Bandit

This is a collection of challenges built around common Linux utilities.

Level 0

To start off we follow the instructions found here https://overthewire.org/wargames/bandit/bandit0.html and ssh to bandit.labs.overthewire.org on port 2220 as the user bandit0. To do this I will be using the Debian Windows Subsystem for Linux (mostly because putty can be annoying to work with). Just as a note, I will not be explaining these solutions, simply showing what commands can be used to find the flag. That being said, I am also using this an an exercise to test my bash-fu a little.

Level 1

The password for the next level is stored in a file called readme located in the home directory. Use this password to log into bandit1 using SSH. Whenever you find a password for a level, use SSH (on port 2220) to log into that level and continue the game.

Solution

cat readme

Flag

boJ9jbbUNNfktd78OOpsqOltutMc3MY1

Level 2

The password for the next level is stored in a file called - located in the home directory

Solution

cat /home/bandit1/-

Flag

CV1DtqXWVFXTvM2F0k09SHz0YwRINYA9

Level 3

The password for the next level is stored in a file called spaces in this filename located in the home directory

Solution

cat ./spaces\ in\ this\ filename

Flag

UmHadQclWmgdLOKQ3YNgjWxGoRMb5luK

Level 4

The password for the next level is stored in a hidden file in the inhere directory.

Solution

cat inhere/.hidden

Flag

pIwrPrtPN36QITSp3EQaw936yaFoFgAB

Level 5

The password for the next level is stored in the only human-readable file in the inhere directory. Tip: if your terminal is messed up, try the “reset” command.

Solution

file inhere/* | grep ASCII | cat $(awk 'BEGIN { FS = ":"} ; { print $1}')

Flag

koReBOKuIDDepwhWk7jZC0RTdopnAYKh

Level 6

The password for the next level is stored in a file somewhere under the inhere directory and has all of the following properties:

• human-readable

• 1033 bytes in size

• not executable

Solution

cat $(find inhere -size 1033c)

Flag

DXjZPULLxYr17uwoI01bNLQbtFemEgo7

Level 7

The password for the next level is stored somewhere on the server and has all of the following properties:

• owned by user bandit7

• owned by group bandit6

• 33 bytes in size

Solution

cat $(find / -size 33c -user bandit7 -group bandit6 2> /dev/null)

Flag

HKBPTKQnIay4Fw76bEy8PVxKEDQRKTzs

Level 8

The password for the next level is stored in the file data.txt next to the word millionth

Solution

grep millionth data.txt | awk ' {print $2} '

Flag

cvX2JJa4CFALtqS87jk27qwqGhBM9plV

Level 9

The password for the next level is stored in the file data.txt and is the only line of text that occurs only once

Solution

cat data.txt | sort | uniq -u

Flag

UsvVyFSfZZWbi6wgC7dAFyFuR6jQQUhR

Level 10

The password for the next level is stored in the file data.txt in one of the few human-readable strings, beginning with several ‘=’ characters.

Note that for this challenge there are multiple lines that match the regular expression ^=+ so we must use the knowledge that the flag is 32 characters long.

Solution

strings data.txt | grep -E '^=+[[:space:]]{1}[[:alnum:]]{32}' | awk '{print $2}'

Flag

truKLdjsbJ5g7yyJ2X2R0o3a5HQJFuLk

Level 11

The password for the next level is stored in the file data.txt, which contains base64 encoded data

Solution

base64 -d data.txt | awk '{print $4}'

Flag

IFukwKGsFW8MOq3IRFqrxE1hxTNEbUPR

Level 12

The password for the next level is stored in the file data.txt, where all lowercase (a-z) and uppercase (A-Z) letters have been rotated by 13 positions

Solution

cat data.txt | tr A-Za-z N-ZA-Mn-za-m | awk '{print $4}'

Flag

5Te8Y4drgCRfCx8ugdwuEX8KFC6k2EUu

Level 13

The password for the next level is stored in the file data.txt, which is a hexdump of a file that has been repeatedly compressed. For this level it may be useful to create a directory under /tmp in which you can work using mkdir. For example: mkdir /tmp/myname123. Then copy the datafile using cp, and rename it using mv (read the manpages!)

