Hack The Box: Impossible Password Reverse Engineering Challenge

This week I’m going to be completing the impossible password reverse engineering challenge from Hack The Box. This challenge is ranked easy, so I’m fairly confident we’ll be able to knock this one out pretty quickly.

What I’ll be using

For this challenge, I’ll be using the following:

  • Linux VM – Any VM that will give you access to a terminal w/ GDB should suffice. I’m personally using Kali Linux 2021.2 in virtualbox on my Windows 10 Machine.
  • GNU Debugger (GDB) – This came with my kali linux vm.
  • GEF (optional) – This is an addon that I use with GDB that you’ll see in my screenshots. It’s really nice to have, but not required. Information on GEF can be found here.
  • Ghidra (optional) – I like to use Ghidra to help orient myself when I’m stepping through assembly. It’s part of my workflow, but definitely not required for this challenge. Information on Ghidra can be found here.

Step 1: Download Challenge Files

Visit the challenge on Hack The Box: https://app.hackthebox.eu/challenges/impossible-password

Once you’re there, download the challenge files and store them in your working directory:

When unzipping the binary into your working directory, the password is hackthebox

Step 2: Poke the file

Now that I have the file in my working directory, I’m just going to start by running it in my linux vm and see what happens.

It seems it’s doing something, but it’s too early to tell exactly what’s going on. Let’s go ahead and boot this thing up into Ghidra and see if we can’t figure out what’s happening under the hood.


Step 3: Open the file in Ghidra and find the executed code

Let’s go ahead and open up the binary in Ghidra with all of the default settings:

After importing the file, let’s open it up and go ahead and click “yes” with the default settings when prompted to analyze the binary:

In the functions view, let’s head over the entry by double clicking it:

Once there, we’ll see some code populating our decompiler. It should look something like this. Let’s navigate over to what looks like will probably be our starting function by double clicking it:


Step 4: Analyze the binary in Ghidra

Now that we’ve entered this function, we should start seeing some interesting things. Let’s take a look at some interesting items that are now populating our decompiler:

The line labeled 52 should stand out to us. That is what we saw at the start of our line above when we executed the binary in our terminal. This tips me off that this is probably the area of code that we experienced previously when we executed the code.

Let’s break down what’s probably happening here in this decompiler with some educated guesses.

  • Line 52: Print the start of the line *
  • Line 53: Probably reading in the user input into variable local_28
  • Line 54: Prints out [local_28] in square brackets with a new line character. We also experienced this behavior when we ran the binary earlier
  • Line 55: This is just a strcmp() from C. So this will compare local_28 and local_10. If they are equivalent, local_14 will equal 0.
  • Lines 56-59: Checks the string comparison result. If they are not equal, then the program will exit.

Analyzing the lines above seem to follow everything that we’ve experienced so far when we first executed this binary. Based on all of this, it would seem that our key is located in local_10 which we can see defined right at the top of my screenshot on line 31 as SuperSeKretKey.

According to the decompiler, if the user enters the correct key the program will proceed forward and then output ** for a second key. Let’s test this hypothesis by executing the program again and entering the key we just found:

It would seem that our hypothesis was correct! Awesome.

Step 5: Finding The Second Key location in Ghidra

Let’s take a look back at the decompiler in Ghidra and focus in on the section of code that we’re now at

Analyzing this it would seem that the following is happening:
  • Line 60: Print out **. This is exactly where we left off in the previous execution.
  • Line 61: This looks like it’s taking user input and storing it in local_28 again.
  • Line 62: Looks like some function is called and storing a string into __s2.
  • Line 63: This is doing a string comparison between the user input, local_28, and the previously mentioned __s2 variable.

Line 63 leads me to believe that __s2 probably will contain our second key, so let’s take a look into function FUN_0040078d by double clicking it in the decompiler.

I’m not going to lie, my brain immediately turned off when I saw this.

Because this is an easy level challenge, I’m sure we can probably cheese this without actually analyzing this too deeply. Let’s backup to our previous location in Ghidra and try grab the address of our strcmp for the second key.

Looking at this, the comparison seems to happen at address 0x00400961. Let’s go ahead and boot this up into GDB and see what we can do from there.

Step 6: Dynamic Analysis with GDB

Inside our linux VM, let’s boot this binary up in GDB

Using the address we just grabbed for our string comparison location, let’s go ahead and add a breakpoint there.

Now run it and enter our first secret key when prompted.
Great. At this point, we can go ahead and enter whatever input we want and press enter. This will continue code execution, but the program will stop as soon as it reaches the string comparison for the second key. This is because of the breakpoint that we set previously.

If you’re using GEF, you’re GDB should look something similar to what’s on my screen as shown above. If you’ll notice here, we can see the last two assembly calls moved some values before the string comparison

Let’s take a look at what’s in register rsi and rdi with the following

Looking at this, I would have to guess that the register rsi probably contains the second key and rdi contains the user input. Let’s see if we can cheese this challenge by just setting the registers equal to each other and then continuing code execution.

Well, I’ll be damned. It actually worked. The flag has been censored because I encourage you to actually try this out for yourself!



As the challenge difficulty suggests, this was an easy reverse engineering problem. I’m not sure if my solution to this problem was what was intended, but it works and for CTF challenges like this that’s all that really matters.

As always, if you liked what you read and you’d like to support me, please consider buying me some coffee! Every cup of coffee sent my way helps me stay awake after my full-time job so that I can produce more high quality blog posts! Any and all support would be greatly appreciated!

Ballad Serial — Trans Arsonist Art Network

6 thoughts on “Hack The Box: Impossible Password Reverse Engineering Challenge

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