Challenge 2 is simple yet interesting. The initial target is a Python 2.2 byte-compiled file, so the first job is to decompile it to get the source code. Fortunately, decompyle just works:
$ decompyle newbie.pyc
Thu Aug 27 02:13:25 2009
# emacs-mode: -*- python-*-
a = 'http://'
dummy = 'http://korea'
b = 'uxcpb.xe'
b = b.encode('rot13')
c = 'co.kr'
cs = '.com'
d = '/vfrp/uxuxux'
dt = '/hackers'
d = d.encode('rot13')
dx = 'coolguys'
ff = urllib.urlopen(((a + b) + d))
f_data = ff.read()
file = open('hkhkhk', 'w')
You can see that the purpose of this script is to download some data from a fixed URL, and save them to a file named hkhkhk. We ran the script, and it indeed downloaded this file. As the script suggests, the content of hkhkhk is encrypted by some cipher.
Opening hkhkhk in a hex editor, one could see that it contains quite a lot of 0x77 characters. A friend of us, Julianor from Netifera, thought that hkhkhk is an executable file, and because excutable file contains a lot of null bytes so 0x77 may be the null byte in the original file. He suggested xoring the content of hkhkhk against 0x77. We did as he suggested, and it worked :-D. hkhkhk turns out to be an ELF executable file:
$ file hkhkhk
hkhkhk: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV),
for GNU/Linux 2.2.5, dynamically linked (uses shared libs), stripped
./hkhkhk [server] [port]
port> 1111, 2222, ..., 9999
Disassembling hkhkhk reveals that this binary is just a simple client that connects to a remote server to get two integers, and send the sum of them back to that server. If the result is correct (which is always), the server will return a congratulation message like below:
$ ./hkhkhk 220.127.116.11 1111At first, we thought we should try to exploit the server to force it to return an error or something, but that didn't work. Then we thought there's something hidden inside hkhkhk, so superkhung and I spent 1 hour to inspect every single instruction of the binary, but we saw nothing weird.
[(867925) + (9792)] = ?
answer is 877717
it's correct. great!, :-)
At this point, a friend suggested us running the binary inside a debugger. He thought that there may be something hidden in the communication between the server and hkhkhk.
The communication? I fired up wireshark, and to my surprise, I saw the answer right away: Pandas likes hkpco XD. It turns out that the congratulation message is something like:
it's correct. great!, :-)\x00Password is "Pandas likes hkpco XD"This message is passed to a printf call, and since printf expects a null-terminated string, one could never see the characters after the null byte if he doesn't run the binary inside a debugger, or sniff the communication like us.
Challenge 9Challenge 9 (IP: 18.104.22.168; port :4600) is a remote stack-based buffer overflow exploitation. It's interesting because WOWHacker doesn't release the binary as other usual exploitation challenges.
While I was banging my head against challenge 8, gamma95 told me that he could crash challenge 9 with 293 bytes. He thought that this challenge is very obvious, and wondered why none was working on it.
Actually we were very short on manpower in the first day of the premilinary round. So we chose to work only on those challenges that we were interested in or had a larger chance of solving them.
When I first saw challenge 9, I thought this challenge should be hard. Blind remote exploitation is supposed to be hard you know. This wrong assumption plus the fact that I haven't practiced software exploitation in the last several months made me decide to leave this challenge for other teamates who might join us in the second day.
But it turns out this challenge is an easy one.
In order to exploit a stack-based buffer overflow vulnerability, one must know which address to return to. Fortunately, WOWHacker gives us a very helpful hint:
Mr.Her give you something "call me~ call me~" : bfbfeaf2So 0xbfbfeaf2 is the return address. Normally this address should point to the beginning of our input buffer which in turn should have this structure:
The next problem is to determine how many bytes we need to control the EIP. The trick is to use \xeb\xfe as the shellcode, and increase the message one byte a time until we see the service hang after it processes our input. If our theory of the structure of the input buffer is correct, this process will succeed eventually because \xeb\xfe means "loop forever":
$ echo -ne '\xeb\xfe' | ndisasm -Using this technique, we can see that we need totally 302 bytes to control the EIP:
00000000 EBFE jmp short 0x0
$ (python -c 'print "\xeb\xfe" * 149 + "\xf2\xea\xbf\xbf"'; cat) | nc 22.214.171.124 4600We use Metasploit to generate a BSD reverse-shell shellcode, and we got the answer: WOWHACKER without beist.
Actually this wasn't as easy as we write here. We made two stupid mistakes: first off, we assumed that this challenge ran on a Linux box; secondly, our connect back box was behind a firewall :-(. Thanks Tora and biest for giving us a hand in resolving them.