Calculating private keys in the RSA cryptosystem The number $43733$ was chosen as base for an implementation of the RSA system.
$M=19985$ is the message, that was encrypted with help of a public key $K=53$.

What is the plaintext text?
What is the private key?

So far, my calculations are:


$n=pq$
$n=101*433$
$\phi(43733) = (101 − 1)*(433 − 1) = 43200$.
The public key is $(n = 43733, e = 53)$.
The private key is $(n = 43733, d = 12343)$.


However, I'm not sure if this is right? Where am I going wrong?
The public key is $K = e = 53$, already given. $n$ (the modulus) must also be given, so you could say that $(e, n)$ is the actual key.

The private key is $d$ which must satisfy $d * e = 1 \mod \phi(n)$.
So you're looking for $d$ for which $(53 * d) \mod 43200 == 1$. A quick brute-force search (with such small numbers it's not a problem) reveals that $17117$ satisfies this equation.

Now $N=43733$, and you're guaranteed that $(M^e)^d = (M^d)^e = M \mod n$.

Your message is $M=19985$:


$19985^{53}  \equiv 17195 \mod 43733$
$17195^{17117} \equiv 19985 \mod 43733 $


This shows that you can encrypt using the public key and decrypt using the private key.
The opposite also works (for signing the message):


$19985 ^{17117}  \equiv 125 \mod 43733$
$125 ^{53} \equiv 19985 \mod 43733$


There are better ways to find $d$ from $e$ if you know $\phi(n)$. But if you don't, you're in trouble, because you need to factorise $n$ to do that.