Is there at least one prime between $n \times 100$ and $(n \times 100) + 100$ where $n \in \mathbb{N}$ Is there at least one prime between the number $n \times 100$ and $(n \times 100) + 100$ for any $n \in \mathbb{N}$ that can be $0$ ?
Question originally formulated by one of my friends.
 A: No.
Consider $n=99!$. $100n = 100!$ is composite, and $100!+1$ is divisible by $101$ (Wilson's theorem) and so is composite. $100n+i$ is divisible by $i$ for $2\leq i \leq 100$.
A: Not necessarily. For example none of the numbers between $200!+100$ and $200!+200$ can be prime, because $200!+k$ is divisible by $k$ when $2\le k\le 200$.
So a concrete counterexample would be $n=\frac{200!}{100}+1$.

For a smaller counterexample one could let $100n$ be the product of all primes up to 109, times 3, plus 10. That gives an $n$ with "only" $43$ digits.

Heuristically, based on the prime number theorem, one would expect counterexamples to start showing up as early as for $n$ in the mid-thousands. The "probability" that a random $n$ works is roughly  $(1-\frac{1}{\ln(100n)})^{100}$, and at $n=4000$ this is more than $1/4000$.
An exhaustive search, however, shows that the smallest counterexample is n=16,718. That's not too far from the above heuristic estimate, considering how crude it is.
A: As a generalization of some of the other answers, there are arbitrarily long sequences of composites: Namely, consider the numbers $N! + 2, N! + 3, N! + 4, ..., N!+N$ which has length $N - 1$. 