Solution

xxd -r ~/data.txt - | gzip -d | bzip2 -d | gzip -d | tar xOf - | tar xOf - | bzip2 -d | tar xOf - | gunzip -d | awk '{print $4}'

Flag

8ZjyCRiBWFYkneahHwxCv3wb2a1ORpYL

Level 14

The password for the next level is stored in /etc/bandit_pass/bandit14 and can only be read by user bandit14. For this level, you don’t get the next password, but you get a private SSH key that can be used to log into the next level. Note: localhost is a hostname that refers to the machine you are working on

Solution

ssh -i sshkey.private bandit14@localhost cat /etc/bandit_pass/bandit14

Flag

4wcYUJFw0k0XLShlDzztnTBHiqxU3b3e

Level 15

The password for the next level can be retrieved by submitting the password of the current level to port 30000 on localhost.

Solution

nc -nv 127.0.0.1 30000 < /etc/bandit_pass/bandit14

Flag

BfMYroe26WYalil77FoDi9qh59eK5xNr

Level 16

The password for the next level can be retrieved by submitting the password of the current level to port 30001 on localhost using SSL encryption.

Helpful note: Getting “HEARTBEATING” and “Read R BLOCK”? Use -ign_eof and read the “CONNECTED COMMANDS” section in the manpage. Next to ‘R’ and ‘Q’, the ‘B’ command also works in this version of that command…

Solution

openssl s_client -ign_eof -connect 127.0.0.1:30001 < /etc/bandit_pass/bandit15

Flag

cluFn7wTiGryunymYOu4RcffSxQluehd

Level 17

The credentials for the next level can be retrieved by submitting the password of the current level to a port on localhost in the range 31000 to 32000. First find out which of these ports have a server listening on them. Then find out which of those speak SSL and which don’t. There is only 1 server that will give the next credentials, the others will simply send back to you whatever you send to it.

Okay, I have to admit that my solution here is ugly, but was able to get it into a form that you could copy and paste. The basic flow of this solution goes:

1. Do an nmap scan for open ports between 31000-32000 on 127.0.0.1, enumerate the service versions on each open port, then output in grepable formate to stdout

2. ignore all lines starting with # or containing Status

3. grab the portion of the output with the open ports results

4. Cut the remaining strings off at the work Ignored

5. change all commas to new lines (this will make it so each port's results are on it's own line)

6. trim all spaces and tabs

7. ignore all lines that contain the word echo (We want to avoid the echo servers)

8. grab the first field, which is the port number of the valid port

9. Put this value into an environment variable named PORT

10. Use $PORT to connect to the correct port and pipe in the correct password from /etc/bandit_pass/bandit16 to get the flag

While technically two commands, I'd say it is pretty compact. This page was super helpful in working with the nmap output: https://github.com/leonjza/awesome-nmap-grep#print-the-top-10-ports

Solution

PORT=$(nmap -sV -p 31000-32000 -oG - 127.0.0.1 | egrep -v "^#|Status" | cut -d ' ' -f4- | sed -n -e 's/Ignored.*//p' | tr ',' '\n' | sed -e 's/^[ \t]*//' | egrep -v "unknown" | cut -d '/' -f1); openssl s_client -ign_eof -connect 127.0.0.1:$PORT < /etc/bandit_pass/bandit16

Flag

-----BEGIN RSA PRIVATE KEY-----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-----END RSA PRIVATE KEY-----

Level 18

There are 2 files in the homedirectory: passwords.old and passwords.new. The password for the next level is in passwords.new and is the only line that has been changed between passwords.old and passwords.new

NOTE: if you have solved this level and see ‘Byebye!’ when trying to log into bandit18, this is related to the next level, bandit19

Solution

diff passwords.old passwords.new | grep '>' | cut -d ' ' -f2

Flag

kfBf3eYk5BPBRzwjqutbbfE887SVc5Yd

Level 19

The password for the next level is stored in a file readme in the homedirectory. Unfortunately, someone has modified .bashrc to log you out when you log in with SSH.

