Shows that there are 1000 consecutive positive integers containing exactly 10 prime numbers As the title says, how can i demonstrate that?
I know how to make a sequence of n consecutive positive integers that are composed.
Let be k the length of the sequence, and consider the following consecutive numbers:
$(k+1)!+2,\ (k+1)!+3,\ ...\ ,\ (k+1)!+(k+1)$ then because $(k+1)!$ is multiple of $2,\ 3,\ ...\ ,\ (k+1)$ all the numbers are compound, can I use it in reverse to generate a sequence with 1000 compound numbers with 10 primes in there?

Hint: Let $f(n)$ be the number of primes in $n,n+1,...,n+999.$ So $f(n)=\pi(n+999)-\pi(n-1).$
What can you say about $f(n+1)-f(n)$?
Now, $f(1)>10.$ What is $f(1001!+2)$?
