Problem 1517

1. We start by considering the factorial of a large number, specifically $2004!$. The factorial $2004!$ is the product of all positive integers up to $2004$. This means that $2004!$ is divisible by every integer from $1$ to $2004$.

2. Consider the sequence of numbers $2004! + 2, 2004! + 3, \ldots, 2004! + 2004$. Each of these numbers can be written as $2004! + k$ where $2 \leq k \leq 2004$. Since $2004!$ is divisible by $k$, $2004! + k$ is also divisible by $k$. Therefore, none of these numbers can be prime because they all have a divisor other than $1$ and themselves.

3. Next, we use the given fact that there are $168$ primes less than $1000$. This means that in the set of the first $1000$ positive integers, there are $168$ primes.

4. Define $p(n)$ as the number of primes in the set $\{n, n+1, \ldots, n+2001\}$. We need to show that there exists an $n$ such that $p(n) = 150$.

5. Note that the function $p(n)$ changes gradually as $n$ increases. Specifically, $|p(n) - p(n+1)| \leq 1$ because the set $\{n+1, n+2, \ldots, n+2002\}$ is almost the same as the set $\{n, n+1, \ldots, n+2001\}$, differing only by the inclusion of $n+2002$ and the exclusion of $n$.

6. We know that $p(1) \geq 150$ because there are $168$ primes less than $1000$, and the set $\{1, 2, \ldots, 2001\}$ includes all these primes.

7. We also know that $p(2004! + 2) < 150$ because the set $\{2004! + 2, 2004! + 3, \ldots, 2004! + 2003\}$ contains no primes, and the number of primes in the set $\{2004! + 2, 2004! + 3, \ldots, 2004! + 2001\}$ is less than $150$.

8. Since $p(n)$ changes by at most $1$ as $n$ increases and $p(1) \geq 150$ while $p(2004! + 2) < 150$, by the Intermediate Value Theorem, there must be some $n$ in the range from $1$ to $2004! + 2$ such that $p(n) = 150$.

Therefore, there exists a set of $2002$ consecutive positive integers containing exactly $150$ primes.

$\blacksquare$