Maruška wrote two different natural numbers on the board. Marta took a card and on one side of it, she wrote the sum of Maruška's numbers, and on the other side, their product. One of the numbers on the card was a prime number, and the sum of the numbers on both sides of the card was 97.

Which numbers did Maruška write on the board?

(L. Hozová)
Let's denote the numbers on the board as $x$ and $y$. On one side of the card, the number $x+y$ was written, and on the other side, the number $x \cdot y$. The total satisfies the equation

$$
x+y+x y=97 \text{. }
$$

The left side of the equation corresponds to three terms in the expansion of the expression $(x+1)(y+1)$, with only 1 missing. By adding 1 to both sides of equation (1), we get

$$
(x+1)(y+1)=98 \text{. }
$$

Since $x$ and $y$ are natural numbers, the factors on the left side of this equation must be at least 2. The number 98 can be expressed as a product of two natural numbers greater than 1 in the following ways (ignoring the order of the factors):

$$
2 \cdot 49=98, \quad 7 \cdot 14=98
$$

The first option corresponds to the numbers $x=1$ and $y=48$, for which $x+y=49$ and $x y=48$. Neither of these two numbers is a prime number, so this option does not satisfy the problem's conditions.

The second option corresponds to the numbers $x=6$ and $y=13$, for which $x+y=19$ and $x y=78$. The number 19 is a prime number, and the number 78 is not, which satisfies the problem's conditions. Maruška wrote the numbers 6 and 13 on the board.

Evaluation. 1 point for expressing (1) and transforming (2); 2 points for the factorizations (3); 2 points for analyzing the options and concluding.

Alternative Solution. Using the same notation as above, we can express $y$ in terms of $x$ from equation (1):

$$
y=\frac{97-x}{1+x} .
$$

By substituting natural numbers for $x$, we can express $y$ and verify whether it is a natural number and whether one of the numbers $x+y$ and $x y$ is a prime number:

| $x$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $y$ | 48 | $\frac{95}{3}$ | $\frac{47}{2}$ | $\frac{93}{5}$ | $\frac{46}{3}$ | 13 | $\frac{45}{4}$ | $\frac{89}{9}$ | $\ldots$ |
| $x+y$ | 49 |  |  |  |  | $\mathbf{1 9}$ |  |  |  |
| $x y$ | 48 |  |  |  |  | 78 |  |  |  |

Among the given options, the only suitable pair of numbers is 6 and 13. Now, the other substitutions for $x$ up to 96 (to ensure $y$ is positive) should be checked. However, this is not necessary because as $x$ increases, the value of $y$ decreases and is certainly less than 9 (if both $x$ and $y$ were simultaneously greater than or equal to 9, then the value of $x+y+x y$ would be greater than or equal to $9+9+81=99$). Given the symmetry of the problem, all solutions must appear in the given table.

Maruška wrote the numbers 6 and 13 on the board.

Evaluation. 1 point for expressing (1) and (4); 2 points for substitution and verification as in the table; 2 points for the completeness of the discussion and conclusion.

Note. The fractional expression (4) can be expressed in the form "integer part plus remainder" as follows:

$$
y=\frac{97-x}{1+x}=\frac{-(1+x)+98}{1+x}=-1+\frac{98}{1+x} .
$$

Since $x$ and $y$ must be natural numbers, $1+x$ must be a positive proper divisor of the number 98, and such divisors are exactly four: $2, 7, 14$, and 49. From this, we get two options (regardless of the order of $x$ and $y$), one of which satisfies all the requirements. This method can be seen as another interpretation of equation (2) and the subsequent reasoning.