Because the .bashrc for this user automatically kicks you, we simply run the command via ssh instead of logging in.

Solution

ssh [e-mail] -p 2220 cat /home/bandit18/readme

Flag

IueksS7Ubh8G3DCwVzrTd8rAVOwq3M5x

Level 20

To gain access to the next level, you should use the setuid binary in the homedirectory. Execute it without arguments to find out how to use it. The password for this level can be found in the usual place (/etc/bandit_pass), after you have used the setuid binary.

Solution

./bandit20-do /bin/cat /etc/bandit_pass/bandit20

Flag

GbKksEFF4yrVs6il55v6gwY5aVje5f0j

Level 21

There is a setuid binary in the homedirectory that does the following: it makes a connection to localhost on the port you specify as a commandline argument. It then reads a line of text from the connection and compares it to the password in the previous level (bandit20). If the password is correct, it will transmit the password for the next level (bandit21)

Solution

# In tmux pane 1:
nc -nvlp 31337 < /etc/bandit_pass/bandit20

# In tmux pane 2:
./suconnect 31337

Flag

gE269g2h3mw3pwgrj0Ha9Uoqen1c9DGr

Level 22

A program is running automatically at regular intervals from cron, the time-based job scheduler. Look in /etc/cron.d/ for the configuration and see what command is being executed.

The cronjob in this challenge is simply writing to the following file:

Solution

cat /tmp/t7O6lds9S0RqQh9aMcz6ShpAoZKF7fgv

Flag

Yk7owGAcWjwMVRwrTesJEwB7WVOiILLI

Level 23

A program is running automatically at regular intervals from cron, the time-based job scheduler. Look in /etc/cron.d/ for the configuration and see what command is being executed.

NOTE: Looking at shell scripts written by other people is a very useful skill. The script for this level is intentionally made easy to read. If you are having problems understanding what it does, try executing it to see the debug information it prints.

Solution

cat /tmp/$(echo I am user bandit23 | md5sum | cut -d ' ' -f1)

Flag

jc1udXuA1tiHqjIsL8yaapX5XIAI6i0n

Level 24

A program is running automatically at regular intervals from cron, the time-based job scheduler. Look in /etc/cron.d/ for the configuration and see what command is being executed.

NOTE: This level requires you to create your own first shell-script. This is a very big step and you should be proud of yourself when you beat this level!

NOTE 2: Keep in mind that your shell script is removed once executed, so you may want to keep a copy around…

Solution

mkdir /tmp/bandit24_results
cd /tmp/bandit24_results
touch password.txt
chmod 666 password.txt
echo <<EOF > ./get_password.sh
#!/bin/bash
cat /etc/bandit_pass/bandit24 > /tmp/bandit24_results/password.txt
EOF
chmod 777 ./get_password.sh
cp ./get_password.sh /var/spool/bandit24
watch cat password.txt

Flag

UoMYTrfrBFHyQXmg6gzctqAwOmw1IohZ

Level 25

A daemon is listening on port 30002 and will give you the password for bandit25 if given the password for bandit24 and a secret numeric 4-digit pincode. There is no way to retrieve the pincode except by going through all of the 10000 combinations, called brute-forcing.

This could certainly be sped up quite a bit with multithreading, but I didn't feel it was necessary.

Solution

python ./solution.py

solution.py

import socket

s=socket.socket(socket.AF_INET,socket.SOCK_STREAM)
bandit24 = open('/etc/bandit_pass/bandit24','r').read().replace('\n','')
print "initiating brute force..."
connect = s.connect(('127.0.0.1',30002))
s.recv(1024)
for x in range(0,10000):
    print('attempting:{} {:04d}'.format(bandit24,x))
    s.send('{} {:04d}\n'.format(bandit24 , x))
    message = s.recv(1024)
    if 'Wrong!' not in message:
        print(message)
        break
s.close()

Flag

uNG9O58gUE7snukf3bvZ0rxhtnjzSGzG

Level 26

Logging in to bandit26 from bandit25 should be fairly easy… The shell for user bandit26 is not /bin/bash, but something else. Find out what it is, how it works and how to break out of it.

