random.md

random

The home folder of this challenge contains the following:

random@ubuntu:~$ ls -l
total 20
-r--r----- 1 random_pwn root     49 Jun 30  2014 flag
-r-sr-x--- 1 random_pwn random 8538 Jun 30  2014 random
-rw-r--r-- 1 root       root    301 Jun 30  2014 random.c

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;
}

In this challenge we need to make sure that key ^ random is equal to 0xdeadbeef to obtain the flag. It is known that the rand() method is insecure for random number generation. The numbers that this method generates are predictable and are just a numerical state of a predetermined formula. When the seed is not given for the rand method, the seed will always be starting on 1. Therefore rand will always give the same answer when the method in only called once in the program.

Not every OS/system has the same implementation of the rand function. Therefore we need to determine which random value will be generated on the server.

/tmp/test/rand.c

#include <stdio.h>

int main(){
        unsigned int random;
        random = rand();

        printf("%d\n", random);
        return 0;
}

random@ubuntu:/tmp/test$ gcc rand.c -o rand
random@ubuntu:/tmp/test$ ./rand
1804289383

Now we have determined the "random" value, we can calculate the key we need to input to obtain the flag. key ^ random = 0xdeadbeef => random ^ 0xdeadbeef = key.

Python 2.7.10
[GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.34)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> 1804289383 ^ 0xdeadbeef
3039230856

We found the key! Now we can use it to obtain the flag.

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