Okay we have a message that is encrypted using some sort of RSA encryption. I think. The public key is given with the message. A hint in https://forum.hackthebox.eu/discussion/689/crypto-infinite-descent says that we should use the files given. Hmmmm. I think we should try to re-create the private key from the public key. Let's try dumping the key parameters using https://github.com/Ganapati/RsaCtfTool
Okay...let's modify the fasterprimes.py script with the new values for p and q and then decrypt the encrypted message using the key created with the pq values. (https://stackoverflow.com/questions/30056762/rsa-encryption-and-decryption-in-python)
What is this number. After some time, I noticed something in AESbootstrap.py (the example at the bottom of that script), it says that it takes a triplet to convert into something.
Hmmm. Let's try separating the decrypted number into 3 digits and feed into the seedval variable.
IT'S THE FLAG!!! FINALLYY!!!
Took me a while to solve this. Don't know if this is the easiest way. If you guys have any other methods, please let me know in the comments. Thanks.
Don't show the flag, please
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