Reading about quadruples of primes, it is clear that these must have the form $(p,p+2,p+6,p+8)$, ($p>5$).
Consider the first three quadruples: $\{11, 13, 17, 19\}, \{101, 103, 107, 109\}, \{191, 193, 197, 199\}$ Now let's add each element of each quadruple: $11+13+17+19=60$, for the second quadruple $101+103+107+109=420$ and for tje third quadruple $191+193+197+199=780$.
Note that:
$60|60\Rightarrow 60*1=60$
$60|420\Rightarrow 60*7=420$
$60|780\Rightarrow 60*13=780$
This property seems to hold forever, where does this property come from? Does the sequence of numbers $1,7,13,...$ have any relationship?
Of the $5$ numbers $$p,\,p+2,\,p+4,\,p+6,\,p+8$$ one must be a multiple of $3$ and one must be a multiple of $5$.
It follows that $3{\,\mid\,}(p+4)$ and $5{\,\mid\,}(p+4)$, so $15{\,\mid\,}(p+4)$, hence $$ p+(p+2)+(p+6)+(p+8)=4p+16=4(p+4) $$ is a multiple of $4{\,\cdot\,}15=60$.