OverTheWire
Over the wire see here, is a website that offers a large variety of wargames designed to teach differents aspect of security, ranging from basic Linux commands in Bandit to more advanced topics in other game.
I’ll propose here my solutions to the first levels of the Bandit challenges which cover the basic Linux commands.
Bandit 0
On this level you just have to log into the game using SSH. basically SSH (Secure Shell) is a cryptographic protocol that is used to surely connect to remote systems over an unsecured network. Each transfer is encrypted to protect against attacks.
Goal
Log into the game with ssh
the server is bandit.labs.overthewire.org
the port is 2220
the username is bandit0
the password is bandit0
Find the readme file located in the home directory and read it.
Theory && Solution
When you work with your favorite terminal, you can probably use ssh to connect to a remote server. If you need to learn more about any command, you can use man <command> to get the documentation.
to connect to a server, the simplest way is to use ssh <username>@<server> and then enter the password. Basically here, it should be ssh bandit0@bandit.labs.overthewire.org and the password is bandit0. But the task is also about connecting on the good port. The solution is then ssh bandit0@bandit.labs.overthewire.org -p 2220 where the flag -p is used to specify the port.
Another fun way to do this is to look up for the IP address of the server with either nslookup <server> or ping <server>. then instead of ssh <username>@<server>, you can use ssh <username>@<ip>.
Then using cat <file> to read the file you can read the password to enter the next level which gives us cat ./readme.
Bandit 1
Goal
In this level, the goal is to :
find the - file (located in the home directory) and read it.
Theory && Solution
the file is supposed to be in the home directory. Then, the first to do is to look at the files located in the home directory. for that, you can use ls. This command basically lists the directory contents. I’ll add a bit more flag to better read and understand the output.
first the flag -a is used to list all the files, including the hidden ones. The second flag -l is used to list the files in a long format. The third is -s, which prints the allocated size of each file in blocks – that’s the number in the leftmost column of the output.
The first command is therefore ls -las.
we obtain this output :
bandit1@bandit:~$ ls -las
total 24
4 -rw-r----- 1 bandit2 bandit1 33 Jun 24 14:58 -
4 drwxr-xr-x 2 root root 4096 Jun 24 14:58 .
4 drwxr-xr-x 150 root root 4096 Jun 24 15:02 ..
4 -rw-r--r-- 1 root root 220 Feb 13 2026 .bash_logout
4 -rw-r--r-- 1 root root 3851 Jun 24 14:50 .bashrc
4 -rw-r--r-- 1 root root 807 Feb 13 2026 .profile
Now we can understand that the file with the most probability of being the one that contains the password is the - file. To be sure we use the command file <file> to get the type of the file. And here this how it renders file ./-. Here I use ./ because the .indicates the current directory and the/-the file in the current directory. It also avoid considering the-as a flag of the commandfile`. Here is the output :
bandit1@bandit:~$ file ./-
./-: ASCII text
now we know that the file is indeed a text file. So now we can just read it with the cat command and go to the next level. the final command is : cat ./-
Bandit 2
Goal
The goal of this level is to :
Find the password in a file called --spaces in this filename-- located in the home directory.
Theory && Solution
Again I begin by using the ls command and here is the result :
bandit2@bandit:~$ ls -las
total 24
4 -rw-r----- 1 bandit3 bandit2 33 Jun 24 14:59 --spaces in this filename--
4 drwxr-xr-x 2 root root 4096 Jun 24 14:59 .
4 drwxr-xr-x 150 root root 4096 Jun 24 15:02 ..
4 -rw-r--r-- 1 root root 220 Feb 13 2026 .bash_logout
4 -rw-r--r-- 1 root root 3851 Jun 24 14:50 .bashrc
4 -rw-r--r-- 1 root root 807 Feb 13 2026 .profile
We can see that the flag is present in the home directory. To read it we just use the cat command. cat ./--spaces\ in\ this\ filename-- is a first way of doing it. Here the \ are used to quote the name of the file. It tells the terminal that the space belongs to the name of the file. If I had use cat ./--spaces in this filename-- the terminal would have consider each word as a file to read. I could also have used cat './--spaces in this filename--' but this time the ' are used to quote the entire name of the file. This option can be preferred because it is more readable.
Bandit 3
Goal
The goal of this level is to :
find the password located in an hidden file in the inhere directory.
