2. The numbers 12 and 60 have an interesting property: their product is equal to their tenfold sum. Determine all such pairs of natural numbers.
Solution. Let $x$ and $y$ be the required numbers. Then $x y=10(x+y)$, from which we sequentially obtain

$$
\begin{aligned}
& x y-10 x-10 y=0 \\
& x y-10 x-10 y+100=100 \\
& (x-10)(y-10)=100
\end{aligned}
$$

But, $100=1 \cdot 100=2 \cdot 50=4 \cdot 25=5 \cdot 20=10 \cdot 10$, so from the last equation, it is easy to see that the required numbers are: 11 and 110, or 12 and 60, or 14 and 35, or 15 and 30, or 20 and 20.