Infinitude of primes in 10 consecutive integers Do there exist infinitely many sets of 10 consecutive positive integers where exactly one is a prime? 
By Dirichlet's Theorem, if $a$ and $d$ are relatively prime, then there infinitely many primes in the arithmetic sequence $a+d, 2a+d, 3a+d, \cdots$.
Let $n+1, n+2, n+3, \cdots, n+10$ be ten consecutive integers, then I want to construct 9 composite integers (say $n+1$ up to $n+9$) and a prime $n+10$, but then I have no idea how to proceed.
 A: A totally different proof: 
Assume there is an M > 0 such that there is no prime p > M with p+1, p+2, ..., p+9 all composite. There are two possibilities: 
Either there is no prime > M at all. That's false because there is an infinite number of primes. 
Or there is a prime p > M, then another prime p1 among p+1, ..., p+9, then another prime p2 among p1+1, ..., p1+9 and so on. Which means lim inf (pi (n) / n) ≥ 1/9, but we know that the limit is indeed 0. 
This also works for any length of the interval. 
A: The answer is yes.
By Dirichlet's theorem, there are infinitely many primes of the form $210k+1$ (this is a special case provable in elementary way). But then $210k+2,210k+4,210k+6,210k+8,210k+10$ are even, $210k+3,210k+9$ are divisible by $3$, $210k+5$ is divisible by $5$ and $210k+7$ is divisible by $7$. Hence $210k+1,...,210k+10$ form a block of 10 consecutive numbers, only one of which is prime.
Let me mention that this method can be extended to blocks of any integer length. The only property of $210$ which was used here is that it is divisible by all the primes up to $10$.
Another proof, based on djechlin's idea in a comment to different answer: for $n\geq 10$, let $p$ be the first prime greater than $n!+1$. Then the gap between $n!+1$ and $p$ contains only composite numbers, and there are at least 9 of them, since $2\mid n!+2,3\mid n!+3,...,10\mid n!+10$, so last 9 composites together with $p$ form a set like the one you asked for.
This gives infinitely many such sets, because we can choose $n$ arbitrarily large.
