow stepping in of functions by setting justMyCode.


After some things were done, the “encrypted” flag was passed into base64 function.


After which requests.post function was used to send a POST request with the encrypted flag.


Once that is over, there no more operations done. Therefore, I made a guess that the flag generation has to be somewhere before the base64 operation. That said, I tried to do a memory dump to see if we can find the flag anywhere on the heap in WinDbg. Turns out after attaching to the python process, and searching in the heap with the following commands, it is possible to find the full flag within the memory space. Commands used were !heap -a and s -a <start address> <end address> “@flare-on.com” which shows positive traces of the flag and eventually, we can dump the flag out.




NGL, but after this challenge, I have invested quite a fair bit of time (24 days) to solving these challenges. I have definitely see some improvements that I have made over the past two to three years and am absolutely delighted with my progress. I would also like to thank the FLARE team for creating challenges like these. Not only is it fun, I did learn quite a lot also especially during challenge 8, including learning C# programming and scripting.