Number of primes in $[30! + 2, 30! + 30]$

Question

How to find number of primes numbers $\pi(x)$ in $[30! + 2$ , $30! + 30]$,

where $n!$ is defined as:

$$n!= n(n-1)(n-2)\cdots3\times2\times1$$

Using Fermat's Theorem: $130=1\mod31$, (since $31 \in \mathbb{P}$). This implies the above is congruent to $17\mod31$.

This is correct, right?

Answer 1

Since $30!$ is divisible by all numbers from $2$ to $30$ obviously,

$30!+2$ is divisible by $2$

$30!+3$ is divisible by $3$

$\vdots$

$30!+30$ is divisible by $30$

No primes.

Answer 2

Observe that

$$n!+m$$ is divisible by $m$ for $2\le m\le n$ and integer $n\ge2$

So, we can have an arbitrarily large sequence of composite numbers for an arbitrary large value of integer $n$