ESET CrackMe Challenge 2015 Walkthrough

The ESET CrackMe Challenge 2015 is divided into 2 parts:

  1. This is the one you download from the ESET website. You are asked to reverse an UPX packed executable and find one password (Drevokokur). Then the application decrypts a message with this password that basically asks you to decrypt in the same way some unreferenced data inside the exe. This unreferenced data, once decrypted, gives you a link to download the 2nd part, which is the subject of this document. I won’t discuss the first part any further.
  2. The second part is made of an EXE and a DLL. The files to analyze are two: EsetCrackme2015.exe and EsetCrackme2015.dll


The aim of this challenge is to find 3 passwords. The application uses various obfuscation techniques and I have set my goal to get things done with the minimum effort. Given the huge amount of obfuscation/hiding techniques used in this challenge, this paper is only meant as a guide to retrieve the three passwords. I won’t describe a lot of things in detail, if you want to deepen some topics you can check this .7z archive. Its structure is as follows:

|                    Is divided in subfolders. Every subfolder’s name is the ID of a resource (see below).
|                    Subfolders contains file related to my analysis, IDA DBs, tools I wrote and, eventually,
|                    the resource itself decrypted (if the original is encrypted).
|                    Contains raw resources dumped from the challenge.
|                    Contains utility I wrote to analyze the .net stuff
|                    An overall description of all the resources.
|                   The challenge itself.

For this challenge I used:

  1. IDA Pro as disassembler/frontend of WinDBG for debugging native code (either for the user mode and the kernel mode part)
  2. D.I.E. (Detect it easy) to analyze the files
  3. dotPeek to reverse the .net stuff
  4. de4dot mainly to rename obfuscated symbols
  5. Other tools I wrote from time to time


DIE Analisys





The DLL is just a collection of resources with a very simple schema:

  • 2 Byte: ID
  • 4 Byte: Size
  • N Byte: data

Some of them are encoded/encrypted, others are not. The encryption/encoding schema varies, see $\Resources.xlsl.

I’ll often reference these resources.


DIE Analysis




Note: the file is small and a large portion of it is made of digital signature (high entropy) so the DIE analysis is distorted.


The application decrypts resource with ID 0x0101 using the key “Irren%20ist%20menschlich” (resource ID 0x0003), then enters in a loop.

In this loop it repeatedly calls the decrypted resource until its result is not zero. At that time it waits for some events to occur (see below), then it starts from the begin.

The application also starts a thread that makes it communicate with every entity that makes a request through the PIPE \\.\pipe\EsetCrackmePipe (ID 0x0002) with a simple protocol:

  1. Request
    1. 1 byte: Type of request
    2. 2 byte: ID
  2. Response
    1. 1 byte: Type of request
    2. 2 byte: ID
    3. 4 byte: size (if request is of type 1)
    4. N byte: data (if request is of type 1)

Type 1 is used to retrieve resources, type 2 is used to signal events.

First password

In the decrypted routine that is called by the application a lot of things can happen.

First time the resources with ID 0x0102, 0x0103, 0x0104 are loaded.

  1. The first is a (simple) virtual machine
  2. The second is the bytecode of a program that uses process hollowing to inject the resource 0x0151 in Svchost.exe.
  3. The third is the bytecode of program that can take the resource data, a path in input, and dump the resource in the path.

I leave out the analysis of the VM for the moment because it is not important for the first password. All we need to know for now is that this program inject something in svchost.exe

This can be easily seen by the context, because if you break at 0x00060AEC (the VM main loop) you’ll see that at some time the process svchost.exe is created suspended, so it’s easy to guess what’s going on. Also, if you look for the processs linked with the GUI that stands out when you open EsetCrackme2015.exe, you’ll see that is svchost.exe and not EsetCrackme2015.exe

We can see what is injected by dumping the resource on disk after decryption ($\Resources\0x0151_Injected\Injected.exe). It doesn’t use any debugging protection, so you can simply attach your debugger to the right instance of svchost.exe and dump process data.

