CTFtime.org / DownUnderCTF 2021 (Online) / Break Me! / Writeup

# **DownUnderCTF 2021**

***

# Table of Contents
* [Miscellaneous](#miscellaneous)
* [Discord](#discord)
* [The Introduction](#the-introduction)
* [Twitter](#twitter)
* [General Skills Quiz](#general-skills-quiz)
* [rabbit](#rabbit)
* [Floormat](#floormat)
* [Cryptography](#cryptography)
* [Substitution Cipher I](#substitution-cipher-i)
* [Substitution Cipher II](#substitution-cipher-ii)
* [Break Me!](#break-me)
* [treasure](#treasure)
* [Reversing](#reversing)
* [no strings](#no-strings)
***

# Miscellaneous

## Discord
How about you visit our help page? You know what they say when you need 'help' discord is there for support :)

**Author:** Crem

**Solution**

The flag is in the `#request-support` channel on the [CTF discord channel](https://discord.gg/vXtuGXy)

**Flag**

**`DUCTF{if_you_are_having_challenge_issues_come_here_pls}`**

## The Introduction
Are you ready to start your journey?

**Author:** Crem

`nc pwn-2021.duc.tf 31906`

**Solution**

With the challenge we're given a command to run, this command (`nc`) connects to a server identified with the domain pwn-2021.duc.tf and interacts with a service running on port `31906`, running it returns the following question from the server:

```
wh47 15 y0ur n4m3 h4ck3r?
```

Entering a name will starts printing the hacker manifesto, this is an essay written in the 80s by "The Mentor" after his arrest and is considered a major piece of the hacker culture, the full essay is linked below and worth a read if you consider yourself part of the hacking community.

After printing the essay, the following line is written:
```
50 4r3 y0u 4c7u4lly 4 h4ck3r?
```
Answering yes will return you the flag (other answers might also work as I didn't check):

```
W3rni0, y0u w1ll n33d 7h15 0n y0ur duc7f j0urn3y 7h3n

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

DUCTF{w3lc0m3_70_7h3_duc7f_7hund3rd0m3_h4ck3r}

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

__
/ \--..____
\ \ \-----,,,..
\ \ \ \--,,..
\ \ \ \ ,'
\ \ \ \ ``..
\ \ \ \-''
\ \ \__,,--'''
\ \ \.
\ \ ,/
\ \__..-
\ \
\ \
\ \
\ \
\ \
\ \
\ \
\ \
\ \
```

**Flag**

**`DUCTF{w3lc0m3_70_7h3_duc7f_7hund3rd0m3_h4ck3r}`**

**Refereces**

* [The Hacker Manifesto](http://phrack.org/issues/7/3.html)

## Twitter
If you have been paying attention to our Twitter, we have been using flags in the background of our graphics this year. However, one of these flags stands out a bit more than the rest! Can you find it?

While you're there, why don't you give us a follow to keep up to date with DownUnderCTF!

**Author:** Crem

**Solution**

This one is pretty guessy but the solution is written in the description so I guess I can't complain, this year the CTF graphics used flags from the previous competition, as can be seen here:

But one of the graphics posted two weeks before the competition included the flag in a brighter color, try finding it in the image:

(hint - look down)

