We have the numbers 21,22,23,6955. This is an encrypted message with AES encryption. The first 3 numbers might be keys and the last is a result of the final encryption. We have no Idea what order the numbers were encrypted. Use knowledge of AES to break each ciphertext to figure out the message.

In a public-key system using RSA, suppose you intercepted a cipher text C = 61 sent to a user whose public key is e = 11, n = 91. What is the plaintext M? Explain the steps that you find the plaintext.

Let's say you generated RSA keys and you chose p=53, q=41, and the smallest encryption exponent that qualifies for this p and q. (You will have to follow the key generation algorithm to fill in the details.) If someone gives you 343 as the ciphertext they created using your public key, what is the plaintext you get when you use your private key to decrypt? Also note that you can test your answer by encrypting it and making sure you get 343 back.

Here's how to secure your bike: Only things with a higher number may be locked again after being unlocked. Locks may be unlocked at any moment. All the locks are X-locks. Demonstrate how serializability is not ensured by the protocol.