Prove that there exist a block of $2002$ consecutive natural numbers. Prove that there exist a block of $2002$ consecutive natural numbers such that exactly $150$ of them are primes.(You may use the fact that there are $168$ Primes less then $1000$.
This is a problem from an high school olympiad so no calculators or computers were allowed. 
This seems to be obvious but I don't know how to prove it. I don't undersand where to use the given fact. 
 A: As you indicate there are more than $150$ primes in the first block of $2002$ numbers.  Also, there is a block of $2002$ numbers with no primes in it, starting at $2003! + 2$.  Every time we move the block forward one step, the number of primes in it changes by $0$ or $\pm 1$.  Since the count starts above $150$ and eventually goes to $0$, and it can't change by more than $1$ in a single step, there must be a block containing exactly $150$ primes.
A: I tried to find the formula with which we can calculate the numbers between which the 150 primes are located. Didn't finish it though. (I hope that I didn't make any big mistakes :-D).
Let's say the primes are between $k$ and $k+l$. For our case $l=2002$. The $n$-th prime will be $m(n)$.
I'll be going by Erastothenes. With each step of Erastothenes $n$ you remove the integer division result between $k+l$ and $m(n)$ or $\left(\mathrm{Floor}\frac{k+l}{m(n)}\right)$ as non primes from the block $k+l$ numbers.
So if the block is of the length $k+l=2002$ you remove a specific number of non primes, you get the following number of possible primes for each step:
$$(2) - \mathrm{Floor}(2002\cdot\frac{1}{2})=1001 
\\ \quad 2002-1001=1001 \text{ possible primes remaining}\\
(3) - \mathrm{Floor}(1001\cdot\frac{1}{3})=333 \\
\quad 1001-333=668\text{ possible primes remaining} $$
We can simplify it:
$$(5) - \mathrm{Floor}(668\cdot\frac{4}{5})=534 \\
(7) - \mathrm{Floor}(534\cdot\frac{6}{7})=457 \text{ possible primes remaining}$$
We see that the number starts to fall of slower with increased $n$. We can simplify the expression and create a formula for finding the number of non-primes in the first 2002 numbers (lets assume that each step yields a integer, so we are not forced to write floor each time):
$$2002\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\dots$$
Now we need to find the numbers with a difference of 2002, between which we have 150 primes or 1852 non primes. We assume that $n_k$ and $n_{kl}=n_k+150$ are the indexes of the first primes smaller than $k$ and $k+l$ respectively. We assume integer divisions in each step (therefore the notation could be misleading).
$$(k+2002)\prod_{n=1}^{n_{kl}}\frac{m(n)-1}{m(n)}-k\prod_{n=1}^{n_k}\frac{m(n)-1}{m(n)}  =150$$
or
$$2002\prod_{n=1}^{n_k+150}\frac{m(n)-1}{m(n)}+k\prod_{n=n_k+1}^{n_k+150}\frac{m(n)-1}{m(n)}  =150
$$
I don't know where to go further, or if it makes any sense to do so:-).
