2. Write down four numbers (not necessarily integers), none of which are the same, such that if a number \( x \) is among the written numbers, then at least one of the numbers \( x-1 \) or \( 6x-1 \) is also among the written numbers. (I. Rubanov, S. Berlov)
Answer. For example, $1 / 5, 6 / 5, 11 / 5, 16 / 5$. Solution. Let the numbers written be $x, x+1, x+2, x+3$. For the last three numbers, the condition of the problem is obviously satisfied. To satisfy it for the first one as well, we choose $x$ so that the equation $x=6 x-1$ holds. Solving the obtained equation, we find that $x=1 / 5$, from which we get the answer given above. Remark. The answer found is far from the only one. Three more sets of four numbers that satisfy the condition of the problem can be obtained by equating $6 x-1$ to the numbers $x+1, x+2$ or $x+3$. Other suitable sets can be found by replacing the sequence $x, x+1, x+2, x+3$ with another one where the condition of the problem is satisfied for all numbers except one for any $x$, for example, with $x, 6 x-1, 6 x-2, 6 x-3$.