Primes number $n,n+2,n+6,n+8,n+12,n+14$ 
Find all natural number $n$ such that all the following numbers are primes : 
  $$n,\;\; n+2,\;\;n+6,\;\;n+8,\;\;n+12,\;\;n+14$$
  are all prime numbers

 A: As $n$ is a prime, $n\ge 2$.
When $n=2$, $n+2=4$ is not prime.
When $n=3$, $n+6=9$ is not prime.
When $n=5$, all of them are prime.
$(n+6)-n\equiv 1$ (mod $5$)
$(n+2)-n\equiv 2$ (mod $5$)
$(n+8)-n\equiv 3$ (mod $5$)
$(n+14)-n\equiv 4$ (mod $5$)
At least one of these integers is a multiple of $5$. 
When $n>5$, at least one of the numbers is not prime.
The only possible answer is $5$.
A: Hint: $$n\equiv_5 n+0$$
$$n+2\equiv_5 n+2$$
$$n+6\equiv_5 n+1$$
$$n+8\equiv_5 n+3$$
$$n+14\equiv_5 n+4$$
So among $n,n+2,n+6,n+8, n+14$ exactly one is divisible by $5$ so one of them is $5$... 
A: Note that 2,6,8,14 are all different mod 5. Furthermore, for each $x=2,6,8,14$, note that if $n+x \equiv_5 0$ then $n+x$ is a proper multiple of 5 i.e., $5|n+x$ but $5 \not = n+x$ so if $5|n+x$ then $n+x$ isn't prime. Thus the only way that $n+x$ can be prime for all  $x \in \{2,6,8,14\}$ is if $n \equiv_5 0$. But the only way $n \equiv_5 0$ and $n$ still also be prime is if $n$ is 5 itself. So $n=5$ is the only possibility.
If you allow negative numbers for $n$ then $n=-19$ will work, and [if you consider $\pm 1$ to be prime] so will $n=-1,-7,-13$. But by the same reasoning as in the above pragraph $|n| < 14+5+1$ [as at least $x \in \{0,2,6,8,14\}$ one must satisfy $|n+x| \le 5$] so those are it.
