1. Semyon has 20 numbers: $1,2,3, \ldots, 19,20$. He formed 10 fractions by writing ten of these numbers in some order as numerators, and the remaining ten in some order as denominators. What is the maximum number of integers Semyon could obtain after simplifying all the written fractions?
Answer: 8 numbers.

## Solution:

For the fractions to have an integer value, the prime numbers $11, 13, 17, 19$ can only be numerators with a denominator of 1. Therefore, to form fractions equal to an integer, no more than 17 numbers can be used, meaning no more than 8 fractions can be formed. Example: 20/10, 19/1, $18 / 9, 16 / 8, 15 / 5, 14 / 7, 12 / 6, 4 / 2$.

## Criteria:

Correct answer without a correct example of fractions - 1 point.

Correct answer with an example of 8 fractions - 3 points.

The idea that each of the numbers 11, 13, 17, and 19 can only be present in one reducible fraction - 2 points.

(If all the above is present, only a correct conclusion remains to be drawn - and to get a full 7 points.)