Pat and Mat each had their favorite natural number, but each had a different one. They added both numbers, subtracted (the smaller from the larger), multiplied, and divided (the larger by the smaller). When they added the results obtained this way, it came to 98.

Which favorite numbers did Pat and Mat have?

(L. Hozová)
If the larger of the two numbers is denoted by $x$ and the smaller by $y$, then the condition from the problem statement is

$$
(x+y)+(x-y)+x y+\frac{x}{y}=98 .
$$

Since the sum, difference, and product of the numbers $x$ and $y$ are natural numbers, and the result is also a natural number, $x$ must be a multiple of $y$, i.e., $x=k y$ for some natural number $k$. Substituting this into the previous equation and further simplifying, we get:

$$
\begin{aligned}
2 k y+k y^{2}+k & =98, \\
k\left(y^{2}+2 y+1\right) & =98, \\
k \cdot(y+1)^{2} & =2 \cdot 7^{2} .
\end{aligned}
$$

From this, it follows that $k=2$ and $y=6$, so $x=12$. (The case $98 \cdot 1^{2}$ has been excluded because $y$ is a natural number, so $y+1>1$.) Pata's and Mat's favorite numbers were 6 and 12.

Evaluation. 1 point for setting up the initial equation; 3 points for justifying $x=k y$, substituting into the equation, and subsequent simplifications; 2 points for the result.

Note. The final simplification (completing the square on the left side and factoring the right side) significantly simplifies the argument. Solutions without this idea require testing possibilities derived from the divisors, or factorizations of the number 98. Evaluate such procedures based on the quality of the accompanying commentary.