Analyzing the injected program we can see that it is a very simple dialog based application:


On the button click events, it computes a hash on the inserted password and then compares them against a pre-loaded hash. These hashes are loaded from the main application through the PIPE, then decrypted (ID 0xBB01). Algorithm which is used to compute hash is SHA-1, so I consider this a dead end.

The application also starts a thread, that replaces the DlgProc at runtime via SetWindowLong.

The new DlgProc is almost useless but it intercepts the OnClick event and it leads us to the first password because it encodes the first inserted password with a slightly modified BASE64, and compares the result to a pre-loaded string (for other details see 0x00072360). This string is the resource 0xBB02.

Applying the algorithm conversely on the loaded string, we can see the first password: Devin Castle, a nice place in Slovakia:


Second Password

Inserting the first password lets us move to the second stage of the challenge: the driver.

When a password is inserted correctly, an event is signaled through the PIPE. The thread that manages the PIPE in the main application, records the event and wakes up the main thread. The main thread re-runs the decrypted DLL’s routine. Basing on the event signaled, it can start the VM program 0x104 to write some resources somewhere. In the case of the 1st password, this resource is the one with ID 0x0152 and the place is the application working directory.

This resource is a zip folder that contains a Win32 legacy driver that you have to install.

This driver is a s, its architecture is more or less the same that you’ll find as example in Microsoft website (see: $\ESET CrackMe_2\Analysis\\RAMDISK). You can read that code to understand the main functionality of this driver.

There are, of course, some important differences that I’ll explain.

The routine that changes most is the DispatchReadWrite: basically, the principle is that you write something in the ramdisk then when you read it the driver decrypt what you wrote on the fly. One of the differences between this driver and the MS example is that this driver does not create any symbolic link, so every read/write is done by creating a handle to the device name that is \\?\GLOBALROOT\Device\45736574\.

The driver in the “add” routine, starts a bunch of threads that use the PIPE to communicate with the application. The application gives the driver the string “ESETConst” (ID 0xAA02) that is used as the name of a registry key and a program for the virtual machine that lies inside the driver (ID 0xAA06).

This program is the one who decrypts what you write inside the ramdisk.

Here we are forced to analyze this VM. I wrote a VM bytecode decompiler in an assembly-like language, you can find it inside $\Analysis\0x0102_VirtualMachine\VMDecompiler (it has some minor bug but should always produce valid disassembly).

The VM that runs inside the driver is slightly different from the one that runs inside the application. In particular, the offset of the header fields are different.

Header contains the following fields:

  1. 2 Byte: Signature
  2. 4 Byte: Size
  3. 4 Byte: MainOffset
  4. 4 Byte: library flag

The signature is compared against a constant (“1337” for app VM, “3713” for driver VM). If the signature does not match, the execution begins from the 1st byte following the header (decryption stub), if it matches, the execution begins from the MainOffset byte following the header.

VM is quite simple. It can support 255 opcodes but only 15 of them are implemented. If the program contains not supported opcocodes, the program terminates.

The VM virtualize a stack, 4MB, and a CPU with 16 registers:

  • Register from 0 to 5 are GP
  • 6 is SP
  • 7 is begin of loaded program
  • 8 is loaded program size
  • 9 is the library flag
  • Register from 10 to X, where 10<X <16 is a constant passed to the VM, contains some DWORD passed to the VM (input data)

Here is a brief description of opcodes, for further reference see the file $\Analysis\0x0102_VirtualMachine\VMOpcodes.xlsl:

