Kimsuky has obfuscated binary strings including the use of XOR encryption and Base64 encoding.[207][208] Kimsuky has also modified the first byte of DLL implants targeting victims to prevent recognition of the executable file format.[209]
byte crypter v3 cracked 56
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During Operation Dust Storm, the threat actors encoded some payloads with a single-byte XOR, both skipping the key itself and zeroing in an attempt to avoid exposing the key; other payloads were Base64-encoded.[275]
Pisloader obfuscates files by splitting strings into smaller sub-strings and including "garbage" strings that are never used. The malware also uses return-oriented programming (ROP) technique and single-byte XOR to obfuscate data.[285]
A common usage scenario for a malware crypter involves an operator purchasing a copy of the crypter in a compiled form (or using a cracked version), then using the crypter to obfuscate the malware executable which is to be distributed. In this scenario the developer of the crypter has no knowledge of what specific malware the threat actor will eventually choose to encrypt with the purchased copy of the crypter.
Iranian Connection: Analysis of the malware and seeding yields a consistent Iranian presence. The binary contains Iranian and Iranian-Persian traces, as do the tools used for obfuscation, which are popular in Iranian cybercrime forums. Similarly, the mailer discovered on the assadcrimes[.]info website is in Persian. There is also the intriguing, but ultimately unproven speculation that the crypter may have been sold to Group5 by a known Iranian malware developer. Furthermore, logs of access to the assadcrimes[.]info site suggest that the operators are working from within Iranian IP space. In addition, the bait content also contains substantial Iranian themes. Finally, the hosting provider (Hostnegar) is Iranian. A final piece of highly circumstantial evidence is that PowerPoint documents containing exploits, albeit often with quite different (and sometimes custom) malware, is a commonly reported feature of many recently-reported Iranian campaigns.
In addition, the code that prints characters stops when a null byte is reached. Investigating the printed plaintext and the ciphertext will reveal that the ciphertext is larger than the plaintext and there is data in addition to the hint seen above, which is not printed.
We only learned later that our LCG reaches a fixed point after a few iterations, meaning that the key stream degraded to only a single byte that was XORed against most of the message. This meant that the challenge could also be solved very quickly by simply XORing the bootloader with every possible byte value and looking for flags in the results.
crypter.py implements two functions, C(K, M) and U(K). C(K, M) is the encryption and decryption routine. For decryption, the decryption key is passed in the first argument, and the encrypted message in the second argument. If the function C is used for encryption, the outcome of U(K) is used as the key and the plaintext is passed in the second argument. The function U(K) computes the mathematical inverse of K where K (a list) is interpreted as a 33 matrix.
Likewise, the first argument K of C(K, M) is interpreted as a matrix as well. The argument M of the function C is a bytes object that gets padded with null bytes to a length that is a multiple of 3, and is then divided into chunks of 3 bytes. Each chunk is interpreted as a vector and multiplied with K using matrix-vector multiplication.
The code reads STDIN byte by byte into the variable c. The integer v12 is set to 1 for the beginning of each line and is reset to 0 after a character is read that is not \t. Therefore, v12 tracks whether the current character inside c could be part of tab-based indentation (if it is \t).
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dorizigf 19191a764c -crypter-v7-cracked-windshield[ -crypter-v7-cracked-windshield ][ -crypter-v7-cracked-windshield ][ -crypter-v7-cracked-windshield ]link= -crypter-v7-cracked-windshieldlink= -crypter-v7-cracked-windshieldlink= -crypter-v7-cracked-windshield
According to this document, the MATRIX record triggers the exploit by setting the field NumberOfRows too high. Only 8 bytes are reserved in eqnedt32.exe for the array RowPartitionLineTypes, but (2 * 0xec + 9) / 8 = 0x3c bytes are copied instead, leading to a stack overflow:
In a nutshell, the function reads the bitmap line by line, and each line pixel by pixel. For every byte of the bitmap, some bits (the lowest significant bits) are extracted and concatenated in order to assemble the final payload. This is textbook steganography. The first line is a bit special since it contains additional info:
So let us try if we got it right. We will open the bitmap BBTREX (which is a DIB bitmap, meaning the BITMAPFILEINFOHEADER is missing) in an hexadecimal editor and try to manually decode the first bytes. We first have to locate the first bitmap row. Good to know: bitmaps are stored upside down, i.e the top-most line is actually the last one in the file. So knowing that our bitmap is 588 pixels wide and is a RGB bitmap (so 3 bytes per pixel), the first line should start at EndOfFile - 588*3 = 0x44ea8:
So first thing first, we will decrypt the first 4 bits integer (aka stegano_num_lsb_bits). The first line starts with the 3 bytes 03 02 02, which gives us the binary number 1100 (in LSB display) = 3. Ok.
Next, the algorithm moves to the second pixel and reads 32 bits. 32 bits / 3 bits per byte means it will read 10 bytes and 2 bits of the 11th byte. The next 11 bytes are: 00 00 00 03 06 02 00 00 00 00 04, which gives us the binary number 000 000 000 110 011 010 000 000 000 000 00(1) (in LSB display) = 91648 ok. The 11th byte contains an additional bit which we did not read which was a (1).
Next we could start reading 2 bytes of the payload, which is 16 bits. Since we still have a bit unread from 04, we just have to read 15 additional bits or 5 bytes. The next five bytes are: 06 04 04 06 02, which gives us the binary numbers (1) 011 001 0--01 011 010 or 0x4d--0x5a ... looks like the start of a PE, great!
And the delimiter ^^Nc can be found as referenced string in the second stage binary at address 0x413f58, so could it be our url? At this point we should look for decrypting functions inside one of the two binaries. But let us be smart. See how the string prefix ammil3(( has repeated characters. Encryption is must likely a weak one-byte cipher. And we know that we are looking for an url, so the plain text string could definitely start with So let us try a few usual cipher:
Knowing that the most frequent x86 function prologue is 55 8B EC (aka push ebp; mov ebp, esp), it looks like all bytes value are just one off. So let us try our hypothesis and just subtract 1 to the complete .text section. This can be done easily using Malcat's transforms, as we can see below:
All DX archives follow the same format, except they may have a unique 12-byte encryption key. The purpose of this program is to allow users to maintain a keystore that the program will use to automatically try each key and attempt to decrypt the archive.
After dumping the file or files, open it in a program like Mad Edit or HxD and try adding the 12 byte string hex code to keys.ini saving it and repeating until you find the correct string. Remember always start from Offset 00000000.
Transition to uint64 values allows reconstruction of the key from the header (there are enough null bytes) and the last 12 bytes of the file (which are all 0 except one 0x40 byte).All files are encrypted with the same key rotated bytewise per file, and the method of rotation can be found from the exe linked above.
C705CA7D8DE3DEF1D90C85F4 Wolf RPG Editor v2.20 betaThis is the base key used to decrypt the header (first 48 bytes of DXv6 archive). All files and the table of content at the end of the archive are encrypted using variations of this base key, 12 in total (including the key itself), 1 per file and TOC. I forgot to add it to the previous message. 2ff7e9595c
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