dont-you-love-banners
Understanding the challenge
We can use nc to connect to the two addresses. The first has leaked a password that we can use when we connect to the second address.
Connecting to the second address prompts us with questions three that we must answer before being granted access to the server.
- What is the password?
My_Passw@rd_@1234 - What is the top cyber security conference in the world? Some light googling allows us to divine the answer is “DEF CON”
- the first hacker ever was known for phreaking(making free phone calls), who was it?
Some further googling reveals the answer is
john draper
Once we are in the server we can look around.
If we use ls -la, we find we have two files: banner and text
Using cat, we see that the banner file contains the banner that we saw at the start of the challenge when initially connecting to the server. The text file simply has the text: “keep digging”.
We do exactly that. We can use ls / to view the contents of the / folder, and ls /root/ to see if we can read into the root folder.
We can see script.py and flag.txt. Attempting to cat /root/flag.txt yields a permission denied, so lets cat /root/script.py.
script.py creates the opening questions when we connect to the server, and importantly has this line:
with open("/home/player/banner", "r") as f:
print(f.read())
After looking up symlinks (thanks to hint 1), we understand they are shortcuts to other files.
With this in mind, we can remove banner from our current directory and recreate it but as a symlink.
Method
rm -rf banner
ln -s /root/flag.txt banner
Now, we can test if this worked. Let’s leave the server and nc again.
Lo and behold, we get the following:
$ nc tethys.picoctf.net 62167
picoCTF{b4nn3r_gr4bb1n9_su((3sfu11y_8126c9b0}
what is the password?
Flag: picoCTF{b4nn3r_gr4bb1n9_su((3sfu11y_8126c9b0}
Related Writeups
basic-mod1
We found this weird message being passed around on the servers, we think we have a working decryption scheme. Download t ...
basic-mod2
A new modular challenge! Download the message here. Take each number mod 41 and find the modular inverse for the result. ...
rsa-oracle
An attacker was able to intercept communications between a bank and a fintech company. They managed to get the message ( ...