rsa3

encrypt.py

from Crypto.PublicKey import RSA
from base64 import b64encode

key = RSA.generate(1024, e=3)

with open('flag.txt', 'rb') as f:
    flag = f.read(36)

c = pow(int.from_bytes(flag.ljust(128, b'\0'), 'big'), key.e, key.n)

print(b64encode(c.to_bytes(128, 'big')))

key.pem

-----BEGIN PUBLIC KEY-----
MIGdMA0GCSqGSIb3DQEBAQUAA4GLADCBhwKBgQC0+b3hWC2gzK8MD6ZUluRlYy+j
GwVHeaVWciWNBTcuDDm6qLq00EbhtlalonDumBLA6OJiqCS8tRFVhnRwf58ywTyk
b18UW6+5u+Ljmx4MZqpKVfNGtpwXsfEkasvUt0/OCBpnZ6JrGX6win00/HjJVXvY
tGRMjVfqtvROvrpZeQIBAw==
-----END PUBLIC KEY-----

rsa3.md

RSA Challenge Redux

Instructions are pretty simple: decrypt the flag

mLteOOUCPNlJc1DBFP9WEg3A2kZNJIUxVcGgluNdV+XQHhmI6C2Glnaya+/HNxycrIEaQa4GDWmBMlnY8C+7HbKRHpiHVgLdZ8wgMDf9JJQCKSnCIIPmCztcA+8IzZj7uJzQBtRuaPJYgX5EAiudZXbLxg3N4HkY/8MeqiukFdU=

Hint 1:

What is 5^3 mod 10000? More generally how can we find x given x^3 mod N when x^3 < N?

Hint 2:

What is the result of decrypting (256^3 * c) mod N for some ciphertext c? Can we undo this operation?