8.2. Four different natural numbers are recorded. It turns out that the sum of their reciprocals is equal to 1. Can the number $2$ be absent among the recorded numbers?
Answer: cannot.

Solution. If at least one of the written numbers is 1, then the reciprocal of it is also 1, then the sum of the reciprocals of the numbers is greater than 1. Therefore, among the written numbers there is no 1. Suppose that among them there is no number 2. Consider the four smallest natural numbers from the remaining ones: $3, 4, 5$ and 6. The sum of the reciprocals of these numbers is $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=0.95<1$. Increasing any of these numbers leads to a decrease in the number that is its reciprocal, so in the absence of the number 2 among the written ones, the sum of the reciprocals will not be equal to 1.

The following sets of numbers satisfy the condition of the problem: (2; 3; 7; 42), (2; 3; 8; 24), (2; 3; 9; 18), (2; 3; $10 ; 15$), ( $2 ; 4 ; 5 ; 20),(2 ; 4 ; 6 ; 12)$. It is not required of students to provide examples of numbers that satisfy the condition.

Grading criteria.

"+" A complete and justified solution is provided

" $\pm$ " Correct overall reasoning is provided, but the case when one of the numbers is 1 is not considered

" $\pm$ " Correct overall reasoning is provided, the case when one of the numbers is 1 and the case 3, 4, 5, 6 are considered, but further reasoning and conclusions are missing

“Ғ” Only the correct answer and the inequality $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}<1$ are written without any explanations

“-" Only the answer is provided

“-" An incorrect solution is provided or it is missing