The flag06 account credentials came from a legacy unix system.
To do this level, log in as the level06 account with the password level06 . Files for this level can be found in /home/flag06.

I had a good idea what I’m looking for here, an easy to crack password hash in /etc/passwd rather than in the shadow file, so:

cat /etc/passwd | grep flag06

shows me the hash is ueqwOCnSGdsuM. I need to “crack” the hash. Time to get john the ripper on the case. At this point I didn’t have any other linux machines to hand, so I went to another tty session on this one and logged in a nebula and installed john (sudo apt-get install john). Then I ran john on the password file (john /etc/passwd) and he showed me the password. I switched over to flag06 account and the password worked as expected.

Check the flag05 home directory. You are looking for weak directory permissions
To do this level, log in as the level05 account with the password level05 . Files for this level can be found in /home/flag05.

The command ll (an alias for ls-alF) showed me that I had read access to ~flag05/.backup, and in there was a backup gzipped tar. Hopefully some goodies in here…

I unpacked the tar and found that it contained a folder called .ssh. This is used for secure shell authentication. That folder contained a private/public key pair and an authorized_keys file. The authorised_keys file is exactly the same as the public key file, so (assuming that the authorised_keys file had not been deleted since the backup) I should be able to ssh in using the private key, as long as it was not encrypted with a passphrase.

I copied the id_rsa file to ~/.ssh and tried to connect using:

ssh flag05@localhost

Bingo!

This level requires you to read the token file, but the code restricts the files that can be read. Find a way to bypass it
To do this level, log in as the level04 account with the password level04 . Files for this level can be found in /home/flag04.

It also contains some source code:

#include 
#include 
#include 
#include 
#include 
#include 

int main(int argc, char **argv, char **envp)
{
    char buf[1024];
    int fd, rc;

    if(argc == 1) {
        printf("%s [file to read]\n", argv[0]);
        exit(EXIT_FAILURE);
    }

    if(strstr(argv[1], "token") != NULL) {
        printf("You may not access '%s'\n", argv[1]);
        exit(EXIT_FAILURE);
    }

    fd = open(argv[1], O_RDONLY);
    if(fd == -1) {
        err(EXIT_FAILURE, "Unable to open %s", argv[1]);
    }

    rc = read(fd, buf, sizeof(buf));

    if(rc == -1) {
        err(EXIT_FAILURE, "Unable to read fd %d", fd);
    }

    write(1, buf, rc);
}

Its fairly clear from looking at the files and the source code (I will admit I had to use a lot of man to help me understand the source code) that I want to read the contents of “token”, but the program won’t allow it. I tried things like ./token and ../flag04/token, but that didn’t work because the program is just searching for the string “token” anywhere in the first argument. Well… how do I get the contents of that file “into” another file without having permission to read the file? Symbolic link! Here’s what I did:

ln -s ~flag04/token /tmp/link
~flag04/flag04 /tmp/link

It turns out that the contents of the token file is the password for the flag04 account so I just did su flag04 and used that password. I ran getflag and violà!

Check the home directory of flag03 and take note of the files there.
There is a crontab that is called every couple of minutes.
To do this level, log in as the level03 account with the password level03 . Files for this level can be found in /home/flag03.

Well looking in ~flag03 there is just one directory (writable.d) file and one file (writable.sh). I’m assuming that the cron job runs writable.sh every couple of minutes so I looked at that script. I can see that the script runs every file in the writable.d folder (which we have write access to), but will kill the process if it takes longer than 5 seconds. It then removes the file.

What we could do is make a quick bash script that will run getflag and save the output like this:

#!/bin/sh
getflag > /tmp/getflag.out

Which works (after we wait for the cron job to run it), but I want shell! So we’re going to borrow a trick from level01 and create a program that will launch a bash shell and get flag03 to set the setuid bit.

My C program looks like this:

#define _GNU_SOURCE
#include 
#include 
#include 

int main() {
    gid_t gid;
    uid_t uid;
    gid = getegid();
    uid = geteuid();

    setresgid(gid, id, gid);
    setresuid(uid, uid, uid);

    system("/bin/bash");
}

Now I just compile it with gcc and drop it in /tmp so that flag03 can access it. All I need the cron job to do now is make a copy and set the setuid bit, so here is the script I dropped in ~flag03/writable.d:

#!/bin/sh
cp /tmp/setuidshell /tmp/setuidshell2
chmod u+s /tmp/setuidshell2

This got me a program (/tmp/setuidshell2) in that gave me shell. From here I was able to run getflag, and also to run crontab -l to see that the cron job is actually called every 3 minutes.

There is a vulnerability in the below program that allows arbitrary programs to be executed, can you find it?
To do this level, log in as the level02 account with the password level02 . Files for this level can be found in /home/flag02.