  1. Terminate the program.
  2. Move data. Destination is a register.
  3. (Native) Call. Make a call to the argument.
  4. Call library. Arguments are the hash of the name of a (loaded) library and the hash of the name of a function exported by the library.
  5. Push argument in the stack (load argument in memory pointed by SP and then decrement SP).
  6. Pop dword from stack into a register (copy dword from memory pointed by SP into register then increment SP).
  7. Logical operation, first operand is a register. Operation could be: not equal, equal, minor. If the condition is met, a flag is set.
  8. Jump. This can be conditional, in that case the IP is updated only if the flag is set.
  9. (Virtual) Call. Push IP, then change it with argument.
  10. Pop dword from stack into IP.
  11. Arithmetic operation, first operand is a register. Operation could be add, sub, left shit, right shift, ROR, ROL, module.
  12. Allocate memory. Operand is the size. Pointer to allocated memory is saved in Register 0.
  13. Deallocate memory. Operand is the pointer.
  14. The program pointed by the argument is loaded in memory takes the place of the loaded one.
  15. Terminate program. This is called when an invalid opcode is found.

The VM iterates through every byte in the program and calls the appropriate handler until it finds an exit opcode or an invalid one. The handlers accept a structure (Context) in input that contains the status of the CPU.

The program that runs inside the driver could be seen in $\ESET CrackMe_2\Analysis\\VM Programs. The odd thing is that in this program the registry key (ESETConst) is checked against null but seeing the dispatch RW routine it is easy to spot that the VM can’t start if the Key is null.

In fact, reversing the algorithm we find the 2nd password: Barbakan Krakowski a building in Krakow:

Eset Crackme Challenge 2015 second password

Eset Crackme Challenge 2015 second password

I wrote a program to emulate the behaviors of the VM in a high-level language. In $\ESET CrackMe_2\Analysis\\DriverVMEmulator there is a software that decrypts the resource 0x0155 as the driver would do. It can also run the algorithm conversely to find what the registry key should be: Reversing is great

The result of the decryption is this bitmap:

Eset Crackme Challenge 2015 decrypted bitmap

Eset Crackme Challenge 2015 decrypted bitmap

Third password

Guessing the 2nd password lets us move to the third part of the challenge.

When you guess the 2nd password, the application writes the resource with ID 0x0153, 0x0154 in its working folder. These are 2 .net applications called PunchCardReader.exe and PuncherMachine.exe

Both of them are windows form applications. Both are obfuscated with a custom obfuscator (I guess). The obfuscation schema is the same and it is very simple:

  1. Most of the obfuscation is based on splitting function bodies into different cases inside a switch (num1^num2). One of the integers does not change among different loops but other can be reassigned inside the various cases. In this way the next time the loop is executed, it can enter in another case. Every case usually ends with a branch to the switch address:

    Eset Crackme Challenge 2015 .Net Disassembly

    Eset Crackme Challenge 2015 .Net Disassembly

  2. Strings are not directly visible. They are merged together and constitute a blob of data. There are some stub methods, one for each string, that know how to extract strings from this blob of data.
  3. As various obfuscation schemas do, invalid C# names are used. In IL you can, for example, re-uses the same name for different types, because in IL you always have to fully-qualify everything you use, so there is no ambiguity.

My first attempt to understand what the software does, was to write a plugin for de4dot. I wrote the skeleton of the obfuscator but then I realized that important functions are very little and it is not hard to understand what they do, so I’ve chosen a different approach:

  1. First of all a “de4dot –un-name !.*” is by default.
  2. So I’ve used dotPeek to decompile the program and recreate the VS project.

With very little changes (mainly moving variable declaration) and some textual parsing (see $\ESET CrackMe_2\Utility\PunchStringDeob) I was able to compile both projects.

In these application, resources loaded from the main app are decrypted with the AES algorithm in ECB mode (that I think could explain a lack of the entropy at the end of DLL). The key is an MD5 hash computed against the body of some methods that has a particular attribute and the attribute itself. Of course when I recreated the project in VS the MD5 hash has changed, so I wrote a program that emulate the hash computation using the original file as input, then I hardcoded the key in the recompiled programs to make things work (see $\Utility\KeyExtractor).

