Problem 11.5. For what least natural $n$ can the numbers from 1 to $n$ be arranged in a circle so that each number is either greater than all 40 following it in the clockwise direction, or less than all 30 following it in the clockwise direction?
Answer: 70.

Solution. If $n \leqslant 39$, then for the number $n$ the condition cannot be satisfied: it cannot be greater than the next 40 numbers (since it is not greater than itself), nor can it be less than the next 30 numbers (since it is the largest).

If $40 \leqslant n \leqslant 69$, then for the number 40 the condition cannot be satisfied: there are neither 40 numbers less than it, nor 30 numbers greater than it.

However, if $n=70$, the numbers can be arranged. For this, they can be placed in the following order clockwise: $1,2,3, \ldots, 40,70,69,68, \ldots, 41$ (Fig. 15). Then the numbers from 1 to 40 will be less than the next 30 following them, and the numbers from 70 to 41 will be greater than the next 40 following them.

![](https://cdn.mathpix.com/cropped/2024_05_06_28b5ece3789ad9441c8ag-3.jpg?height=498&width=495&top_left_y=615&top_left_x=479)

Fig. 15: to the solution of problem 11.5