Problem 4. Five numbers $2,0,1,9,0$ are written in a circle on the board in the given order clockwise (the last zero is written next to the first two). In one move, the sum of each pair of adjacent numbers is written between them. For example, such an arrangement of numbers (on the right) will be after the first move:

![](https://cdn.mathpix.com/cropped/2024_05_06_edfa979e23ca6aeb44cfg-3.jpg?height=226&width=511&top_left_y=93&top_left_x=447)

After 5 moves, Polina calculated the sum of all numbers from the first zero (the one initially between 2 and 1) to the second zero (the one initially between 9 and 2) when moving around the circle clockwise, while Alina calculated the sum of all other numbers. What is the difference between Alina's and Polina's numbers?
Answer: $8 \cdot 3^{5}=1944$.

Solution. Let's see how the numbers between the first and second zero (which Polina will count at the end) change. On each move, each "old" number is included as an addend in two "new" numbers. This means that if all the "new" numbers are added together, each "old" number will be summed twice, so the sum of the "new" numbers is twice the sum of the "old" numbers. Therefore, the total sum of the numbers between the first and second zero triples on each move.

The same thing happens with the other numbers. It remains to note that their difference is initially 8, and it triples with each move. After five moves, it will reach the value $8 \cdot 3^{5}=1944$.

## Criteria

## The largest suitable criterion is used:

7 p. Any complete solution to the problem.

## 4 p. It is proven that the sum of the numbers on the segment between two zeros triples on each move.

## 3 p. The correct answer is given.