For this challenge, we were given a ransomware (VIKING_CRYPT.exe and VIKING_DECRYPT.exe) along with a LOCKER.HTML file and IMPORTANT_NOTE.txt.vkg.

Viking locker ransom message

Let’s take a look at that encrypted note:

$ file Perso/IMPORTANT_NOTE.txt.vkg
Perso/IMPORTANT_NOTE.txt.vkg: ASCII text, with no line terminators
$ cat Perso/IMPORTANT_NOTE.txt.vkg
8a9b600b5a745d39cb7c7e7890e816848a9fe77280b42ca0751d150bb849

It doesn’t really look like any encryption I know.

We can try to run VIKING_DECRYPT.exe:

Viking locker decrypt executable

As expected, we’re asked for a decryption key. Let’s encrypt a few files in a VM, and compare the plaintext and the .vkg files.

Note: the challenge description tells us that, for safety, the malware will only encrypt files in C:\Users\nMnHkfcWUjE\Desktop\Perso.

$ echo 'hello world' > normal.txt
$ echo 'h3110 w0r1d' > leet.txt

PS C:\Users\nMnHkfcWUjE\Desktop> .\VIKING_CRYPT.exe

$ cat normal.txt.vkg
a1a74e2267006c2df04f5e

$ cat leet.txt.vkg
a1f1137f38006c72f0125e

Now this is promising. The matching h has the same value. Let’s try a simple xor between the plaintext and the encrypted value, to see if we get a matching key:

>>> import codecs
>>> print([a ^ b for a, b in zip(b'hello world', codecs.decode('a1a74e2267006c2df04f5e', 'hex'))])
[201, 194, 34, 78, 8, 32, 27, 66, 130, 35, 58]
>>> print([a ^ b for a, b in zip(b'h3110 w0r1d', codecs.decode('a1f1137f38006c72f0125e', 'hex'))])
[201, 194, 34, 78, 8, 32, 27, 66, 130, 35, 58]

The key is exactly the same! Let’s try decrypting our important note:

>>> key = [a ^ b for a, b in zip(b'h3110 w0r1d', codecs.decode('a1f1137f38006c72f0125e', 'hex'))]
>>> print(bytes([a ^ b for a, b in zip(key, codecs.decode('8a9b600b5a745d39cb7c7e7890e816848a9fe77280b42ca0751d150bb849', 'hex'))]))
b'CYBERTF{I_D'

Well that’s only the start of the flag. Let’s encrypt something longer, to get more bytes of the key:

$ echo 'Lorem ipsum dolor sit amet, consectetur adipiscing elit. Sed non enim quam. Morbi luctus tortor dolor, sed pellentesque urna vestibulum eget. In egestas diam tempus risus molestie, id consectetur ante euismod. Integer eget erat sapien.' > long.txt
$ # Run VIKING_CRYPT.exe
$ cat long.txt.vkg
85ad502b65007232f15657689af325bbccc6cb4f9ca7129f24074a1bb65bc8c30162703e3f36149f38c754567af424eb60682eda379edab1b7fe5b44c4df29c471f802c8d8716192291c829b89c3c49695ec48abf2f3721e5b229a20afb2b7c751779c51ff63f378a7a53c27113b306c26fc2b87562768cc82b70cfdbd3fc93730f890234ac890c6e6b9e052f65341eaaf3f7e56d9476c170f17dd5e3830867656c0fd885a68a55ae58bcb182a09ee0a840a03511fb3222d25637b1d344340168dfebdd97a8bc327a36e6d9703f9e6427e9dfd07df86f0140a067a3a843aeef6f4782df2f1e56e380957c2

>>> cleartext = b'Lorem ipsum dolor sit amet, consectetur adipiscing elit. Sed non enim quam. Morbi luctus tortor dolor, sed pellentesque urna vestibulum eget. In egestas diam tempus risus molestie, id consectetur ante euismod. Integer eget erat sapien.'
>>> cipher = '85ad502b65007232f15657689af325bbccc6cb4f9ca7129f24074a1bb65bc8c30162703e3f36149f38c754567af424eb60682eda379edab1b7fe5b44c4df29c471f802c8d8716192291c829b89c3c49695ec48abf2f3721e5b229a20afb2b7c751779c51ff63f378a7a53c27113b306c26fc2b87562768cc82b70cfdbd3fc93730f890234ac890c6e6b9e052f65341eaaf3f7e56d9476c170f17dd5e3830867656c0fd885a68a55ae58bcb182a09ee0a840a03511fb3222d25637b1d344340168dfebdd97a8bc327a36e6d9703f9e6427e9dfd07df86f0140a067a3a843aeef6f4782df2f1e56e380957c2'
>>> key = [a ^ b for a, b in zip(cleartext, codecs.decode(cipher, 'hex'))]
>>> print(bytes([a ^ b for a, b in zip(key, codecs.decode('8a9b600b5a745d39cb7c7e7890e816848a9fe77280b42ca0751d150bb849', 'hex'))]))
b'CYBERTF{I_D0nt_P4y_Th3_R4ns0m}'

Here we go, the flag is CYBERTF{I_D0nt_P4y_Th3_R4ns0m}.

By the way, the key doesn’t seem to repeat (at least for the first 235 values), so it’s probably made from a RNG.

You can view the sources on github or read other writeups.