Cryptography - 200 points
Bob is extremely paranoid, so he decided that just one RSA encryption is not enough. Before sending his message to Alice, he forced her to create 5 public keys so he could encrypt his message 5 times! Show him that he still is not secure... rsa.txt.
The text is encrypted multiple times using the same n but different e
Take a look at: Is it safer to encrypt twice with RSA? - StackExchange
It is said that for 2 exponents: c = (m^e1)^e2 (mod n)
Hence, our e in this case is to multiply all 5 exponents together.
e = 11 * 41 * 67623079903 * 5161910578063 * 175238643578591220695210061216092361657427152135258210375005373467710731238260448371371798471959129039441888531548193154205671
And we have our new e to work with
e = 27587468384672288862881213094354358587433516035212531881921186101712498639965289973292625430363076074737388345935775494312333025500409503290686394032069
Looking at some lists of exploits to do...
We notice our e is huge. (Reference: PlaidCTF CTF 2015 Curious)
When e is large, d is small.
Hence, Wiener's attack can be used when d is small.
python3 solve.py
easyctf{keblftftzibatdsqmqotemmty}