If $n$ is a natural number such that $ n \geq 2$, then the numbers $n! + 2, n! + 3, n! + 4... n! + n$ are all composite. (Thus, for any n greater than or equal to 2, one can find n consecutive composite numbers)
I started with just plugging in numbers to see if they were composite and they were. But I don't know how to prove it for all natural numbers.
For $ 2 \leq i \leq n, \ \ \ n! + i = n(n-1)(n-2) \ldots i \ldots (3)(2)(1) + i $
$ = i( \ (n)(n-1)\ldots (i+1)(1)(i-1)\ldots (3)(2)(1) + 1 \ )$
which is clearly composite
The statement you have to prove is that if $ 2 \leq k \leq n$ then $n! + k$ is composite.
Isn't $k$ a factor of $n!$?
Notice that if $k \le n $ then $k $ divides $n! $. So $k $ divides $n! +k $. So $n! +k $ is composite if $k >1$.
let $\overline n_k=n!/k = 1*2*.... (k-1)*(k+l)*...*n $. Then $n!+k =k (\overline n_k +1)$.
$n ! + k = k\cdot(n\cdot(n - 1)\cdot...\cdot (k + 1)\cdot(k - 1)\cdot... \cdot 2\cdot 1 + 1), 2\leq k \leq n.$
