7
$\begingroup$

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?

$\endgroup$
  • 9
    $\begingroup$ First step: see if you can tell whether $13!+2$ is prime. Second step: see if you can tell whether $13!+3$ is prime. $\endgroup$ – Erick Wong Dec 4 '16 at 18:19
  • 1
    $\begingroup$ In this small of a range instead of using probable prime formulae, you can manually use primality tests. $\endgroup$ – Eli Sadoff Dec 4 '16 at 18:21
11
$\begingroup$

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 $n$
$\endgroup$
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $13$.

$\endgroup$
  • 4
    $\begingroup$ While in base 10 your observations over numbers multiple of 2 and 5 are correct, you should review your statements about multiples of 3 and 7. $\endgroup$ – Lorenzo Dec 4 '16 at 23:40
  • $\begingroup$ I was going to edit it, cos I knew it might be confused. $6227020803$ is divisible by $3$ - this what I was meant and so on. Perhaps I shouldn't use plural form and write a whole number. To find out if a number is divisible by $7$, 'take the last digit, double it, and subtract it from the rest of the number' should be added as well. $\endgroup$ – usiro Dec 5 '16 at 0:01
  • $\begingroup$ More interesting thing is why I haven't included 6227020810 as divisible by $2$ rather then by $5$. There is an important reason for that, but not really important for the question. $\endgroup$ – usiro Dec 5 '16 at 0:17
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.