**Flag**

**`DUCTF{EYES_ON_THE_PRIZES}`**

## General Skills Quiz
QUIZ TIME! Just answer the questions. Pretty easy right?

**Author:** Crem

`nc pwn-2021.duc.tf 31905`

**Solution**

This challenge requires you to decode and encode using different popular formats such as base64, binary and ROT13. Not enough time is given to do those manually, so I ended up writing the following python script that interacts with the server and performs the conversions (the tasks required are commented):
```python
from pwn import *
from urllib.parse import unquote
from base64 import b64decode, b64encode
from codecs import encode,decode

s = remote('pwn-2021.duc.tf', 31905)

# Ready to start
s.sendline()
# 1+1=?
s.recvuntil(': ')
s.sendline('2')
# Decode an hex string to decimal
s.recvuntil(': ')
s.sendline(str(int(s.recvline(),16)))
# Decode an hex string to ASCII letter
s.recvuntil(': ')
s.sendline(str(chr(int(s.recvline(),16))))
# Decode a URL encoded string
s.recvuntil(': ')
s.sendline(unquote(s.recvline(keepends=False).decode()))
# Base64 decode
s.recvuntil(': ')
s.sendline(b64decode(s.recvline()))
# Base64 encode
s.recvuntil(': ')
s.sendline(b64encode(s.recvline(keepends=False)))
# ROT13 decode
s.recvuntil(': ')
s.sendline(decode(s.recvline(keepends=False).decode(), 'rot_13'))
# ROT13 encode
s.recvuntil(': ')
s.sendline(encode(s.recvline(keepends=False).decode(), 'rot_13'))
# Binary decode
s.recvuntil(': ')
s.sendline(str(int(s.recvline(),2)))
# Binary encode
s.recvuntil(': ')
s.sendline(bin(int(s.recvline())))
# Best CTF competition
s.recvuntil('?')
# s.sendline('picoCTF')
s.sendline('DUCTF')
s.interactive()
```
I'll add a description for each encoding used later on, after doing all those tasks we're given the flag:
```
Bloody Ripper! Here is the grand prize!

.^.
(( ))
|#|_______________________________
|#||##############################|
|#||##############################|
|#||##############################|
|#||##############################|
|#||########DOWNUNDERCTF##########|
|#||########(DUCTF 2021)##########|
|#||##############################|
|#||##############################|
|#||##############################|
|#||##############################|
|#|'------------------------------'
|#|
|#|
|#|
|#|
|#|
|#|
|#|
|#|
|#|
|#|
|#|
|#| DUCTF{you_aced_the_quiz!_have_a_gold_star_champion}
|#|
|#|
|#|
//|\\
```
**Flag**

**`DUCTF{you_aced_the_quiz!_have_a_gold_star_champion}`**

## rabbit
Can you find Babushka's missing vodka? It's buried pretty deep, like 1000 steps, deep.

**Author:** Crem + z3kxTa

[flag.txt](challenges//flag.txt)

**Solution**

This type of challenge is really common and I've covered a variation of it before in [my writeup for TJCTF 2020](https://github.com/W3rni0/TJCTF_2020#zipped-up).

With the challenge we're given a file falsely named `flag.txt`, this file is actually a compressed file of either `bzip2`, `zip`, `gzip` or `xz` format. And, as the description for the challenge suggests, it contains another compressed file of the same type and so on. decompressing each file manually is possible but will require a lot of work, so we can automate the process by scripting.

For that I wrote the following bash script that for each file in the working directory checks its type, and if it of one of the above compression file formats, the script extract the content of the file and removes it. This process terminates only when there aren't any compressed files in the directory:

```bash
#!/bin/bash
get_flag() {
for filename in $(pwd)/*; do
echo "$filename"
if [[ $(file --mime-type -b "$filename") == "application/x-bzip2" ]]
then
bunzip2 "$filename"
rm "$filename"
get_flag
return
elif [[ $(file --mime-type -b "$filename") == "application/zip" ]]
then
mv $filename flag.zip
unzip flag.zip
rm flag.zip
get_flag
return
elif [[ $(file --mime-type -b "$filename") == "application/x-xz" ]]
then
mv $filename flag.xz
unxz -f flag.xz
rm flag.xz
get_flag
return
elif [[ $(file --mime-type -b "$filename") == "application/gzip" ]]
then
mv $filename flag.gz
gunzip flag.gz
rm flag.gz
get_flag
return
else
cat "$filename"
return
fi
done
}

get_flag
```

And by letting the script run for a while I got a text file containing the following string:
```
RFVDVEZ7YmFidXNoa2FzX3YwZGthX3dhc19oM3IzfQ==
```
I inferred by the symbols in the string that this is a base64 encoded message, and decoding it from base64 using [CyberChef](https://gchq.github.io/CyberChef/) I got the flag:

**Flag**

**`DUCTF{babushkas_v0dka_was_h3r3}`**

## Floormat
I've opened a new store that provides free furnishings and floormats. If you know the secret format we might also be able to give you flags...

