Elementary number theory(number of prime numbers) Let $n= 2016!+1$. Then the number of primes among $n+1,n+2,\ldots, n+2015 $ will be? 
I am confused about how to begin solving this. Could anyone please teach me the concept I need to know to achieve this problem's answer. And also please let me know if there is any shortcut or logic behind these kind of problems. 
 A: Let $1\leq k\leq 2015$. Then, note that
$$
n+k=2016!+(k+1) 
$$
is divisible by $k+1$ since $k+1$ divides both $k+1$ and $2016!$, since $k+1\leq 2016$. 
A: The number $A = 1\times2\times3\times4\times5\times\cdots\times 2016$ is divisble by $1$ and by $2$ and by $3$ and by $4$ and by $\ldots\ldots\ldots$ and by $2016$.
Since $A$ is divisible by $7$, we conclude that $A+7$ is also divisible by $7$, and therefore is not prime. Similarly $A$ is divisible by $43$, and we conclude that $A+43$ is divisible by $43$ and is therefore not prime.
And similarly, for the same reason, every one of the numbers in your list, $A+2, A+2, A+3, \ldots, A+2016$ fails to be prime.
A: Added: I had not noticed that the initial number given was already $2016! + 1.$ Therefore, the bit about Wilson's Theorem does not come into play for the range of numbers specified. Cute, though.
Critical that $2017$ is prime. Wilson's Theorem says that $2016! + 1$ is divisible by $2017.$
However: if Ibrahimović were unable to do this problem, would he be playing for Manchester United right now? I don't think so.
