ctf-writeup/Cyber Apocalypse 2023 The Cursed Mission/Small StEps

Small StEps

As you continue your journey, you must learn about the encryption method the aliens used to secure their communication from eavesdroppers. The engineering team has designed a challenge that emulates the exact parameters of the aliens' encryption system, complete with instructions and a code snippet to connect to a mock alien server. Your task is to break it.

About the Challenge

We were given a zip file (You can download the file here). And if we unzip the file, there are another file called solver.py and server.py. Here is the content of server.py

from Crypto.Util.number import getPrime, bytes_to_long

FLAG = b"HTB{???????????????}"
assert len(FLAG) == 20


class RSA:

    def __init__(self):
        self.q = getPrime(256)
        self.p = getPrime(256)
        self.n = self.q * self.p
        self.e = 3

    def encrypt(self, plaintext):
        plaintext = bytes_to_long(plaintext)
        return pow(plaintext, self.e, self.n)


def menu():
    print('[E]ncrypt the flag.')
    print('[A]bort training.\n')
    return input('> ').upper()[0]


def main():
    print('This is the second level of training.\n')
    while True:
        rsa = RSA()
        choice = menu()

        if choice == 'E':
            encrypted_flag = rsa.encrypt(FLAG)
            print(f'\nThe public key is:\n\nN: {rsa.n}\ne: {rsa.e}\n')
            print(f'The encrypted flag is: {encrypted_flag}\n')
        elif choice == 'A':
            print('\nGoodbye\n')
            exit(-1)
        else:
            print('\nInvalid choice!\n')
            exit(-1)


if __name__ == '__main__':
    main()

As we can see, the public exponent is very small (3) and we only have an information about the ciphertext, public exponent, and the public key

How to Solve?

Im using RsaCtfTool to solve the chall, here is the command I used

python3 RsaCtfTool.py -n 7079390432974855345319643397277546809126929622683749677273378219032261712283081531451428558847051155395790169332117706417782515882054955482729948765182401 -e 3 --uncipher 70407336670535933819674104208890254240063781538460394662998902860952366439176467447947737680952277637330523818962104685553250402512989897886053

Wait for a while and you will get the flag

HTB{5ma1l_E-xp0n3nt}