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If $p$ is a $3$-digit prime, then there always exist $p$ consecutive composite numbers.

How to approach this?

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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 – Robert Israel Jul 8 '12 at 6:38
up vote 4 down vote accepted

$P_k =(s + 1)! + k$ for $k = 2$ to $(s + 1)$

are $s$ consecutive positive integers

as $P_k$ is always divisible by $k$, hence the statement is true

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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 $\rm n =\,$ __

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The statement is false!
The first 3 digit prime is 101, the next prime is 103; difference is 2 not 101! Take another 3 digit prime, 131 the next prime is 137; difference is 6 not 131! You can look up any of the 3 digit primes and by inspection the difference is well under 100. Ron

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It seems to me that you have grossly misread the question. -1 – mixedmath Jul 8 '12 at 10:45

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