Pwn2Win CTF 2016 Secret Accounts (Py 80) Writeup


Description: Through many months of sniffing, we discovered a server running a software which the Club uses to manage information about secret bank accounts abroad. We even obtained its source code. We need to obtain access to the system in order to discover the real name of the owner of the account possessing the greater amount of money, in which bank it is, and the real amount. As you might expect, it seems that the Club has hunkered down to assert only authorized people, which really know what they are doing, are able to operate this system and to interpret information provided by it. Rumors exist that the Mentor himself manages it, as amazing as it may seem (he is old but not deciduous!).

You will need good “python-fu” to win this challenge!

Submit the flag in the following format: CTF-BR{info1_info2_info3}.

Hint: Who is Fideleeto (Cuba!) in real life? Take this into account. :)


The provided Python file was somewhat obfuscated with lots of unconventional things like octal numbers, unnecessary arithmetics and so on. The first part of the challenge was getting past the intial password prompt. The password check algorithm was very convoluted, so we started off by translating octal to to decimal, and simplified it:

name=raw_input("Login name: ")

if ((name[:9] == master[:9]) and
	(name[9:13] == master[9:13]) and
	(len(name[13:]) == len(master[13:]))
	and (name[13:] > master[13:])) == False:
	status = False
	status = True

So the password is 9 characters, appended by a birthyear and lots and lots of lowercase and uppercase Z. There was a hint in the description referencing Cuba. We used that to figure out the birthyear as the birthyear of Cuban leader Fidel Castro. Since the assignment is following the theme of the CTF, the first nine characters are likely Fideleeto. The remaining part of the password was easy to construct. Just add the Z, and make sure the string input string is “larger” than that in the master variable (we did this by changing one of the uppercase Z to lowercase).

Once we got past the password, we were shown a prompt with different account-related actions:

Available Options:

1 - View Owners of Accounts
2 - View Banks of Accounts
3 - Modify Sum
4 - Exit

Simply entering the number here is not sufficient, since the input numbers are altered in the function toption(opt). To figure out what input that gave what option value, we just bruteforced it. We found the following mappings:

toption(110) = 1
toption(132) = 2
toption(144) = 3
toption(162) = 4

So now that we can provide the menu with real options, we need to enter another password. Looking at the code, we can see that things get interesting here. They are using the input() function. The Python documentation states this this is equivalent to eval(raw_input(prompt)). That sounds very dangerous. Let’s see if it can give us shell ;)

We enter the following in the input prompt: os.system('/bin/sh'), and we’re provided a shell as expected. From there we can read the infos file, containing bank account information:

{"Eduardo Cunha" :
	["Credit Suisse","5002005"],

"Dilma Roussef" :
	["Coop Bank","22582599"],

"Lul4 M0lusc0" :
	["Compagnie Bancaire Helvetique","25987369"],

"Delubio Soares" :
	["Compagnie Bancaire Helvetique","2100125"],

"Delcidio Amaral" :
	["Graubundner Kantonalbank","1780695"],

"Jose Dirceu" :
	["WIR Bank","8528314"],

"Renan Calheiros" :
	["Julius Baer", "11254255"]}

Here we can see that Lul4 M0lusc0 has the highest balance, so we enter his information into the flag format, and get our points.