Looking at the /etc/passwd entry for bandit26 we see that it is calling a shell script instead of an actual shell. This script simple calls more on a file. If we change the size of our terminal so that the file cannot fit, then we can abuse the v command in more. This will open an editor for the file, in this case the default editor is vim, and we can use that editor to view the password file with the following vim command:

Solution

:e /etc/bandit_pass/bandit26

Flag

5czgV9L3Xx8JPOyRbXh6lQbmIOWvPT6Z

Level 27

Good job getting a shell! Now hurry and grab the password for bandit27!

The solution to this challenge is nearly identical to the last. Only this time we spawn a shell instead of editing a file.

Solution

# getting a shell from vim
:set shell=/bin/bash
:shell

# using the suid binary
./bandit27-do /etc/bandit_pass/bandit27

Flag

3ba3118a22e93127a4ed485be72ef5ea

Level 28

There is a git repository at ssh://bandit27-git@localhost/home/bandit27-git/repo. The password for the user bandit27-git is the same as for the user bandit27.

Clone the repository and find the password for the next level.

Solution

git clone ssh://bandit27-git@localhost/home/bandit27-git/repo
cat repo/README

Flag

0ef186ac70e04ea33b4c1853d2526fa2

Level 29

There is a git repository at ssh://bandit28-git@localhost/home/bandit28-git/repo. The password for the user bandit28-git is the same as for the user bandit28.

Clone the repository and find the password for the next level.

For this challenge the password is hidden in a previous commit, which we can see with git log

Solution

git clone ssh://bandit28-git@localhost/home/bandit28-git/repo
cd repo
git log
git checkout 186a1038cc54d1358d42d468cdc8e3cc28a93fcb
cat README.md

Flag

bbc96594b4e001778eee9975372716b2

Level 29

There is a git repository at ssh://bandit29-git@localhost/home/bandit29-git/repo. The password for the user bandit29-git is the same as for the user bandit29.

Clone the repository and find the password for the next level.

For this challenge the password is hidden in a previous commit, which we can see with git log

Solution

git clone ssh://bandit29-git@localhost/home/bandit29-git/repo
cd repo
git checkout dev
git log -p

Flag

5b90576bedb2cc04c86a9e924ce42faf

Level 30

There is a git repository at ssh://bandit30-git@localhost/home/bandit30-git/repo. The password for the user bandit30-git is the same as for the user bandit30.

Clone the repository and find the password for the next level.

Solution

git clone ssh://bandit30-git@localhost/home/bandit30-git/repo
cd repo
git tag
git show secret

Flag

47e603bb428404d265f59c42920d81e5

Level 31

There is a git repository at ssh://bandit31-git@localhost/home/bandit31-git/repo. The password for the user bandit31-git is the same as for the user bandit31.

Clone the repository and find the password for the next level.

For this challenge we simply need to push a file with specific contents to the repo, but it is originally ignored due to the .gitignore. We get past this with the -f flag.

Solution

git clone ssh://bandit31-git@localhost/home/bandit31-git/repo
echo 'May I come in?' > key.txt
git add -f key.txt
git push origin master

Flag

56a9bf19c63d650ce78e6ec0354ee45e

Level 32

After all this git stuff its time for another escape. Good luck!

It looks like we are stuck in yet another jail shell. This one seems to translate every character to uppercase, but by using characters that can't be translated to uppercase we can escape the shell.

Solution

# this command doesn't work
ls

# But after executing this we have a real shell
$0

cat /etc/bandit_pass/bandit32

Flag

c9c3199ddf4121b10cf581a98d51caee