This is a question on cryptography — secure communication.
Let $p= 11$, $q= 13$, and $e = 17$.
Bob’s public key is given by the two numbers $p × q$ and $e$.
Bob’s private key, the number $d$, satisfies
$$ d ≡ e^{−1} \pmod {(p− 1) (q− 1)} $$
(i) Determine Bob’s public key numbers and the private key number.
(ii) Alice encrypts a message $M = 12$ by forming
$$ M^e ≡ a \pmod {pq} $$
Bob recovers the message by using his private key $d$ such that
$$M ≡ a^d \pmod {pq}$$
Find $a \pmod {pq}$ and show that
$$ M ≡ a^d \pmod {pq}$$
(i) Bob's public key numbers are $p \times q = 11 \times 13 = 143$ and $17$.
Bob's private key number $d$ satisfies
$$ \begin{align} d & \equiv e^{−1} \pmod {(p− 1) (q− 1)} \\ \\ & \equiv e^{-1} \pmod {120} \\ \\ 17d & \equiv 1 \pmod {120} \end{align} $$
Here $d=113$ satisfies this, and so Bob's private key is 113.
(ii) We have $a$ as
$$ \begin{align} a & \equiv M^e \pmod {pq} \\ \\ & \equiv 12^{17} \pmod {143} \\ \\ & \equiv (12^2)^8 \times 12 \pmod {143} \\ \\ & \equiv (-1)^8×12 \pmod{143} \\ \\ a & \equiv 12 \pmod {143} \end{align} $$
We now recover $M$
$$ \begin{align} a^d & \equiv 12^{113} \pmod {143} \\ \\ & \equiv (12^2)^{56} \times 12 \\ \\ & \equiv (-1)^{56} \times 12 \\ \\ & \equiv 12 \pmod {143} \\ \\ & \equiv M \pmod {143} \end{align} $$