6. Find all such triples of distinct prime numbers $p, q, r$ and a natural number $k$, such that $p q - k$ is divisible by $r$, $q r - k$ is divisible by $p$, $r p - k$ is divisible by $q$, and the number $p q - k$ is positive.
Answer: $2,3,5$ and all permutations, $k=1$

Solution. Since $p q-k$ is divisible by $r$, the number $n=p q+q r+r p-k$ is also divisible by $r$. Note that $n \geqslant p q-k>0$. Similarly, the number $n$ is divisible by $p$ and by $q$. Since $p, q$ and $r$ are different prime numbers, they are coprime, and thus $n$ is divisible by $p q r$. Therefore, $n \geqslant p q r$. Hence, $p q+q r+r p-k \geqslant p q r$ or, equivalently, $p q r\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-1\right)=$ $p q+q r+r p-p q r \geqslant k>0$. Therefore, $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-1>0$. We need to find all triples of distinct prime numbers, the sum of whose reciprocals is greater than 1. Let's assume for definiteness that $p<q<r$. If $p \neq 2$, then $p \geqslant 3, q \geqslant 3$ and $r \geqslant 3$, and such numbers are clearly not suitable. Therefore, $p=2$. If $q \neq 3$, then $r \geqslant q \geqslant 5$ and such numbers are also not suitable. Therefore, $q=3$. If $r \neq 5$, then $r \geqslant 7$ and this triple is also not suitable. Thus, $p=2, q=3$ and $r=5$. Then $n=2 \cdot 3+3 \cdot 5+5 \cdot 2-k=31-k$ and we need to find such $k$ for which $31-k$ is divisible by $2 \cdot 3 \cdot 5=30$, so $k=1$.