**Author:** todo#7331

`nc pwn-2021.duc.tf 31903`

[floormat.py](challenges//floormat.py)

**Solution**

The function str.format() is used to replace a string we control with an object, which gives us access to object's attributes, one of them is a dictionary of global variables in the init of the object, which contains the flag:
```python
from pwn import *

s = remote('pwn-2021.duc.tf', 31903)
# Ask for a custom format
s.sendline("my l33t format")
# Send Payload
s.sendline("{f.__init__.__globals__[FLAG]}")
s.sendline("F")
# Ask for one of the furnishing (not important which)
s.sendline("flutter")
s.interactive()
```
**Flag**

**`DUCTF{fenomenal_flags_from_funky_formats_ffffff}`**

***

# Cryptography

## Substitution Cipher I
Just a simple substitution cipher to get started...

**Author:** joseph#8210

[substitution-cipher-i.sage](challenges//substitution-cipher-i.sage) | [output.txt](challenges//output_i.txt)

**Solution**

The encryption is known and works letter by letter, thus we can find the encryption of each letter and use that to inverse the ciphertext:
```python
from string import printable

def encrypt(msg, f):
return [chr(f.substitute(c)) for c in msg]

def decrypt(enc, f):
subtitution_table = encrypt(printable.encode('utf-8'), f)
return ''.join(printable[subtitution_table.index(c)] for c in enc)

P.<x> = PolynomialRing(ZZ)
f = 13*x^2 + 3*x + 7

enc = open('./output.txt', 'r').read().strip()
FLAG = decrypt(enc, f)
print(FLAG)
```
**Flag**

**`DUCTF{sh0uld'v3_us3d_r0t_13}`**

## Substitution Cipher II
That's an interesting looking substitution cipher...

**Author:** joseph#8210

[substitution-cipher-ii.sage](challenges//substitution-cipher-ii.sage) | [output.txt](challenges//output_ii.txt)

**Solution**

Similar to the previous challenge, only difference is that we need to find the polynomial now, we can do that by knowing that the flag starts with '`DUCTF{`' and ends with '`}`', and use lagrange interpolation to find the polynomial:

```python
from string import ascii_lowercase, digits, printable
CHARSET = "DUCTF{}_!?'" + ascii_lowercase + digits
n = len(CHARSET)

def encrypt(msg, f):
ct = ''
for c in msg:
ct += CHARSET[f.substitute(CHARSET.index(c))]
return ct

def decrypt(enc, f):
subtitution_table = encrypt(CHARSET, f)
return ''.join(CHARSET[subtitution_table.index(c)] for c in enc)

P.<x> = PolynomialRing(GF(n))
enc = open('./output.txt', 'r').read().strip()

X = [0, 1, 2, 3, 4, 5, 6]
Y = [CHARSET.index(c) for c in enc[:6]] + [CHARSET.index(enc[-1])]
points = zip(X,Y)
f = P.lagrange_polynomial(points)

FLAG = decrypt(enc,f)
print(FLAG)
```
**Flag**

**`DUCTF{go0d_0l'_l4gr4fg3}`**

## Break Me!
AES encryption challenge.

**Author:** 2keebs

`nc pwn-2021.duc.tf 31914`

[aes-ecb.py](challenges//aes-ecb.py)

