I was given two natural numbers. Then I rounded both of them to the tens. Determine which numbers I had, if you know that:

- the ratio of the rounded numbers is the same as the ratio of the original numbers,
- the product of the rounded numbers is 295 greater than the product of the original numbers,
- the sum of the rounded numbers is 6 greater than the sum of the original numbers.

(L. Simünek)
A natural number is rounded to the tens by adding a suitable integer from -4 to 5. According to the third condition in the problem, the sum of two rounded numbers is 6 greater than the sum of the original numbers. Therefore, both original numbers increase during rounding and must satisfy one of the following assumptions:

a) One ends in the digit 5, the other in the digit 9. (If we denote them as $p, q$, rounding to the tens gives $p+5, q+1$.)

b) One ends in the digit 6, the other in the digit 8. (If we denote them as $r, s$, rounding to the tens gives $r+4, s+2$.)

c) One ends in the digit 7, the other in the digit 7. (If we denote them as $t$, $u$, rounding to the tens gives $t+3, u+3$.)

According to the second condition in the problem, the product of the rounded numbers is 295 greater than the product of the original numbers. Since the product of the rounded numbers must end in the digit 0, it follows from this condition that the product of the original numbers ends in the digit 5. The product of the numbers according to assumption a) indeed ends in the digit 5, because $5 \cdot 9=45$. However, the product of the numbers according to assumption b) ends in the digit 8, because $6 \cdot 8=48$, and the product of the numbers according to assumption c) ends in the digit 9, because $7 \cdot 7=49$. We will therefore proceed only with assumption a), as it is the only one that can lead to the correct result. According to the first and second conditions in the problem, we arrive at a system of two equations:

$$
\begin{aligned}
& \frac{p+5}{q+1}=\frac{p}{q} \\
& (p+5)(q+1)=p q+295
\end{aligned}
$$

By manipulating the first equation, we obtain the relationship $p=5 q$. Using this relationship, we solve the second equation:

$$
\begin{aligned}
p q+p+5 q+5 & =p q+295 \\
p+5 q & =290 \\
5 q+5 q & =290 \\
q & =29
\end{aligned}
$$

By substitution:

$$
p=5 \cdot 29=145 .
$$

The resulting numbers $p$ and $q$ have the digits in the units place as we determined in our assumption. The problem has a unique solution: the sought numbers are 29 and 145.

Another solution. By the same method, we arrive at the establishment of assumptions a), b), c). Instead of discussing the digit in the units place of the product of the original numbers, we set up a separate equation for each assumption according to the second condition in the problem:
a) $(p+5)(q+1)=p q+295$, after rearrangement $p+5 q+5=295$,
b) $(r+4)(s+2)=r s+295$, after rearrangement $2 r+4 s+8=295$,
c) $(t+3)(u+3)=t u+295$, after rearrangement $3 t+3 u+9=295$.

We see that the rearranged equation b) cannot have a solution in the set of integers; substituting any integers on the left side results in an even sum, which contradicts the number 295 on the right. Similarly, the rearranged equation c) cannot have a solution in the set of integers, because substituting any integers on the left side of the equation results in a sum divisible by three, which contradicts the number 295 on the right. Only equation a) can have a solution in the set of integers, and we arrive at it using the relationship $p=5 q$ as described above.

Yet another solution. As in the previous solutions, we first conclude that one of the original numbers has the digit 9 in the units place, and the other has the digit 5. The first can be written as $10 a+9$, and after rounding to the tens, we get $10 a+9+1$, i.e., $10(a+1)$. The second can be written as $10 b+5$, and after rounding to the tens, we get $10 b+5+5$, i.e., $10(b+1)$. The numbers $a, b$ are natural numbers or zero (the problem does not state that no number can be a single-digit number). According to the second condition in the problem, we have:

$$
\begin{aligned}
10(a+1) \cdot 10(b+1) & =(10 a+9) \cdot(10 b+5)+295, \\
100 a b+100 a+100 b+100 & =100 a b+90 b+50 a+45+295, \\
50 a+10 b & =240, \\
5 a+b & =24 .
\end{aligned}
$$

We list all possible solutions in a table:

| $a$ | 0 | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $b$ | 24 | 19 | 14 | 9 | 4 |

The corresponding pairs of numbers are

| $10 a+9$ | 9 | 19 | 29 | 39 | 49 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $10 b+5$ | 245 | 195 | 145 | 95 | 45 |

We verify for each of these pairs whether the first condition is satisfied:

- $\frac{9}{245} \neq \frac{10}{250}$, so the pair 9 and 245 is not a solution,
- $\frac{19}{195} \neq \frac{20}{200}$, so the pair 19 and 195 is not a solution,
- $\frac{29}{145}=\frac{30}{150}=\frac{1}{5}$, so the pair 29 and 145 is a solution,
- $\frac{39}{95} \neq \frac{40}{100}$, so the pair 39 and 95 is not a solution,
- $\frac{49}{45} \neq \frac{50}{50}$, so the pair 49 and 45 is not a solution.

Only the pair 29 and 145 satisfies all the given conditions.