CyberX CTF

Can't believe we won Champion for our first CTF tournament!
My writeups for the cryptography CTFs.

Math?

We are given

python

n = {LARGE_VALUE}
e1 = {LARGE_VALUE}
e2 = {LARGE_VALUE}
c1 = {LARGE_VALUE}
c2 = {LARGE_VALUE}

and equations in png:

This can actually be solved in 3 lines lol (or even less XD)

python

q = math.gcd(n, pow(c1, e2, n) * pow(5, e1 * e2, n) - pow(c2, e1, n) )
q // 16
p = n // q

Remember to check the values:

python

if p * q == n:
    print("yes")
else:
    print("no")

Here's the full code: math.py

For those who are actually curious on how I solved it?

I think we need to change the math equation into simpler and more efficient.
So, I just follow the reference I found online, and try to modify the equations into more efficient ones.

References:

XOR II

I found an article of XOR Crypto: link

Here's how XOR worked:

When you construct a XOR cipher, you need to enter 2 values: the key and the flag

We don't know the key, but we know the flag starts with "CyberX{"

flag can be used as a key too to

note: "CyberX{" consists of 7 letters, the output will also be 7 letters.

So we use the output "Pokemon" as the key

Bingo!

ChessMaster

This one is a bit tough since I have little knowledge in chess.

We are given a python code, and 18 chessmoves images

The python code looks like cryptography decryption with chessmoves!

python

import random

mapping = [
    'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H',
    'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P',
    'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X',
    'Y', 'Z', 'a', 'b', 'c', 'd', 'e', 'f',
    'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n',
    'o', 'p', 'q', 'r', 's', 't', 'u', 'v',
    'w', 'x', 'y', 'z', '1', '2', '3', '4',
    '5', '6', '7', '8', '9', '0', '{', '}'
]

# Create a scrambled version of the mapping
scrambled_mapping = mapping.copy()
random.seed(12423915)
random.shuffle(scrambled_mapping)

def encrypt(flag):
    value = []
    for i in range(len(flag)):
        z = scrambled_mapping.index(flag[i])  # Find the index in scrambled mapping
        chars = "abcdefgh"

        filemodulus = (z % 8)  # Calculate the "file" (a-h)
        file = chars[filemodulus]

        rank = int(z / 8) + 1  # Calculate the "rank" (1-8)

        value.append(''.join(f'{file}{rank}'))
    return value

def decrypt(encrypted_values):
    chars = "abcdefgh"
    decrypted_message = []

    for value in encrypted_values:
        file = value[0]  # Extract the file (a-h)
        rank = int(value[1])  # Extract the rank (1-8)

        # Find the index corresponding to the file
        filemodulus = chars.index(file)

        # Calculate the original index of the scrambled_mapping
        z = (rank - 1) * 8 + filemodulus
        decrypted_message.append(scrambled_mapping[z])

    return ''.join(decrypted_message)

# Example encrypted chess positions (this should match the output of `encrypt` function)
encrypted_values = ['']

# Decrypt the message
decrypted_message = decrypt(encrypted_values)
print("Decrypted:", decrypted_message)

# Example encryption of a message
message = "ILOVEUTM"
try:
    encrypted_message = encrypt(message)
    print("Encrypted:", encrypted_message)
except:
    print("Error: Message contains characters not in the mapping.")

We just need to figure out what is the encrypted values

python

encrypted_values = ['?'] # ?????

When you look at the chessmoves images, you will see they can all be solved with just 1 move checkmate.

Now write all the values in their respective chess positions

python

encrypted_values = ['g5', 'h2', 'h5', 'd2', 'd1','g3','d7', 'a5', 'e5', 'e3', 'd2', 'b5', 'a6', 'a7', 'd1', 'g4', 'f2', 'c6']

There is also some tricky in the last image (the white and black position were swapped):

Finally, run the Code and Voila!

APT APT APT APT

This challenge is actually very easy but just annoying.

First, it looks like a cipher. So, I just went to decode.fr to search for music related cipher:

I ciphered everything just in case lol (pain)

CyberX{Ch4mp10n}

CyberX CTF

Math?

XOR II

ChessMaster

APT APT APT APT

Simon & Bob