Meet PuncherMachine.exe

Once open, it shows this image:

Eset Crackme Challenge 2015 Puncher Machine

Eset Crackme Challenge 2015 Puncher Machine

It uses PIPE to check if EsetCrackme2015.exe is open. If it isn’t, it won’t start.

Of course here I made an “ignorant” attempt to make the application load the file extracted from driver. Of course it didn’t work.

There are some odd things inside PuncherMachine.exe that promptly stand out.

  1. It tries to refer a resource named “resource0” and there is no such resource in the assembly.
  2. It tries to retrieve a resource from the main application with ID 0xFF03 and there is no such resource in the DLL.

You can make the application find resource0 into the assembly or you can perform a “man in the middle” attack through the PIPE to inject the resource, the result is pretty much the same: a hash is computed on this resource and is checked against the hash of the file that you choose from the file dialog. If it matches, the software lead you to the 2nd stage, if it doesn’t match an “Invalid punchcard!” message is shown.

If the application doesn’t find one or the other, it checks the hash of the input file against an unknown hash (resource ID 0xFF02): 95eceaa118dd081119e26be1c44da2cb. Then the check fails because I don’t have the file that has that MD5 hash, but maybe I’m missing something here. In fact I’m quite sure that there is far more than it seems in the bitmap. Of course a bitmap full of noise seems very strange in a challenge that is stuffed with mysteries. My guess is that there is another bitmap that can be decoded starting from the first, and that bitmap has the required hash. By the way I haven’t spent time on this because the bitmap plays a minor role in this process and it is not functional to retrieve the 3rd password. Also, as I’ll explain later, this bitmap although I think is decoded well, does not allow to complete the challenge.

At this point I made the recompiled software load “resource0”, that is my bitmap.

Selecting a valid bitmap make the GUI change:

Eset Crackme Challenge 2015 Puncher Machine App

Eset Crackme Challenge 2015 Puncher Machine App

I’m asked to insert 2 strings: calibration code 1 and calibration code 2.

Seeing the handler to the “Calibrate!” button click I can see that “Calibration code 1” is given in input to a method of an assembly.

The assembly is obtained through PIPE (resource ID 0xFF04) and the method is called “createMethod”.

After dumping the assembly on disk I could analyze it (see $\Analysis\0xFF04_CalibrationDynMethod.dll\CalibrationDynMethod for full code). The assembly is very simple and luckily is not obfuscated. It can create the skeleton of a method with Reflection.Emit classes. In this skeleton some instructions are missing. The method uses the input constants as hashes to decide which opcode to emit so understanding what the generated function should do is the first step to decide which opcode should be there. It is not that hard because it contains some hardcoded constants and searching these constants on Google it comes out that this is the “Knuth Hash” algorithm and the missing instructions are then: ldarg.0 and mul.

Basing on this, the Calibration code 1 should be 0364ABE72D29C96C (see $\Analysis\0xFF04_CalibrationDynMethod.dll\CalibrationDynMethod_HashTable.txt).

The created method is then called in a loop. Every char of the calibration code 2 is added to a char in a list, then the Knuth Hash is computed basing on this input. Every hash in output, then, should be the key of an hashtable for the “calibration process” to be successful. This hashtable is the resource 0xFF00.

It comes out that the only string that met this condition is “Infant Jesus of Prague” (that it is also the third password), a statue in Prague:

Eset Crackme Challenge 2015 Infant Jesus of Prague

Eset Crackme Challenge 2015 Infant Jesus of Prague

4th Condition

Honestly, at this point my interest in the Eset CrackMe Challenge 2015 really dropped down. I found the 2nd password interesting to find (with my actual skills) and I was expecting some kind of escalation, but the third one is actually way too boring and way too easy compared to the 2nd, also bearing in mind that I’m a complete newbie with the .net stuff (the VM decompiler is actually my hello-world in C#) and so my expectations were greatly disillusioned.

