pwnable.kr - random

4 minute read

Lets connect to the server

ra@moni~/P/p/random> ssh random@128.61.240.205 -p 2222
random@128.61.240.205's password:
 ____  __    __  ____    ____  ____   _        ___      __  _  ____
|    \|  |__|  ||    \  /    ||    \ | |      /  _]    |  |/ ]|    \
|  o  )  |  |  ||  _  ||  o  ||  o  )| |     /  [_     |  ' / |  D  )
|   _/|  |  |  ||  |  ||     ||     || |___ |    _]    |    \ |    /
|  |  |  `  '  ||  |  ||  _  ||  O  ||     ||   [_  __ |     \|    \
|  |   \      / |  |  ||  |  ||     ||     ||     ||  ||  .  ||  .  \
|__|    \_/\_/  |__|__||__|__||_____||_____||_____||__||__|\_||__|\_|

- Site admin : daehee87@gatech.edu
- IRC : irc.netgarage.org:6667 / #pwnable.kr
- Simply type "irssi" command to join IRC now
- files under /tmp can be erased anytime. make your directory under /tmp
- to use peda, issue `source /usr/share/peda/peda.py` in gdb terminal
You have new mail.
Last login: Fri Jun  4 17:29:35 2021 from 82.196.111.219
random@pwnable:~$

Lets list the files using ls -la,

random@pwnable:~$ ls -la
total 40
drwxr-x---   5 root       random 4096 Oct 23  2016 .
drwxr-xr-x 115 root       root   4096 Dec 22 08:10 ..
d---------   2 root       root   4096 Jun 30  2014 .bash_history
-r--r-----   1 random_pwn root     49 Jun 30  2014 flag
dr-xr-xr-x   2 root       root   4096 Aug 20  2014 .irssi
drwxr-xr-x   2 root       root   4096 Oct 23  2016 .pwntools-cache
-r-sr-x---   1 random_pwn random 8538 Jun 30  2014 random
-rw-r--r--   1 root       root    301 Jun 30  2014 random.c

As usual there are some privilege restrictions to access the flag,we have to read it through our binary

Lets analyze the file type of our binary using file command,

random@pwnable:~$ file random
random: setuid ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/l, for GNU/Linux 2.6.24, BuildID[sha1]=f4eac0a1434a84aef72dfabfc1f889e6f6f73023, not stripped

So it is a not stripped binary

Lets view the source code of the binary,

random@pwnable:~$ cat random.c
#include <stdio.h>

int main(){
	unsigned int random;
	random = rand();	// random value!

	unsigned int key=0;
	scanf("%d", &key);

	if( (key ^ random) == 0xdeadbeef ){
		printf("Good!\n");
		system("/bin/cat flag");
		return 0;
	}

	printf("Wrong, maybe you should try 2^32 cases.\n");
	return 0;
}

Lets try running our binary,

random@pwnable:~$ ./random
12
Wrong, maybe you should try 2^32 cases.
random@pwnable:~$ ./random
1234
Wrong, maybe you should try 2^32 cases.
random@pwnable:~$ ./random
3735928559
Wrong, maybe you should try 2^32 cases.

So its expecting a different input

To bypass this we need to find that rand() is producing random output or fixed one

ra@moni~/P/p/random> cat test.c
#include <stdio.h>

int main(){
	unsigned int random;
	random = rand();
	printf("Random is %d",random);
	return 0;
}
ra@moni~/P/p/random> gcc -o test test.c
test.c: In function ‘main’:
test.c:5:11: warning: implicit declaration of function ‘rand’ [-Wimplicit-function-declaration]
    5 |  random = rand();
      |           ^~~~
ra@moni~/P/p/random> ./test
Random is 1804289383                                                                              
ra@moni~/P/p/random> ./test
Random is 1804289383                                                                              
ra@moni~/P/p/random> ./test
Random is 1804289383                                                                               
ra@moni~/P/p/random> ./test
Random is 1804289383                                                                               
ra@moni~/P/p/random> ./test
Random is 1804289383

NOTE

rand() is not a secure function to implement in programs

So this rand() produces same output because the seed in it is not initialized,if there is different seed each time it would create a different random number

Here rand() is called with default seed (1),thus it produces the same output for random number

To prevent this type of security failure by rand() for pseudo random number generation, use srand()

We know that the condition to pass the program is,

if( (key ^ random) == 0xdeadbeef )

Now random gives 1804289383,

key is our input,

If random performs xor operation with key it should give 0xdeadbeef as result,

On converting int to hex,

>>> hex(1804289383)
'0x6b8b4567'

So 0x6b8b4567 ^ key == 0xdeadbeef,

Its time for mathematical calculation (Converting into binary format),

0x6b8b4567 = 1101011100010110100010101100111

0xdeadbeef = 11011110101011011011111011101111

To find the key just XOR these,

I used Online XOR Calculator

So our key value should be ,

In hex —> 0xb526fb88

In int —> 3039230856

Now if we pass this int type key in the input,we get our flag

random@pwnable:~$ ./random
3039230856
Good!
Mommy, I thought libc random is unpredictable...

Done! We got our flag

Flag: Mommy, I thought libc random is unpredictable...

Gokul AP

NOTE