Prime Between $n$ and $n!$

This statement is True . This can be solved two different ways:

a) Consider a number $x$ between $1$ and $n+1$ . This number must divide $n!$ , since $n!$ is the product of all numbers between $1$ and $n+1$ . However, this number $x$ cannot divide $n! - 1$ since there will be a remainder left over when dividing by any number between $1$ and $n+1$ . This can either mean that $n! - 1$ is prime, or there is some number, which is prime, that is greater than or equal to $n+1$ that divides $n! - 1$ . In either case, we have found a prime $p$ and showed that this statement is true!

b) One can use Bertrand's Theorem , which states that for $n>1$ , there is always a prime $p$ between $n$ and $2n$ . This can be remembered by the rhyme: "Bertrand said it and I'll say it again, there's always a prime between n and 2n." (other variants of this rhyme exist too). Notice that $n!=n(n-1)(n-2)\cdots(2)(1)$ , so $n! \geq 2n$ for $n \geq 3$ , therefore there is always at least one prime $p$ between $n$ and $n!$ .

Can you come up with a general forumla for the number of primes between n and n !

For example:

Between 2 and 2 ! there are no primes.

Between 3 and 3 ! there is 1 prime number -> 5

Between 4 and 4 ! there are 5,7,11,13,17,19,23 -> 7 prime numbers...

So how many primes are between n and n! in terms of n ?

By Bertrand'e postulate, there is always a prime between n and 2n if n > 3. Ed Gray

Consider a number $x$

$x=n! -1$

Consider any prime divisor $p$ of $x$ .

Then $p$ does not divide $n!$

Hence $p > n$

Also $n!-1 \geq p$

Hence $n!>p>n$

Proved.....

Let $P(n)$ equal the greatest prime number less than or equal to $n$ and let $S_n = \{2, 3, 5, 7 \cdots P(n)\}$ . Note that since $S_n$ is essentially a set containing all the prime numbers less than or equal to $n$ , if we take the product of all the elements in $S_n$ and add $1$ it follows that the result is not divisible by any of the numbers in $S_n$ and therefore this result is either prime itself or is the product of other primes which are not contained in $S_n$ or in other words it is the product of primes greater than $n$ . Since for any $n > 3, \prod{S_n} + 1< n!$ it must mean that $n< p < n!$ , where $p$ is a prime, is $\boxed{\text{True}}$ .

Even though this reasoning does not work for $3$ we can easily work out that there is a prime number between $3$ and $3!$ .

The prime number theorem states that for sufficiently large $N$ , the distribution of prime numbers less than or equal to $N$ is approximately $\dfrac{N}{\ln{N}}$ . Since this is less than $N$ for all $N \geq 3$ , it is clear that the next prime number will be less than $2N$ . As such, for $N \geq 3$ , there is always at least $1$ prime number between $N$ and $N!$ .