Zukov G.

Given an infinitely increasing arithmetic progression. The first few of its terms were added and the sum was declared the first term of a new sequence, then the next few terms of the original progression were added and the sum was declared the second term of the new sequence, and so on. Could the new sequence turn out to be a geometric progression?
Example 1. $1,2+3+4,5+\ldots+13, \ldots, 1 / 2\left(3^{n}+1\right)+\ldots+1 / 2\left(3^{n+1}-1\right), \ldots=1,9,81, \ldots, 9^{n}, \ldots$

Example 2. $3,5+7, \ldots,\left(2^{n}+1\right)+\ldots+\left(2^{n+1}-1\right), \ldots=3,12, \ldots, 3 \cdot 4^{n-1}, \ldots$

## Answer

Could.

|  | [Numerical tables and their properties $]$ | Difficulty: 4- |
| ---: | :--- | ---: |
| Problem $\underline{65508}$ Topics: |  |  |
|  | $[$ Evenness and oddness $]$ | Grades: $7,8,9$ |

In each cell of a $13 \times 13$ table, one of the natural numbers from 1 to 25 is written. A cell is called good if among the twenty-five numbers written in it and in all cells of the same row and the same column, there are no identical ones. Can all the cells of one of the main diagonals be good?

## Solution

For each cell of one of the main diagonals, consider the cross, which is a set of twenty-five cells: the cell itself and all cells that are in the same row or column with it.

Consider all crosses formed by the cells of the selected main diagonal. Note that each cell of this diagonal is part of only one cross (its own), while any other cell of the table is part of exactly two such crosses.

There are different ways to reason further.

First method. Consider a natural number from 1 to 25 that is missing from the selected main diagonal (such a number must exist, as there are only 13 cells on the diagonal). If all cells of the main diagonal are good, then this number appears in each of the thirteen crosses exactly once. However, any number outside the main diagonal must appear in two crosses, so all crosses should be paired, which is impossible for thirteen crosses. Contradiction.

Second method. For the number 1 to appear in each of the 13 crosses, it must be written in the table at least seven times. The same can be said about each of the twenty-five given numbers. Therefore, for all cells of the considered main diagonal to be good, at least $7 \cdot 25=175$ cells need to be filled. But the table has only $13 \cdot 13=169$ cells. Contradiction.

Thus, all cells of the main diagonal cannot be good.

## Answer

They cannot.

![](https://cdn.mathpix.com/cropped/2024_05_06_0e910282974b56b80e3ag-01.jpg?height=180&width=2056&top_left_y=2517&top_left_x=1)[^0]

For which $N$ can the numbers from 1 to $N$ be rearranged so that the arithmetic mean of any group of two or more consecutive numbers is not an integer?

## Solution

Odd $N$ do not work: in this case, the arithmetic mean $1 / 2(N+1)$ of all numbers is an integer.

For even $N=2 m$, a suitable arrangement is, for example, this: $2,1,4,3,6,5, \ldots, 2 m, 2 m-1$.

Indeed, a group of $2 k$ consecutive numbers can be divided into pairs of numbers symmetrically located relative to the center of the group (for example, the group $1,4,3,6,5,8$ is divided into pairs (1,8), (4,5), (3,6)). In each pair, the arithmetic mean is the same half-integer number (the pair contains numbers of different parity). Therefore, the arithmetic mean of the entire group is also the same non-integer number.

A group of $2 k+1$ consecutive numbers consists of $2 k$ consecutive numbers and one number at the edge. If this number is swapped with its neighbor in the arrangement, not included in the group, we get $2 k+1$ consecutive numbers, whose arithmetic mean is an integer. Therefore, the arithmetic mean of the numbers in the group differs from this integer by $1 / 2 k+1$.

## Answer

For all even $N$.