Analyzing Ransomware - Completing a FULL Analysis

In this video, we complete a FULL analysis of a single ransomware - from reversing the keygen, to discovering the encryption algorithm, and figuring out the file format of encrypted files.
*Note: In the video I mention I thought the 1 byte written is a checksum of the filesize; I actually just realized it the number of padding bytes added to the last block, so the ransomware can remove that many bytes after decrypting. It's basically using ANSI X9.23 padding.
Sample: www.hybrid-analysis.com/sampl...
VirtualBox: www.virtualbox.org/
DIE: ntinfo.biz/
PeID: www.aldeid.com/wiki/PEiD
IDA Free: www.hex-rays.com/
x64dbg: x64dbg.com/
CryptoTester: download.bleepingcomputer.com...
More information on the ransomware analyzed: / 1093226825772384257

[@OALABS]

Nice! I'm finding these tutorials very interesting... always curious to see different folks RE process. Keep them coming : )

[@rianray5296]

Thank you so much mate for analyzing the ransomware malware and help the victims. now a days this kind of people is very rare who helping others without any reason. a hearty wishes and good luck for you "my friend".

[@nasmRE]

Very beatiful analysis !!

[@xthukuh]

Brilliant! Learning a lot!!

[@2008pomsin]

This man is true genius!

[@yunietpiloto4425]

good stuff buddy, thanks for sharing

[@jurdanesh5383]

Thank you for your hard work=)

[@Siy1995]

Great work 💥
So Sir I have .tfudet extension ransomeware how can I dectypt my file ?

[@akmalhasbi2400]

My computer have problem. Reco file .reco. How to fix it?

[@albaromamani7393]

Hi Michael I have a ransomeware .Clop, you have any.idea how is encrypted? We are more than 5 computers

[@Demonslay335]

I took a look at that one, and it is not decryptable. On too of that, the criminals make a mistake that makes it so they cannot decrypt... So I wouldn't recommend paying them even if you were thinking about it.

[@albaromamani7393]

@@Demonslay335 thanks Michael for your response. We have tried everything with all published methods but it seems that any methods gets worse. 1 computer doesn't recognize the hard disk. There will be nothing but format everything.

[@saurabhjoshi94]

Hi, did you get the answer to this query ?
And @Michael Gillespie can you please post a video on analysis of CLOP malware ?

[@sent4dc]

10:02 Small usability tip: To Ctrl+V into cmd window use: dennisbabkin.com/clc/

[@mallikarjunfp23]

Kindly do full analysis of encrypted pe\dll malware.

[@Demonslay335]

I could maybe try, but I honestly only focus on ransomware in particular. Maybe if one happens to have an encrypted payload though. ;)

[@misharatkevich9808]

Just stumbled upon this video -- am I missing something, or does this particular conversion to lowercase ASCII _drastically_ reduce the number of possible TEA keys? Instead of properly using something like base64 to keep all of the original options, the creators of this specific malware are doing it through mod 25 (and then add 97 to get it into the right ASCII) range, which... hang on. Each byte can now only be a lowercase letter... so 26 options for each byte, and there are 16 bytes total in the TEA key.
16*26=416
If that's _really_ it, simple bruteforce of lowercase ASCII letters would work very, very quickly to recover the TEA key used for file encryption.
Or am I missing something here? Is there any big reason this type of targeted bruteforce wouldn't actually work? I'm a total noob, so I barely understand what's happening, LOL.
EDIT: Went to do my groceries and literally minutes later I realized I typed the operation wrong into calculators, it's 26^16, not 26*16, LMAO. I knew that about regular keys (it's 2^128, not 2*128) but somehow forgot and mistyped it like this for this version.
They're still not using their entire keyspace, so maybe 26^16 is feasible enough for a brute force attack, but probably not. :P Wolfram Alpha is giving a number with 39 digits for 2^128, and with 23 digits for 26^16, so it's slightly above 2^75 in terms of security. (2^75 is 37 sextillion and something, 26^16 is 43 sextillion and something, 2^76 is 75 sextillion and something, so it's between 2^75 and 2^76. The keyspace reduction is vast, but this is still a hell of a number to crunch.

[@upliftingspirit6873]

The only gate that could result in a vulnerability is the fact that they encrypt a 128 bit buffer with 1024 bit modulo. This results in the rest 1024-128=896 bytes being null bytes and one could crack it using the Coppersmith's Short Pad Attack. However this works for small values of e, the public exponent, so no luck here :(