581 words

3 minutes

CyberX{Internal_CTF}

2024-12-22

CTF

Crypto

/

Misc

CyberX CTF#

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

Math?#

We are given

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

and equations in png:

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

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

Remember to check the values:

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if p * q == n:
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    print("yes")
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else:
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    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!

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import random
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mapping = [
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    'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H',
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    'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P',
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    'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X',
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    'Y', 'Z', 'a', 'b', 'c', 'd', 'e', 'f',
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    'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n',
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    'o', 'p', 'q', 'r', 's', 't', 'u', 'v',
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    'w', 'x', 'y', 'z', '1', '2', '3', '4',
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    '5', '6', '7', '8', '9', '0', '{', '}'
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]
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# Create a scrambled version of the mapping
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scrambled_mapping = mapping.copy()
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random.seed(12423915)
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random.shuffle(scrambled_mapping)
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def encrypt(flag):
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    value = []
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    for i in range(len(flag)):
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        z = scrambled_mapping.index(flag[i])  # Find the index in scrambled mapping
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        chars = "abcdefgh"
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        filemodulus = (z % 8)  # Calculate the "file" (a-h)
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        file = chars[filemodulus]
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        rank = int(z / 8) + 1  # Calculate the "rank" (1-8)
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        value.append(''.join(f'{file}{rank}'))
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    return value
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def decrypt(encrypted_values):
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    chars = "abcdefgh"
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    decrypted_message = []
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    for value in encrypted_values:
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        file = value[0]  # Extract the file (a-h)
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        rank = int(value[1])  # Extract the rank (1-8)
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        # Find the index corresponding to the file
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        filemodulus = chars.index(file)
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        # Calculate the original index of the scrambled_mapping
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        z = (rank - 1) * 8 + filemodulus
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        decrypted_message.append(scrambled_mapping[z])
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    return ''.join(decrypted_message)
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# Example encrypted chess positions (this should match the output of `encrypt` function)
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encrypted_values = ['']
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# Decrypt the message
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decrypted_message = decrypt(encrypted_values)
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print("Decrypted:", decrypted_message)
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# Example encryption of a message
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message = "ILOVEUTM"
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try:
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    encrypted_message = encrypt(message)
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    print("Encrypted:", encrypted_message)
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except:
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    print("Error: Message contains characters not in the mapping.")

We just need to figure out what is the encrypted values

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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

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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)

Simon & Bob#

CyberX{Internal_CTF}

https://fuwari.vercel.app/posts/cyberxctf-2024/

Author

Bakayang

Published at

2024-12-22

License

CC BY-NC-SA 4.0

CSLU{Cyber_Skills_Level_Up_@_UPM}

Google Beginners Quest