$p$ is a $3$-digit prime, then there always exist $p$ consecutive composite numbers. True/False
If $p$ is a $3$-digit prime, then there always exist $p$ consecutive composite numbers.

How to approach this?
$P_k =(s + 1)! + k$ for $k = 2$ to $(s + 1)$

are $s$ consecutive positive integers

as $P_k$ is always divisible by $k$, hence the statement is true
The statement is false!
The first 3 digit prime is 101, the next prime is 103; difference is 2 not 101!
Take another 3 digit prime, 131 the next prime is 137; difference is 6 not 131!
You can look up any of the 3 digit primes and by inspection the difference is well under 100.
  Ron