Theory && Solution
First we go in the inhere directory and with the help of the ls command we list the files in the directory> This gives us the following output :
bandit3@bandit:~$ cd inhere/ && ls -las
total 12
4 drwxr-xr-x 2 root root 4096 Jun 24 14:59 .
4 drwxr-xr-x 3 root root 4096 Jun 24 14:59 ..
4 -rw-r----- 1 bandit4 bandit3 33 Jun 24 14:59 ...Hiding-From-You
Now we can see that the hidden file is called ...Hiding-From-You. Now we just have to read it using cat ./...Hiding-From-You
Here, it is important to notice the that the flag -a of the ls command is used to list all the files, including the hidden ones and that is the key to solve this level.
Bandit 4
Goal
The goal here is to :
Find the password stored in the only human-readable file in the inhere directory.
Theory && Solution
Here I begin by going in the inhere directory. Then I use the ls command to list the files in the directory. Here is the output :
bandit4@bandit:~$ cd inhere/ && ls -las
total 48
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file00
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file01
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file02
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file03
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file04
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file05
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file06
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file07
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file08
4 -rw-r----- 1 bandit5 bandit4 33 Jun 24 14:59 -file09
4 drwxr-xr-x 2 root root 4096 Jun 24 14:59 .
4 drwxr-xr-x 3 root root 4096 Jun 24 14:59 ..
Here I can see that the inhere directory has different files and that I cannot distinguish which one of them is the human readable file. For that I’ll use the file <file> command to get the information about each file in the current directory. Here is the command and the output :
bandit4@bandit:~/inhere$ file ./* | grep text
./-file07: ASCII text
./-file09: Motorola S-Record; binary data in text format
Let’s decrypt the command. The command file <file> gives the type of the file and. I use the ./* to apply the command to all the files in the current directory. It is precisely the * that tells the terminal to apply the command to all the files, it basically means “everything inside”. The | is called a pipe and is there to take the output of the command on the left and give it as input to the command on the right. The next command is grep <pattern> <file> where the pattern is the string that we want in the file. Here, with the pipe, we only use grep <pattern> because the pipe send the output of the previous command as the input of the grepcommand.
Now we can see that the only file that is human readable is the -file07 file. The final command is therefore cat ./-file07.
Bandit 5
Goal
The goal here is to : find the password stored in a file somewhere under the inhere directory and that has all of the following properties :
human-readable, 1033 bytes in size, not executable
Theory && Solution
Again we begin by going in the inhere directory. Then I use the ls command to list the files in the directory. Here is the output :
bandit5@bandit:~$ cd inhere/ && ls
maybehere00 maybehere02 maybehere04 maybehere06 maybehere08 maybehere10 maybehere12 maybehere14 maybehere16 maybehere18
maybehere01 maybehere03 maybehere05 maybehere07 maybehere09 maybehere11 maybehere13 maybehere15 maybehere17 maybehere19
Let’s check the maybehere00 directory :
bandit5@bandit:~/inhere$ cd maybehere00 && ls -las
total 72
4 -rwxr-x--- 1 root bandit5 1039 Jun 24 14:59 -file1
12 -rw-r----- 1 root bandit5 9388 Jun 24 14:59 -file2
8 -rwxr-x--- 1 root bandit5 7378 Jun 24 14:59 -file3
4 drwxr-x--- 2 root bandit5 4096 Jun 24 14:59 .
4 drwxr-x--- 22 root bandit5 4096 Jun 24 14:59 ..
4 -rwxr-x--- 1 root bandit5 551 Jun 24 14:59 .file1
8 -rw-r----- 1 root bandit5 7836 Jun 24 14:59 .file2
8 -rwxr-x--- 1 root bandit5 4802 Jun 24 14:59 .file3
8 -rwxr-x--- 1 root bandit5 6118 Jun 24 14:59 spaces file1
8 -rw-r----- 1 root bandit5 6850 Jun 24 14:59 spaces file2
4 -rwxr-x--- 1 root bandit5 1915 Jun 24 14:59 spaces file3
By doing this again, I understand that each directory contains many files, and the one that I want is located somewhere. To find it let’s use the find <starting point> <filters> command that scans through the starting directory and all its subfolders to locate a file based on specific criteria.
