# $p$ 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?

• Hint: forget about "3-digit prime", just prove that for any positive integer $P$ there exist $P$ consecutive composite numbers. – Gerry Myerson Jul 8 '12 at 3:14
• Green-Tao theorem? – Hyperbola Jul 8 '12 at 3:16
• @Hyperbola: That theorem says that there are arbitrarily long sequences of primes in arithmetic progression. It is neither related to, nor necessary, to prove this. HINT: $(n+1)!$ is composite whenever $n\gt 1$. Can $(n+1)!+2$ be prime? What about $(n+1)!+3$? – Arturo Magidin Jul 8 '12 at 3:16
• @Hyperbola: Consider posting your solution as an answer. People can then check it, and after some time you can even accept it. It will also prevent this question from appearing as "unanswered". – Arturo Magidin Jul 8 '12 at 3:25
• A more difficult question is whether there exist $p$ consecutive composite numbers preceded and followed by primes. Without the "3 digit" restriction this is not known. With it, it is a matter for explicit computation (see trnicely.net/gaps/gaplist.html#MainTable) – Robert Israel Jul 8 '12 at 6:38

$P_k =(s + 1)! + k$ for $k = 2$ to $(s + 1)$
are $s$ consecutive positive integers
as $P_k$ is always divisible by $k$, hence the statement is true
Hint $\rm\ 1<a,b,c\:|\:n\:\Rightarrow\:n\!+\!a,\,n\!+\!b,\,n\!+\!c\,$ are composite. Now put $\rm\,a,b,c = 2,3,4\ldots$ and $\rm n =\,$ __