The truth is that it was hard to stay in the right mood in order to continue the challenge and I continued until I could do things that required limited effort, but I gave up at the really end because I thought it wasn’t fun anymore.

By the way, if you want I have paved the way and you have all you need in order to finish the challenge, maybe you can prove I was wrong.

Ok, enough bla-bla, back to the challenge.

If you look inside 0x10000CDD you’ll see that there are 4 conditions to meet in order to make the application show you a greeting message. 3 of them are events fired when you enter the right passwords. The 4th condition is an event fired by PunchCardReader.exe, we’ll get there.

What matter for the moment is that when you enter the valid calibration codes, the GUI changes:

Eset Crackme Challenge 2015 Puncher Machine Calibration

Eset Crackme Challenge 2015 Puncher Machine Calibration

As you can see on the “PunchIt!” button handler, it can encode the string that you insert in the bitmap that then are written on disk with name punch_card_XXX.bmp (where XXX is in range 000-999) with a fancy animation and a very relaxing background noise.

Next step is to figure out what should be encoded in bitmaps.

Meet PunchCardReader.exe

Once opened, it shows this dialog:

Eset Crackme Challenge 2015 Punch Card Reader

Eset Crackme Challenge 2015 Punch Card Reader

This is another windows form application. It just has one button in the interface: “Read punch card”.

The principle, of course, is to produce some “punch cards” from PuncherMachine.exe then read these “punch cards” from PunchCardReader.exe

Seeing the disassembled program, it could be easily seen that punch cards are decoded into string. These strings are passed to the method of an assembly.

The assembly is the resource 0xFF05 and it is very similar to the former. It now accepts a string array that is used as IL instructions inside the skeleton of method. So inside these bitmap, created by PuncherMachine.exe there should be encoded a list of IL instruction.

Seeing the assembly, it could easily be spotted what these instructions should be. The code does pretty much this:

((x(0xDEAD,0xBEEF) ^ y(0xCAFE, 0xBABE)) ^ 0xFACE) ^ -229612108 == ToUInt32(“ESET”)

You have to spot what functions x and y are. There is another unknown instruction at the end, but it is clear from the signature of the returned method that it is a ret.

So it is very easy to spot the instructions:

  • x = mul
  • y = add
  • z = ret

Punching the bitmap with mul, add and ret does not let PunchCardReader.exe validate the bitmaps although these seem to be the right strings. I spent very little time on this but I think that it is due to the fact that the original bitmap is wrong, and this does not let the decoding process to work well.

If this assumption is correct, I think that 2 hypotesis are the most plausible:

  1. The re-implementation in C# of the VM program that run inside the driver is bugged. At first glance this seems unlikely to me because it produces an overall valid bitmap, but who knows.
  2. There is another bitmap that can be extracted from the 1st in some way (like trying to decrypt the content of the bitmap, XOR it or whatever). That bitmap is the one that has the unknown hash and will allow the decode process to work well.


That’s all!
After all it is a nice crackme. I think the VM part is the best because it is very well written and stable. Kudos to the authors.

Looking forward for the 2016 challenge, bye.

>>> Download Archive <<<


  1. Nice analysis! 🙂 I also solved the ESETCrackme2015 (IDA + WinDbg, didn’t use a VM emulator, nice one of you!), but probably there are 2 challenges (and I solved the first one) because I noticed some differences: the decrypted bitmap (PunchCard.bmp) is a valid image in my case (and not an entropy like one); there are some minor differences also in the driver VM (nothing important). This is my results:

    • chaplin89 says:

      Hi Matteo,
      as I said it could simply be the fact that my re-implementation of the driver VM program is bugged. Onestly I didn’t cared too much. I noticed the differences in the driver VM too (I also said something about this in the article).
      I’ve some question:
      1) What means “didn’t use a VM emulator”? How did you have analyzed the VM?
      2) How did you have extracted the bitmap?
      3) Did you injected the bitmap through the PIPE (or directly in the assembly) or your bitmap has the hash 95eceaa118dd081119e26be1c44da2cb?

