# modulo RSA decrypt question

Given the following RSA generated public key: $P(3, 55)$. Which integer value should be chosen for $d$ to decrypt messages encrypted with $P$?
Check your answer with $M = 8$ and $C = 17$.

$P(3,55)$ is $p(e,n)$

So through trial and error i have got $d = 7$ but how do you calculate this the proper way?

To check this we can use the encryption formulas:
$p(M) = M^e \bmod n$
so $p(8) = 8^3 \bmod 55 = 17 = c\\ s(c)= c^d \bmod n \\ s(17)=17^d \bmod 55 = M = 8$

so basically we have $17^d \bmod 55=8$

Now how do we find out $d$ from that equation other than trial and error, which was how I found $d=7$

Because we (as decrypters) know that $55=5\times 11$, we also know that Carmichael's function $\lambda(55) =\text{lcm}(\phi(5), \phi(11)) = \text{lcm}(4,10) = 20$
So we need to find the multiplicative inverse of $3 \bmod 20$, that number $b$ such that $3b \equiv 1 \bmod 20$, which is obvious enough by inspection for these small numbers $(3\times 7 = 21)$ but otherwise can be found by the extended Euclidean algorithm, which I explain here.
$$\begin{array}{c|c|c|c} \large{n}\atop{ = ms+at} & \color{gray}{s} & t & q \\ \hline 20 & \color{gray}{1} & 0 & \\ 3 & \color{gray}{0} & 1 & 6 \\ 2 & \color{gray}{1} & -6 & 1 \\ 1 & \color{gray}{-1} & \color{red}{7} & 2 \\ \end{array}$$ and $3^{-1} \equiv \color{red}7 \bmod 20$. $\quad 3\times 7 = 21 \equiv 1 \bmod 20 \quad \checkmark$
And when we have that, that gives us $(8^3)^7=8^{21} = 8^{k\lambda(55)+1} \equiv 8^1 \bmod 55$, so decrypting the message.