Austria2019

The four prime numbers have to fulfill $p>5$ and $s<p+10$ and hence they must be among the five consecutive odd numbers $p$, $p+2$, $p+4$, $p+6$ and $p+8$. As we have to choose 4 out of the five numbers $p$, $p+2$, $p+4$, $p+6$, $p+8$, we have to omit exactly one of these numbers. If we omit one of the numbers $p$, $p+2$, $p+6$ or $p+8$, three subsequent odd numbers remain, one of which has to be divisible by 3, which is excluded. Therefore, we have to omit $p+4$. Hence the four prime numbers have to be $p$, $q=p+2$, $r=p+6$ and $s=p+8$. Exactly one of the five consecutive integers $p$, $p+2$, $p+4$, $p+6$ and $p+8$ is divisible by 5. By construction, none of the chosen number $p$, $q$, $r$, $s$ can be divisible by 5. This implies that $p+4$ is divisible by 5. So $p+4$ is divisible by 15. The fact that $$p+q+r+s=p+(p+2)+(p+6)+(p+8)=4p+16=4(p+4)$$ yields that the sum is divisible by 60.