How many prime numbers are there in between $1000!+1$ and $1000!+1000$, inclusive? I know $1000!$ is not a prime number as any number $1000$ or less is a divisor, but how would I know if $1000!+1$ is prime? Any hints?
Also, use the above question to prove that you can find $n$ consecutive composite numbers.
 A: Hint: $k$ divides $1000!+k$, for every natural $k\le 1000$.
A: All the integers from $1000!+2$ up to $1000!+1000$ are clearly not prime, and a simple check verifies that $6563$ is a factor of $1000!+1$, so there are no primes in the list.
A: The only non-trivial case is $n=1000!+1$.
However, you can easily check with a computer that $2^{n-1} \not \equiv 1 \pmod n$, thus it's not a prime number (it's just an instance of Fermat primality test). If you want to try this yourself, use an efficient modular exponentiation method.
You may also have a look at FactorDB, which will give you a partial factorization:
$$1000! + 1 = 6563 \cdot 1190737 \cdot 115205557790605547 \cdot C_{2541}$$
where $C_{2541}$ is a composite number with 2541 decimal digits.
A: I cannot give a proof, but the OEIS entry on factorial primes claims that  $1000!+1$ is not prime, as 1000 isn't listed. This is likely a proof-by-computer, though.
As for the rest of the numbers in that range, everyone else has already mentioned that $k\mid n!+k$ for $1\leq k\leq n$.
A: You can use wolfram alpha to check. Not really elegant but effective :-)
http://www.wolframalpha.com/input/?i=Is+1000%21%2B1+prime%3F
