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$n \cdot x ≡ k \mod{20} \\n,k \in \mathbb{Z}$
Choose $n,k$ so there is only $(a)$ one, $(b)$ no solution in $\mathbb{Z_{20}}$.
If $n = 0$, then two solution exist, and that is $k=20$ and $k=-20$, as we are in $\mathbb{Z_{20}}$
If $n,k = 0$, then there is one solution, because $0 ≡\mod{20} = 0$
Stuck on the question how to find no solution. Basically you could just put $n \geq 21$ and $\mod{20}$ will never be able to generate a number $\geq 20$
For one solution, say $n=1$, $k=1$. Clearly, $x\equiv 1\pmod{20}$ is the only solution.
For no solution, $n=2$, $k=1$. If $n=2$, then $nx$ is always even, so cannot be congruent to any odd number modulo $20$.
For more than $1$ solution, $n=4$, $k=0$. Note that $x\equiv 0\pmod{20}$ is a solution, but so is $x\equiv 5\pmod{20}$. Also, $10$ and $15$ work.
For exactly two solutions, $n=2$, $k=0$ works, the only solutions are $x\equiv 0\pmod{20}$ and $x\equiv 10\pmod{20}$.
Surely you could have found these, or ones like them. For all the questions, there are several answers that work.
Your calculation was not correct, $20$ and $-20$ and $0$ are all the same modulo $20$. In fact, if we take $n=0$, $k=0$, there are $20$ solutions (any $x$ works).
Hint: $ax \equiv b \pmod{m}$ has a unique solution if and only if $\gcd(a,m) = 1$.
More generally, using Linear congruence theorem, this equation is solvable $\iff$ $(20,n)\mid k$
In that case,we have exactly $(20,n)$ solutions.
