I was given two two-digit natural numbers. Then I rounded both of them to the tens. Determine which numbers I had, given that at the same time:

- the difference of the rounded numbers is the same as the difference of the original numbers,
- the product of the rounded numbers is 184 greater than the product of the original numbers.

(L. Simünek)
Natural numbers are rounded to the tens by adding a suitable integer from -4 to 5 to them. If the difference between the original and rounded numbers is the same, it means that we added the same number to both original numbers during rounding, i.e., the original numbers have the same digit in the units place.

The product of the rounded numbers has the digit 0 in the units place. The product of the original numbers is 184 less according to the problem, so it has the digit 6 in the units place. The value of this digit is influenced only by the digit in the units place of the sought numbers. Therefore, the sought numbers could have the digit $4 (4 \cdot 4 = 16)$ or the digit $6 (6 \cdot 6 = 36)$ in the units place. However, from the second condition in the problem, it is clear that the sought numbers were rounded up. They must therefore end in the digit 6.

If we denote the sought numbers as $p$ and $q$, then according to the second condition in the problem, we can set up the equation

$$
(p+4) \cdot (q+4) = pq + 184
$$

which we can rearrange as

$$
\begin{aligned}
pq + 4p + 4q + 16 & = pq + 184 \\
4(p + q) & = 168 \\
p + q & = 42.
\end{aligned}
$$

The only two-digit numbers ending in the digit 6 and satisfying this equation are 16 and 26.

Another solution. Using the same method as above, we determine that both sought numbers have the digit 6 in the units place. The sought numbers, which are two-digit according to the problem, can be written as $10a + 6$ and $10b + 6$, where $a$ and $b$ represent the digits in the tens place. The numbers have the value $10a + 6 + 4 = 10(a + 1)$ and $10b + 6 + 4 = 10(b + 1)$ after rounding. The equation according to the second condition in the problem is then

$$
(10a + 6) \cdot (10b + 6) + 184 = 10(a + 1) \cdot 10(b + 1)
$$

and after rearranging, we get $a + b = 3$. The unknowns $a$ and $b$ are the digits in the tens place of the two sought numbers. The digit 0 in the tens place is not allowed by the problem, as the numbers must be two-digit. The sum 3 can only be achieved by the digits 1 and 2, so the sought numbers are 16 and 26.

Evaluation. 1 point for recognizing that the sought numbers end in the same digit; 2 points for justifying that this digit is $6; 1$ point for setting up the equation; 2 points for the correct conclusion.