1. Find all prime numbers $p, q, r$ and $s$ such that

$$
6 p+7 p q+8 p q r+9 p q r s=2018
$$
1. Factoring the number 2018 into prime factors: $2018=2 \cdot 1009$. Since the expression on the left side of the given equation is equal to $p(6+7 q+8 q r+9 q r s)$, and since $p$ is a prime number, it is possible that $p=2$ or $p=1009$. The second case is invalid, because then the expression in parentheses would have to be equal to 2, which is clearly greater than 2. Therefore, we are left with $p=2$. Then we have $6+7 q+8 q r+9 q r s=1009$, which simplifies to $q(7+8 r+9 r s)=1003=17 \cdot 59$. From this, it follows that $q=17$ (it is impossible for $q=59$ because the expression in parentheses is greater than 17). Further, we get $7+8 r+9 r s=59$, which simplifies to $r(8+9 s)=52=2^{2} \cdot 13$. From this, it follows that $r=2$ (it is impossible for $r=13$ because the expression in parentheses is greater than 4). Finally, we are left with $8+9 s=26$, which simplifies to $9 s=18$, so we get $s=2$.

Therefore, the only solution is: $(p, q, r, s)=(2,17,2,2)$.