Assume that $p = 5$ and $q = 11$, and all other variables are defined as per the RSA theorem
(a) Suppose we consider $e = 3$. Would $(e, n)$ be a suitable public key?
(b) Prove that if $d = 27$, then $(d, n)$ is indeed the private key for the given public key $(3, n)$.
(c) Suppose you receive the cipher-text $C = 47$ (note $0 \leq C < n$). Decrypt the message using your private key.
I tried part (a), well as much as I could. I got
$n=pq= 55$
$z = (p-1)(q-1) = 40$
Then the public keys would be $40$ and $7$, I'm not sure if $3$ is an option for a suitable key or not.
Confused with b and c.
(a) A suitable public key is any number $e$ such that $1<e<z$ and $e$ and $z$ have no common prime factors. Is this true for $e=3$ and $z=40$? (I'm not sure where your $7$ came from, but it too would be a suitable public key).
(b) Given $e$, the choice of $d$ is determined by the equation $ed\equiv 1 \pmod z.$ Does $(3)(27)$ satisfy this equation?
(c) If $C$ is the ciphertext, it is decrypted by calculating $C^d \pmod n$.