A correspondence mathematical competition takes place in three rounds, whose difficulty increases. Only those solvers who were successful in the first round advance to the second round, and only successful solvers of the second round advance to the third round. The winner is anyone who is a successful solver of the last, i.e., the third round. In the last year of this competition, exactly $14 \%$ of the solvers were successful in the first round, exactly $25 \%$ of the solvers of the second round advanced to the third round, and exactly $8 \%$ of the solvers of the third round won.

What is the smallest number of participants who could have taken part in the first round? How many winners would there be in such a case?

(M. Petrová)

Idea. All intermediate results must be natural numbers.
Let's denote the number of all solvers in the first round as $x$. The number of successful solvers in the first round (and thus the number of all solvers in the second round) is $14\%$ of $x$, which is $0.14x$. The number of successful solvers in the second round (and thus the number of all solvers in the third round) is $25\%$ of $0.14x$, i.e., $0.25 \cdot 0.14x = 0.035x$. The number of successful solvers in the third round (and thus the number of winners) is $8\%$ of $0.035x$, i.e., $0.08 \cdot 0.035x = 0.0028x$.

Since all calculations are exact (without rounding), the numbers $x$, $0.14x$, $0.035x$, and $0.0028x$ must be natural numbers. Let's start with the last one:

$$
0.0028x = \frac{28}{10000}x = \frac{7}{2500}x
$$

Thus, the number $x$ must be a multiple of 2500. Since we are looking for the smallest solution, we will try multiples of 2500 until all mentioned numbers are natural:

| $x$ | $0.14x$ | $0.035x$ | $0.0028x$ | conclusion |
| :---: | :---: | :---: | :---: | :---: |
| 2500 | 350 | 87.5 | 7 | does not satisfy |
| 5000 | 700 | 175 | 14 | satisfies |

The smallest number of participants who could have participated in the first round is 5000. In this case, the number of winners would be 14.

Alternative solution. Let's denote the number of all solvers in the first round as $x$. The number of successful solvers in the first round (and thus the number of all solvers in the second round) is $14\%$ of $x$, which is

$$
\frac{14}{100}x = \frac{7}{50}x
$$

The number of successful solvers in the second round (and thus the number of all solvers in the third round) is $25\%$ of the previous number, i.e.,

$$
\frac{25}{100} \cdot \frac{7}{50}x = \frac{7}{200}x
$$

The number of successful solvers in the third round (and thus the number of winners) is $8\%$ of the previous number, i.e.,

$$
\frac{8}{100} \cdot \frac{7}{200}x = \frac{7}{2500}x
$$

All the above expressions must be natural numbers, so the number $x$ must be a common multiple of the numbers 50, 200, and 2500. Since we are interested in the smallest possible number of participants in the first round of the competition, we are looking for the least common multiple of the mentioned numbers, which is 5000.

The smallest number of participants in the first round is therefore 5000, and the number of winners in this case would be

$$
\frac{7}{2500} \cdot 5000 = 14
$$