Have following approach, which either is inadequate or am unable to amend for this problem:

  1. By definition, if $n\le m$, then $n \,|\, m!$.
  2. Thus for all integers $2\le n\le (m + 1)$, $n \,|\, (m + 1)!$.
  3. For all integers, $n \,| \,n$ trivially.
  4. Also, if $n\, |\, a$ and $n\, | \,b\implies n\, | \,(a + b)$, with $a = n, b= (m+1)!$.
  5. Hence, have : $$2\,|\, 2 +(m+1)!$$ $$3\,|\, 3 +(m+1)!$$ $$\vdots$$ $$n+1\,|\, n +1+(m+1)!$$
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    $\begingroup$ Good idea, but you are using the letter $n$ for two different things. $\endgroup$ – Lord Shark the Unknown Jun 16 '18 at 16:49
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    $\begingroup$ And for the $1+(m+1)!$ don't forget the Wilson's theorem $\endgroup$ – rtybase Jun 16 '18 at 16:56
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    $\begingroup$ @jiten you need an $n=m!, m \geq 200$, then $m!+2,m!+3,...,m!+200$ are composite ... you need to be careful with $m!+1$, for that see my previous comment ... so you need a prime $p> 200$ and $n=(p-1)!$ $\endgroup$ – rtybase Jun 16 '18 at 17:02
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    $\begingroup$ jiten: "to prove that n|n+200 is not obvious still." Sorry but what? Where do you see that one would need to prove that n divides n+200? $\endgroup$ – Did Jun 16 '18 at 17:14
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    $\begingroup$ The stretch of (at least) 200 composite numbers is known as a prime gap. See the Wikipedia article for more information. Simple number theory considerations suffice to establish the existence of arbitrarily long prime gaps in the integers, as study of the Answers below will show. We don't need to go out as far as $200!+2$ to begin a prime gap of length 200 or more. With a bit of programming one finds that following prime $20831323$ there is a prime gap of length $209$. This gives something of an indication of what is known by simple reasoning versus what requires "tools" to work out. $\endgroup$ – hardmath Jun 17 '18 at 5:29

What you were doing is correct. What you need is a clever choice for $n$. Choose $n=202!+2$ and it'll work out.

  • $\begingroup$ Please elaborate, am not clear. $\endgroup$ – jiten Jun 16 '18 at 17:17
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    $\begingroup$ It is clear that $n \mid 202!+n$ for $2 \leq n \leq 202$. That's what you need. $\endgroup$ – SinTan1729 Jun 16 '18 at 17:19
  • $\begingroup$ But, then why need a 'clever choice'. It is not obvious to me - the significance of that. For me, it means : if $2\le n \le m+1\implies n\,|\, (m+1)!$. $\endgroup$ – jiten Jun 16 '18 at 17:24
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    $\begingroup$ Yeah. But that's exactly what's needed. You need at least $201$ such numbers. Hence $n>201$. Anything above $202$ works too. 'Clever' because it doesn't involve any checking, just simple algebraic trickery. $\endgroup$ – SinTan1729 Jun 16 '18 at 17:53


$k_1 = 2+(n)!, k_2 =3+(n)!,....$

$...k_{n-1} = (n) +(n)!$.

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

  • $\begingroup$ If the question is good, then please up-vote to prevent from deletion by system ( as I feel negative vote would cause automatically, after some fixed interval). $\endgroup$ – jiten Jun 19 '18 at 4:40
  • $\begingroup$ jiten +1.Good question .Good effort.Keep it up. $\endgroup$ – Peter Szilas Jun 19 '18 at 5:38

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