Finding how many prime numbers lie in a given range

Question

How many prime numbers $p$ are there which satisfy this condition?

$$13! +1 \lt p \leq 13! +13$$

Which method should I use to solve this, or could you help with the first steps?

Answer 1

None, since $13!+n$ is divisible by $n$ for every $n\in[2,13]$.

This technique is also used for proving that there is no finite bound on the gap between two primes, since for every $n\in\mathbb{N}$, there is a consecutive sequence of (at least) $n-1$ numbers, none of which is prime:

• $n!+2$, which is divisible by $2$
• $n!+3$, which is divisible by $3$
• $\dots$
• $n!+n$, which is divisible by $n$

Answer 2

Notice that if $2 \le k \le 12$ then $k \mid 13!$ because $13! = 1 \cdot (k-1) \color{red} {\cdot k} \cdot (k+1) \cdot \dots 13$, therefore $k \mid 13! + k$ whenever $2 \le k \le 12$, so none of the numbers $13! + k$ with $2 \le k \le 12$ is prime.

Answer 3

I think this question has been answered well above. When it comes about numbers:

$13! = 6227020800$

So the range you are looking for is: $6227020802 - 6227020813$

Numbers ending up $ ...802, ...804, ...806, ...808, ...812$ are divisible by $2$.

Numbers ending up $ ...803, ...809$ are divisible by $3$.

$ ...805, ...810$ are divisible by $5$.

$...807$ are divisible by $7$.

$...811$ and $...813$ are good candidates for prime numbers. You could check if they are semiprimes or composite numbers. Unfortunately $...811$ is divisible by $11$ and $...813$ is divisible by $13$.

Answer 4

What you have to remember is that $n!$ is divisible by each prime number $p \leq n$. Then $n! + p$ is also divisible by $p$.

In the specific case of $n = 13$, it follows that $13!$ is divisible by $2, 3, 5, 7, 11, 13$, and consequently, $13! + 2$ is divisible by $2$, $13! + 3$ is divisible by $3$, you get the idea.

So yeah, no primes there.