Primes in consecutive numbers

The answer is $\boxed{yes}$ .

For each positive integer $n$ let $f(n)$ be the number of prime numbers among $n, n+1, \dots, n+2014$ . I will show that $f(k) = 15$ for some positive integer number $k$ .

First of all we note that

• $f(1) ≥ 15$ since $\text{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}$ are prime numbers.
• $f(2016!+2) = 0$ since for each $2 ≤ l ≤ 2016$ the number $2016!+l$ is not a prime number.

Now note that by the definition for each positive $n$ the difference $f(n + 1) - f(n)$ is equal to $\text{0, -1 or 1}$ . In other words, while $n$ increases by $1$ , $f(n)$ can change only by $1$ .

Thus, when n changes from $1$ to $2016! + 2$ , $f(n)$ smoothly (at most by $1$ ) changes from some number exceeding $15$ to $0$ . Therefore, for some integer $1 < k < 2016! + 2$ we have $f(k) = 15$ .