## Task B-1.1.

Determine all pairs of natural numbers such that their eightfold greatest common divisor is greater by 6 than their least common multiple.
## Solution.

Let $x$ and $y$ be the sought natural numbers, $V$ their least common multiple, and $M$ their greatest common divisor.

It holds that $V+6=8 M \Rightarrow V=8 M-6$.

For the numbers $M$ and $V$, the equality $x \cdot y=M \cdot V$ always holds. Therefore,

$x \cdot y=M \cdot(8 M-6)$.

Let $n, m \in \mathbb{N}$ be such that $x=n \cdot M$ and $y=m \cdot M$.

$n \cdot m \cdot M^{2}=M \cdot(8 M-6)$.

$n \cdot m=8-\frac{6}{M}$.

Thus, it can be that $M \in\{1,2,3,6\}$.

For $M=1$, we get $V=2, n \cdot m=2, n=1, m=2, x=1, y=2$.

For $M=2$, we get $V=10, n \cdot m=5, n=1, m=5, x=2, y=10$.

For $M=3$, we get $V=18, n \cdot m=6, n=1, m=6, x=3, y=18$ and $n=2, m=3, x=6, y=9$.

For $M=6$, we get $V=42, n \cdot m=7, n=1, m=7, x=6, y=42$.

All pairs of the sought natural numbers are $\{1,2\},\{2,10\},\{3,18\},\{6,9\}$ and $\{6,42\}$.