**Solution**

AES-ECB one byte at a time attack, I covered it before in my writeup for [H@cktivityCon CTF 2020](https://github.com/W3rni0/HacktivityCon_CTF_2020#a-e-s-t-h-e-t-i-c):

```python
from pwn import *
import re
from string import printable
from Crypto.Cipher import AES
from base64 import b64decode

# Part 1 - Retrieving the key
s = remote('pwn-2021.duc.tf', 31914)
s.recvuntil(":")
key = '' # !_SECRETSOURCE_!

for i in range(len(key) + 1,16):
s.sendline('a' * (16 - i))
base_block = re.findall('[A-Za-z0-9+\/=]{64}', s.recvuntil(":").decode('utf-8'))[0]
print(base_block)
for c in printable:
s.sendline('a' * (16 - i) + key + c)
block = re.findall('[A-Za-z0-9+\/=]{64}', s.recvuntil(":").decode('utf-8'))[0]
if block == base_block:
key = key + c
print(key)
break

s.close()

# Part 2 - Getting the flag
s = remote('pwn-2021.duc.tf', 31914)
s.sendlineafter(':\n', ' ' * 16)
ct = b64decode(s.recvuntil("\n").decode().strip())
cipher = AES.new(key.encode(), AES.MODE_ECB)
pt = cipher.decrypt(ct)
print(pt)

```

**Flag**

**`DUCTF{ECB_M0DE_K3YP4D_D474_L34k}`**

## treasure
You and two friends have spent the past year playing an ARG that promises valuable treasures to the first team to find three secret shares scattered around the world. At long last, you have found all three and are ready to combine the shares to figure out where the treasure is. Of course, being the greedy individual you are, you plan to use your cryptography skills to deceive your friends into thinking that the treasure is in the middle of no where...

**Author:** joseph#8210

`nc pwn-2021.duc.tf 31901`

[treasure.py](challenges//treasure.py)

**Solution**

We can easily retrieve the secret by first sending 1 as our share, which means that the assumed secret is only the product of the other two shares raised to the power of two, and using our own share we can calculate the secret similarly to the way the combiner does that.

Using the secret and the fake_coords we will calculate the 3rd root of the division between them, and create a new share which is the product of the real share and the result, this will guarantee that the result of the combiner is `secret * fake_secret / secret = fake_secret` and so the result is the fake coordinates, and we can safely use our real coordinates, I wrote the following sage script to perform all of those calculation and interact with the server:
```python

from Crypto.Util.number import long_to_bytes
from pwn import *
from gmpy2 import iroot

FAKE_COORDS = 5754622710042474278449745314387128858128432138153608237186776198754180710586599008803960884
p = 13318541149847924181059947781626944578116183244453569385428199356433634355570023190293317369383937332224209312035684840187128538690152423242800697049469987
K = GF(p)

s = remote('pwn-2021.duc.tf', 31901)
share = int(s.recvline().split(b' ')[-1])
s.sendlineafter(': ', '1')
product = int(s.recvline().split(b' ')[-1])
REAL_COORDS = pow(share, 3, p) * product % p

FAKE_DIV_REAL = K(FAKE_COORDS) / K(REAL_COORDS)
FAKE_DIV_REAL_ROOT = FAKE_DIV_REAL.nth_root(3)
new_share = share * FAKE_DIV_REAL_ROOT

s.sendlineafter(': ', str(new_share))
s.sendlineafter(': ', str(REAL_COORDS))
s.interactive()
```

**Flag**

**`DUCTF{m4yb3_th3_r34L_tr34sur3_w4s_th3_fr13nDs_w3_m4d3_al0ng_Th3_W4y.......}`**

***

# Reversing

## no strings
This binary contains a free flag. No strings attached, seriously!

**Author:** joseph#8210

[nostrings](challenges//nostrings)

**Solution**

With the challenge we're given an ELF binary file, running it prompt us to give the flag:

So this is a flag checker, we can look at the disassembly to try and understand how it works, for disassembly in this challenge I used IDA free. The first thing that pops while looking at the disassemly is the main function:

Even without looking at the instructions themselves we can figure out that it is looping over something, presumably our input, at further inspection we can see that is does just that while comparing each letter to a string it has in it's data section (annotated to make it more understandable):

And we can get the full flag from the data section:

The reason that strings don't normally will pick up this flag is because it is encoded with 16-bit where strings uses 7-bit (ASCII) by default, we can choose another encoding using the `-e` flag:

**Flag**

**`DUCTF{stringent_strings_string}`**

Original writeup (https://github.com/W3rni0/DownUnderCTF_2021#break-me).

Break Me!

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