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Encryption using a block cypher such as AES by passing plaintext blocks directly to the encryption function is known as Electronic Code Book mode (ECB) and is not CPA secure as (as you say in your question) it is entirely deterministic and two identical plaintext blocks will result in two identical ciphertext blocks. To prevent this an initialisation ...

To be secure against a chosen-plaintext attack, an encryption scheme must be non-deterministic — that is, its output must include a random element, so that e.g. encrypting the same plaintext twice will result in two different ciphertexts. Indeed, if that was not the case, an attacker could easily win the IND-CPA game just by using the encryption ...

In base 10 we write for example $133$ when we mean $$133 = 1 * 10^2 + 3*10^1 + 3*10^0. $$ If we want to write $49$ in base $2$ then note first that: $$49 = 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 1*2^0. $$ Because of this $49$ is $110001$. Noe obviously, you "don't know this", but I wanted to write it down so that you can see what happens as you divide by ...