      • 1) I didn’t analyze the VM till the end, I simply followed the execution flow using WinDbg + IDA to understand what conditions were necessary to decrypt the bitmap. So, your solution of using a VM decompiler & rewrite the interesting part of the decryption routine is better then mine but you probably have to fix something to get the valid image.
        2) As said, I simply met the conditions inside the driver and let it decrypt the bitmap inside the RamDisk, BUT I probably misunderstood something, because during a call to a driver routine I modified on the fly a pushed value, otherwise the decryption would have failed.
        3) My bitmap has the correct hash and in fact is a valid image; it shows an old punch card.

  2. mehspizdy4ebura6ki says:

    Hi chaplin89! Thanks for your article, it looks exact at many details. Since the moment to reveal the internals of this crackme has arrived i would like to contribute a bit concerning the punchcard bitmap: the resource ID_0155 is the arc4-encrypted bmp-image (by 512-byte chunks), and the VM-bytecode executed by the driver implements the arc4 algo, while the keyword “Barbak……” is used to initialize a permutation table.

    • Hi mehspizdy4ebura6ki,
      thank you for the info, very precious!
      I’m not a cryptogeek so I didn’t recognize the ARC4 algorithm, I simply tried to emulate the VM behaviour in C#. Maybe when I’ll have some spare time I can try to figure out what’s wrong with my C# implementation and update the article. Thanks!

      • mehspizdy4ebura6ki says:

        having a glance look taken at your sources it seems you reconstructed a working arc4 impl. But guess, probably you are trying to decode the ID_0155 as a whole piece. While it’s necessary to decode it by 512 byte chunks. Look, the first 512 byte of the file Analysis\\DriverVMEmulator\ConsoleApplication4\bin\Debug\punchcard_1.bmp are decrypted correctly, but the next bytes are junk. Althouh it’s a stream cypher the driver ends up using it as a block one: 512 byte block / 512 byte key. You are supposed to simulate this too.

    • I know nearly nothing about encryption algos, only solved 3 keygenme with RSA some weeks ago. I was curious about the algorithm used in the crackme (I simply ignored it and let it decrypt the image), how did you get it was RC4? I think one can identify it only if he knows how it is implemented, right?

      • mehspizdy4ebura6ki says:

        Matteo, certainly it’s necessary to know some details for to recognize the algo. But the basic arc4 impl for home usage is simple enough and takes only several codelines in C. When reversing the most evident indicators are: a permutation table using; an input text is just xored by a key to obtain a cypher (and vice versa).

        BTW the driver also uses the arc4 to decrypt the VM-engine (.data:00405020), VM-bytecode (.data:004071D0), ID_AA02 (dl from pipe), ID_AA06 (from pipe). You can found the native x86 arc4 routines in the driver at:
        .text:00403190 void ARC4_crypt_buffer(uint8_t* data, size_t data_size) // uses the password “3531_4ever”
        .text:004032F0 void ARC4_init_key_stream(arc4_context* ctx, uint8_t* password, size_t password_size)
        .text:00403230 void ARC4_crypt_impl(arc4_context* ctx, uint8_t* data, size_t data_size)

        And man, i’m just wondering how freely you are concatenating the “I know nearly nothing about encryption algos” with “only solved 3 keygenme with RSA”, that’s awesome 🙂

        • Thanks for the fantastic explanation, as soon as I’ll have my commented IDA db I’ll look into the implementation. Believe me the KeygenMe I solved where implemented using MIRACL and really simple (I read about RSA vulnerabilities on a book and tested the implementation against them, it worked). I will read about RC4 for sure 🙂 also, this challenge gave me so much to learn, from driver analysis, VM & .NET.