A number theory problem by Ilham Saiful Fauzi

The six numbers $\pmod{5}$ are:

$n+1,n+3,n+2,n+4,n+3,n$

As these give five different residues $\pmod{5}$ it follows one of them is divisible by $5$ as they are prime, one of the equals $5$ .

Using $n+1 \geq 2 \Rightarrow n \geq 1$ and $n+1 \leq 5 \Rightarrow n \leq 4$ gives us $n=1,2,3,4$ to test:

$n=1 \Rightarrow 2,4,8 \cdots \quad \quad \color{#3D99F6}{2 \vert 4} \\ n=2 \Rightarrow 3,5,9, \cdots \quad \quad \color{#3D99F6}{3 \vert 9} \\ n=3 \Rightarrow 4,6, \cdots \quad \quad \color{#3D99F6}{2 \vert 4} \\ n=4 \Rightarrow 5,7,11,13,17,19$

All these six numbers are prime so our only solution in $n=4$ giving the sum of possible values of n as:

$\color{#3D99F6}{\boxed{\boxed{4}}}$