CSICTF-Writeups/Crypto/Modern Clueless Child

MODERN CLUELESS CHILD

The main idea finding the flag is decryption using XOR keys.

Step-1:

After reading the given message:

I was surfing the crimson wave and oh my gosh I was totally bugging. I also tried out the lilac hair trend but it didn't work out. That's not to say you are any better, you are a snob and a half. But let's get back to the main question here- Who am I? (You don't know my name)

Ciphertext = "52f41f58f51f47f57f49f48f5df46f6ef53f43f57f6cf50f6df53f53f40f58f51f6ef42f56f43f41f5ef5cf4e" (hex) Key = "12123"

Step-2:

I quickly removed the f from cipher text as looked like it was used for space. So I wrote a script sub.py to replace f with ''.

ciphertext = "52f41f58f51f47f57f49f48f5df46f6ef53f43f57f6cf50f6df53f53f40f58f51f6ef42f56f43f41f5ef5cf4e"
sub = ciphertext.replace('f','')
print(sub)

On running python3 sub.py this, it gave me 52415851475749485d466e5343576c506d53534058516e425643415e5c4e.

Step-3:

I had to check if I am not missing any cipher text so I cross check the flag by XOR checks. So, I wrote this xor1.py script get the csictf{ code:

from pwn import xor
flag = bytes.fromhex('52415851475749485d466e5343576c506d53534058516e425643415e5c4e')
print(xor(flag, 'csictf{'.encode()))

Output:

b"1212312+./\r'%,\x0f#\x040'&#2\x1d+57'%?="

Step-4:

Since we got the key 1212312 means we are right path as key has cyclic property key (12123). Now it was just replacement on the key with ASCII.

exlpoit.py to get flag:

from pwn import xor
flag = bytes.fromhex('52415851475749485d466e5343576c506d53534058516e425643415e5c4e')
print(xor(flag, '12123'.encode()))

On running python3 exploit.py, Voila! I got the flag.

b'csictf{you_are_a_basic_person}'

Step-5:

Finally the flag becomes: csictf{you_are_a_basic_person}