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Posts Tagged ‘C’

HACKvent 2016: Day 5

05 Dec 2016
CTF: Hackvent 2016
Link to challenge: http://hackvent.hacking-lab.com
Date Completed: 5 December 2016

Challenge

everybitisimportant

 

Solution

This seems like a series of boolean logical operators. As the hint tell use to use 32 bits, we will solve this problem with a quick C++ program so we can guarantee the data type used is 32 bits. Furthermore, we will try both signed and unsigned variants, it turns out that we need to use signed integers for this problem.

We come up with C++ code (splitting up the operations into 3 steps):

We run the program and the the printed result is:

I enter this into the ball-o-matic and get the daily QR code and daily flag!

Day 5 Solution Ball

 

Flag:  HV16-2wGq-wOX3-T2oe-n8si-hZ0A

 
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Posted in Hackvent 2016

 

HACKvent 2015: Day 14

14 Dec 2015
CTF: Hackvent 2015
Link to challenge: http://hackvent.hacking-lab.com
Date Completed: 14 December 2015

Challenge

The following Windows binary was also provided: Download EXE File

 

Solution

I download the binary and run it and am presented with the following program:

Hackvent Day 14 Program

It turns out that this program will tell you (via a messagebox) if you enter in the correct daily nugget or not!
So all we have to do is check the binary to see what causes the successful message box to appear.

Note: You can do this challenge using IDA or a .NET disassembler like ILSpy (link).

If using IDA, its useful to be familiar with CIL instructions.

ILSpy Approach
I decided to use ILSpy as it is apparently a very good .NET disassembler. I open the program and load the binary and it disassembles it into various classes as you would expect.
We are mainly interested in the hv15 class. By searching for strings like yes, that is the key! we realise the only important functions we need to look at are Button1_Click and  Encrypt .

This is the code for both:

Button1_Click

 

Encrypt

 

It becomes super simple to solve this challenge at this stage. The input parameter is just the text we enter into the textbox and the pass parameter is  Form1.GlobalVariables.assembly which is defined to be the string  __ERROR_HANDLER. All we have to do is reverse the encryption starting with an input that equals  zV5/UFU8PUD3N2T49IBuCwvGzCLYz39tkMZts7rfBU4=. We first decode the base64 string into a byte array and then run the program again but with  rijndaelManaged.CreateEncryptor() changed to rijndaelManaged.CreateDecryptor().

I wrote a small C# program that accomplishes what we want to do:

We run the above program and get our flag!

Flag:  HV15-uQEJ-4HPX-Qcau-Xvt7-NAlP

 
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HACKvent 2015: Day 12

12 Dec 2015
CTF: Hackvent 2015
Link to challenge: http://hackvent.hacking-lab.com
Date Completed: 12 December 2015

Challenge

The following C code is also provided:

 

Solution

Clearly the issue here is that this program is very inefficient.
So what does the program do? First it sets the unsigned 64 bit integer variables  val and i  to 0. Then 0xC0DE42 iterations occur and the val is recomputed each calculation. The previous val  and i  values are used to recompute the new val  value.

After all this, it appears as if the nugget we need is printed out! The 4 missing blocks in the nugget are based on 4 groups of 16 bits which come from the final val value.
So we begin to optimize the code!

 

foo and bar

We start with the foo and bar  functions which are very similar to each other.
It turns out that they simplify to:

So we have just optimised these two functions slightly. The next step is go through the code and replace all calls to foo  and bar  with simply increments or decrements. I’m not going to show the changes I made to each call as that would make this post too long but you get the idea.

Optimized functions:

 

baz

baz  should look like this at this stage:

It is clear from the code that baz  can be optimised. First the variable r  is set to zero and then a 1 is added to r , a times. Then 1 is added to rb times.
This is simply the same as adding a  to b  and storing the result in r .

Optimized function:

 

spam

Spam should look like this at this stage:

This function is simply setting r  to a and then taking away 1 from rb times before returning the result in r .
This is the same as returning a - b .

Optimized function:

 

eggs

eggs  should look like this at this stage:

Well all that is happening here is r  is increasing by b, a times. That means we are adding a number of b’s to r .
This is the same as returning a * b .

Optimized function:

 

merry

merry  should look like this at this stage:

So at this stage, we are taking away b  from a  while a  is still bigger than b . The value of  i is then returned which is the number of times we took away b  from a .
This is clearly a simple division operation: a / b

Optimized function:

 

xmas

xmas  should look like this at this stage:

This one is a little more tough. a  is divided by b  then multiplied by b  and this result is subtracted from a . You may be tempted to think that this returns 0 all the time but it does not.
This is because the division operation that occurs is integer division and multiplying that result by b  may not result in the original a  being restored.
A small example: 10/3 = 3.33 but is stored as 3 in an integer. However, 3*3 = 9 and not 10!

In this case, a modulo operation is happening: a % b

Optimized function:

 

hackvent

hackvent should look like this at this stage:

This is a fun one. Notice how r is set to 1 initially, that is important. r is then multiplied by ba number of times.
This is the same as calculating b to the power of a.
However, we cannot just use the pow function defined in <math.h>.
We are dealing with uint64_t data types and must thus get a power function that can handle these larger numbers.
I decide to find one online and use it.

Optimized function ( ipow  and hackvent):

 

Putting it all together

At this stage you can do two things.
You can either run your program as is to find the answer or you can inline your function or convert the function calls in main so you save making all those function frames.
In this case it doesn’t really matter but this is what the calculation of val in main should look like with no function calls:

Then you run your program and within a few seconds you get the flag!

Flag:  HV15-mHPC-067e-751e-f50e-17e3

Note: You can download my final solution C file here:
Download hv15-d12-solution.c

 
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Posted in Hackvent 2015