It also contains some source code:

#include 
#include 
#include 
#include 
#include 

int main(int argc, char **argv, char **envp)
{
    char *buffer;

    gid_t gid;
    uid_t uid;

    gid = getegid();
    uid = geteuid();

    setresgid(gid, gid, gid);
    setresuid(uid, uid, uid);

    buffer = NULL;

    asprintf(&buffer, "/bin/echo %s is cool", getenv("USER"));
    printf("about to call system(\"%s\")\n", buffer);

    system(buffer);
}

This is similar to the Level 01. An environment variable $USER is being used to construct a string that is printed to the screen before being run. If we can edit that environment variable, we can inject a malicious command.

Initially I changed $USER so that running the program would execute getflag. The command I used was:

USER=;getflag;echo

I’ll break this down:
; – end the command and start a new one
getflag – run the getflag program
; – end the command and start a new one
echo – start a new echo command so that the following arguments don’t cause an error

This results in the following command being run:

/bin/echo ;getflag;echo is cool

I got a success message from get flag, but I wanted shell, so I changed my command to:

USER="Opening escalated shell...;bin/bash;echo Closing pwned shell, now that"

This time I got shell, and some cool text when going into the shell and when coming out (after typing exit)

There is a vulnerability in the below program that allows arbitrary programs to be executed, can you find it?
To do this level, log in as the level01 account with the password level01 . Files for this level can be found in /home/flag01.

It also contains some source code:

#include 
#include 
#include 
#include 
#include 

int main(int argc, char **argv, char **envp)
{
    gid_t gid;
    uid_t uid;
    gid = getegid();
    uid = geteuid();

    setresgid(gid, gid, gid);
    setresuid(uid, uid, uid);

    system("/usr/bin/env echo and now what?");
}

I’m not all that familiar with C (I’m more of a scripter), but I can understand enough; this appears to basically sets all uids for the process to the effective uid (presumably the setuid bit is present) and then calls a command line of:

/usr/bin/env echo and now what?

I wasn’t familiar with the env command so used a bit of googling until I learned that env is used to launch programs in a different environment. It also also sometimes used because a script needs to start with a shebang and followed by an interpreter directive, which must be an absolute path. Because some interpreters are not always installed at the same location, env is sometimes used to launch the correct interpreter by file name rather than full path (e.g. #!/usr/bin/env/ python). To do this, env searches through the list of paths in in the environment variable $PATH in order until it finds a correctly named file that it can execute in one of them. Presumably (for some unknown reason) env is being used here to invoke echo, but it means we can make a different echo program run by creating a malicious script and changing $PATH to point to it first.

I changed the path to include /tmp at the beginning by running the follwing command:

PATH=/tmp:$PATH

and then created a new symbolic link called echo to the getflag program:

ln -s /bin/getflag /tmp/echo

Now when I ran the vulnerable program I got a success message, but I wanted to go one further. I wanted shell…

I tried creating a symbolic link to bash, but now running flag01 failed due to the invalid arguments (“and now what?” are valid arguments for echo, but not bash), so I removed the symbolic link and created an executable shell script that ignored all arguments, and saved it as echo in /tmp. It contained the following two lines of code:

#!/bin/bash
/bin/bash

This, I hoped, would cause the vulnerable program to spawn a shell. I tested it and it worked. I then ran whoami to confirm that I was flag01 and then getflag to get a success message.

This post is about Nebula Level 00. The information about this level says:

This level requires you to find a Set User ID program that will run as the “flag00” account. You could also find this by carefully looking in top level directories in / for suspicious looking directories.
Alternatively, look at the find man page.
To access this level, log in as level00 with the password of level00 .

This is a pretty simple challenge, but did mean I had to learn all about normal unix filesystem permissions and the more advanced setuid/setguid/stickybit permissions I also learned how to suppress errors from the find command and how to better use the find and man command.

The command I used was

find / -perm -u=s 2>/dev/null

I’ll break down what this does:

• find – search for files in a directory hierarchy
• / – start at the root of the filesystem
• -perm -u=s – find files that have the setuid bit set in their permissions
• 2>/dev/null – discard all errors (mostly about not having permission to scan directories)

One of the results was /bin/…/flag00. This (…\) is a suspicious looking directory! Running ll /bin/…/flag00 showed me that the owner was flag00 and the setuid bit was indeed set so I ran the file which told me to now run getflag then changed the user to flag00. Running getflag gave me a success message.

What I liked about this was that I had a shell running as the flag00 user so I could run other commands like whoami before typing exit to get out of the shell. At the time, I had no idea how I was put into a new shell, but it all becomes clearer in the next level…