Prove that there is some positive integer $n$ such that $n, n +1, n+2,... , n + 200$ are all composite. Have following approach, which either is inadequate or am unable to amend for this problem:


*

*By definition, if $n\le m$, then $n \,|\, m!$. 

*Thus for all integers $2\le n\le (m + 1)$, $n \,|\, (m + 1)!$. 

*For all integers, $n \,| \,n$ trivially. 

*Also, if $n\, |\, a$ and $n\, | 
   \,b\implies n\, | \,(a + b)$, with $a = n, b= (m+1)!$. 

*Hence, have : $$2\,|\, 2 +(m+1)!$$  $$3\,|\, 3 +(m+1)!$$ $$\vdots$$  $$n+1\,|\, n +1+(m+1)!$$

 A: What you were doing is correct. What you need is a clever choice for $n$. Choose $n=202!+2$ and it'll work out.
A: Rephrasing:
$k_1 = 2+(n)!, k_2 =3+(n)!,....$
$...k_{n-1} = (n) +(n)!$.
All of the above numbers $k_i, i=1,2,...,(n-1)$ are composite. 
To have $201$ consecutive composite numbers choose 
$n= 202$.
This way you can have arbitrarily large gaps between consecutive primes.
Ref:Wiki: Prime gaps
A: Let $n=(200!)^{3} \in \mathbb{N}$ (absurd number, I know). Then
\begin{alignat}{3}
200 &| n &&= & & (200!)^{3} \\
  &      &&= & & 200(199!)(200!)^{2} \\
& && & & \\
m &| n+1 &&= & & (200!)^{3}+1 \\
  &      &&= & & (200!)^{3}-(-1)^{3} \\
  &      &&= & & [200!-(-1)][(200!)^{2}+(-1)^{1}(200!)^{1}+(-1)^{2}] \\
  &      &&= & & [200!+1][(200!)^{2}-200!+1] \qquad \qquad \qquad \qquad \qquad \text{ (and so let } m=200!+1\in \mathbb{N}) \\
& && & & \\
3 &| n+3 &&= & & (200!)^{3}+3 \\
  &      &&= & & 3[(1 \cdot 2 \cdot 4 \cdot … \cdot 200)(200!)^{2}+1] \\
&... && & & \\ 
199 &| n+199 &&= & & (200!)^{3}+199 \\
 &       &&= & & 199[(1 \cdot 2 \cdot … \cdot 198 \cdot 200)(200!)^{2} + 1] \\
& && & & \\
200 &| n+200 &&= & & (200!)^{3}+200 \\
 &       &&= & & 200[(199!)(200!)^{2} + 1]. \\
\end{alignat}
QED.
