2016/068) Show that there exists 2016 consecutive numbers that contains exactly 100 primes.

we know that number of primes less than 1000  $= 168$
now let f(x) be number of  primes in a sequence of 2016 primes starting at x.
$f(1) > 100$.
now when we move to next number the number of primes increases/decreases by 1 or remains unchanged
$f(2017!+2) = 0$ as 2016 numbers starting from this number  all are composite
so from 1 going upto 2017!+2 the starting number ( $>100$) remains unchanged or increases by 1 or decreases by 1
going to 0.
hence at some point it is 100.