# 9.1. (7 points)

Two different natural numbers are subjected to four operations:

a) their sum is found;

b) the smaller is subtracted from the larger;

c) their product is found;

d) the larger number is divided by the smaller.

The sum of the results of all four operations is equal to $3^{5}$. What are these numbers?
Answer: 54 and $2 ; 24$ and 8.

Solution: Let $x$ and $y$ be the required natural numbers, $x>y$ :

$$
\begin{gathered}
x+y+x-y+x y+\frac{x}{y}=3^{5} \\
2 x+x y+\frac{x}{y}=3^{5} \\
2 x y+x y^{2}+x=3^{5} y \\
x\left(y^{2}+2 y+1\right)=3^{5} y \\
x(y+1)^{2}=3^{5} y \\
x=\frac{3^{5} y}{(y+1)^{2}}
\end{gathered}
$$

Since $x$ is a natural number, $3^{5}$ must be divisible by $(y+1)^{2}$. Therefore, $(y+1)^{2}=3^{2}$ or $(y+1)^{2}=3^{4}$. From the first equation, $y=2$, and from the second, $y=8$. We obtain two pairs of values for $x$ and $y: 54$ and $2 ; 24$ and 8.