For $n \in \Bbb N$ we define the set $\Bbb Z_n = \{1, 2, \ldots , n - 1\}$ and on this set we define the modular product as follows:
For $x, y, z \in \Bbb Z_n : (x\cdot y = z) ⇔ (x\cdot y \equiv z \pmod{n})$. In other words, we get the number $z$ by computing the product of $x$ and $y$ as the ordinary product of two natural numbers and then calculating the remainder of this product after dividing by $n$.
Examples for $n = 5$ and different values of $x$ and $y$: in $\Bbb Z_5 : 3\cdot 4 = 2 , 2\cdot 3 = 1 , 2\cdot 4 = 3 \ldots$
Problem: We construct a graph $G$ such that its vertices are elements of the set $\Bbb Z_{101}$ and the two vertices corresponding to the elements $x$ and $y$ are connected by an edge if and only if $x\cdot y = 1$ in the sense of the modular product defined above.
a) Is the graph $G$ ordinary?
b) Is the graph $G$ continuous?
c) Is the graph of $G$ a tree?
d) What will be the sum of all the numbers in the adjacency matrix of graph $G$?
Suppose $[x] \in \mathbb{Z}_n$ such that $gcd(x,n) = 1$.
By Bézout’s Identity there exists $a,b \in \mathbb{Z}$ such that $xa + nb = 1$. Therefore $xa \equiv 1 \pmod {n}$, and $[x][a] = [1]$ in $\mathbb{Z}_n$.
Now if there exists $[a],[b] \in \mathbb{Z}_n$ such that $[x][a] = [x][b] = [1]$, then we have
$$ a \equiv a(xb) \equiv (ax)b \equiv b \pmod{n} $$
Hence $[a] = [b]$.
In summary, if $[x] \in \mathbb{Z}_n$ with $gcd(x,n) = 1$, there exists a unique element $[x]^{-1} \in \mathbb{Z}_n$ such that $[x]^{-1}[x] = [x][x]^{-1} = [1]$. We call $[x]^{-1}$ the multiplicative inverse of $[x]$.
Therefore if $p$ is a prime number, then every non-zero element of $\mathbb{Z}_p$ has a unique multiplicative inverse. Since $101$ is a prime number we can use this fact to understand the structure of the graph $G$.
The vertices of $G$ are the non-zero elements of $\mathbb{Z}_{101}$ and the edges of $G$ are just the multiplicative inverse pairs of the vertices. That is, $$V(G) = \{[x] \in \mathbb{Z}_{101} \mid [x] \neq [0] \} $$ $$E(G) = \{ \{[x],[x]^{-1} \} \mid [x] \in V(G) \} $$
Notice that $[1]$ and $[p-1]$ are multiplicative self-inverses in $\mathbb{Z}_p$. Furthermore, none of the remaining elements of $\mathbb{Z}_p$ are multiplicative self-inverses. To see this, let $x \in \mathbb{Z}$ such that $0 < x < p$ and $[x]^2 = [1]$. Then $x^2 \equiv 1 \pmod{p}$, and by definition $p \mid x^2 - 1 = (x-1)(x+1)$. Since p is prime, $p \mid (x-1)$ or $p \mid (x+1)$. But this is only possible if $x = 1$ or $x = p-1$.
Therefore $G$ has exactly two loops, and they are at the vertices [1] and [100]. The remaining $98$ vertices are adjacent to exactly one other distinct vertex. This gives a complete picture of the structure of $G$. The answers to your questions can be readily deduced from this.