Let’s use the following command :
bandit5@bandit:~/inhere$ find ./ -type f -size 1033c ! -executable
./maybehere07/.file2
Bingo ! We managed to find only one compatible file. Let’s explain the command. The ./ is used to indicate the starting directory, the -type f means that we are looking for a file, the -size 1033c means that the must has a size of 1033 bytes, and the ! -executable means that the file is not executable. We use the ! to indicate that we want to exclude the executable files. Normally, I would have used :
bandit5@bandit:~/inhere$ find -type f -size 1033c ! -executable -exec file {} +
./maybehere07/.file2: ASCII text, with very long lines (1000)
Here, we used the -exec <command> {} + option to execute a command on each file sent back with the command find. In this case, the file command is used to determine the type of the file. We can see that the file returned is human-readable. If there were many files returned we could have used | grep to filter only the files that are human-readable.
The final command is to read the file : cat ./maybehere07/.file2.
Bandit 6
Goal
The goal here is to :
Find the password stored somewhere on the server and has all of the following properties:
owned by user bandit7, owned by group bandit6, 33 bytes in size
Theory && Solution
For this level, no need to go in the inhere directory, we have to find the password somewhere on the whole server. With this, goal in mind let’s directly use the find command :
bandit6@bandit:~$ find / -type f -size 33c -user bandit7 -group bandit6
find: ‘/snap’: Permission denied
find: ‘/lost+found’: Permission denied
...............
find: ‘/sys/fs/bpf’: Permission denied
find: ‘/tmp’: Permission denied
As we can see, we mostly obtain error messages telling us that we cannot access the directory. So, we’ll just shut them down to see if we can at least extract something from the find command. Here, the <starting point> is basically the root of the system /, and I add two new filters : The -user filter for the user and the -group filter for the group.
bandit6@bandit:~$ find / -type f -size 33c -user bandit7 -group bandit6 2> /dev/null
/var/lib/dpkg/info/bandit7.password
Bingo ! Without any error message, there is only one file compatible with all the options. But what did I do ? Well a StackOverflow answer tells us the answer, and I’ll just copy paste the answer here :
As the other answers state, you can use command 2> /dev/null to throw away the error output from command
But what is going on here?
> is the operator used to redirect output. 2 is a reference to the standard error output stream, i.e. 2> = redirect error output.
/dev/null is the 'null device' which just swallows any input provided to it. You can combine the two to effectively throw away output from a command.
Full reference:
- > /dev/null throw away stdout
- 1> /dev/null throw away stdout
- 2> /dev/null throw away stderr
- &> /dev/null throw away both stdout and stderr
Perfect, but I would normally use the following command :
bandit6@bandit:~$ find / -type f -size 33c -user bandit7 -group bandit6 2> /dev/null -exec file {} + | grep text
/var/lib/dpkg/info/bandit7.password: ASCII text
This command applies for every file that matches the filters, (and without error message), the command file. The output of this file command is then piped to the grep command to filter only the files that are human-readable.
The final command is therefore cat /var/lib/dpkg/info/bandit7.password.
Bandit 7
Goal
The goal here is to find the password stored in the file data.txt next to the word millionth.
Theory && Solution
Here, I’ll begin by using the ls command to confirm that the file is in the current directory.
bandit7@bandit:~$ ls -las
total 4108
4 drwxr-xr-x 2 root root 4096 Jun 24 14:59 .
4 drwxr-xr-x 150 root root 4096 Jun 24 15:02 ..
4 -rw-r--r-- 1 root root 220 Feb 13 2026 .bash_logout
4 -rw-r--r-- 1 root root 3851 Jun 24 14:50 .bashrc
4 -rw-r--r-- 1 root root 807 Feb 13 2026 .profile
4088 -rw-r----- 1 bandit8 bandit7 4184396 Jun 24 14:59 data.txt
And there it is ! If we try to read the file data.txt we can see that the file is too long. The first I did it I had to stop the command with Ctrl+C… Instead of that I’ll use again the pipe and the grep command. Here is the command and the output :
bandit7@bandit:~$ cat data.txt | grep millionth
millionth password here
With this command we give all the text of the file to the grep command and we ask it to take the line with the word millionth and give it to us. This way we can read the password. We could also have more simply used grep millionth data.txt. I just prefer the first one.