6,7 |  |
| $[\underline{\text { Prime numbers and their properties }}]$ |  |  |

## Author: Raskina I.V.

A group of tourists is dividing cookies. If they evenly divide two identical packs, one extra cookie remains. But if they evenly divide three such packs, 13 extra cookies remain. How many tourists are in the group?
Distribute three times two packs, 3 cookies will remain. But the same six packs of cookies can be distributed differently - three and another three, and then $2 \cdot 13=26$ cookies will remain. This means that $26-3=23$ cookies can be divided equally among the tourists. Since the number 23 is prime, this is possible only if there are 23 tourists.

## Answer

## [Chess boards and chess pieces]  Problem 66013 Topics:  $[\quad$ Parity and oddity $\quad]$

Difficulty: $3+$

Grades: $9,10,11$

Vasya thinks of eight cells on a chessboard, none of which lie in the same row or column. On a move, Petya places eight rooks on the board, none of which attack each other, and then Vasya indicates all the rooks standing on the thought cells. If the number of rooks indicated by Vasya on this move is even (i.e., 0, 2, 4, 6, or 8), then Petya wins; otherwise, all the pieces are removed from the board and Petya makes the next move. In how few moves can Petya guarantee a win?

## Solution

We will show how Petya can win in two moves. On the first move, he places eight rooks along the diagonal of the board. If he does not win on this move, then there is an odd number of Vasya's thought cells on the diagonal. In particular, there is both a cell $A$, thought of by Vasya, and a cell $B$, not thought of by him, on the diagonal.

On the second move, Petya places rooks on six diagonal cells, except for $A$ and $B$, and on cells $C$ and $D$, which lie in the same rows as $A$ and $B$ respectively, and in the same columns as $B$ and $A$ respectively. Each new rook stands either in the same row or in the same column as $A$, so Vasya could not have thought of their cells. Thus, in the new rook configuration, the number of cells thought of by Vasya will decrease by exactly one, i.e., it will become even, and Petya will win.

It remains to show that Petya cannot guarantee a win in one move. Suppose he managed to do so. By rearranging the columns of the board, we can assume that he made the first move as shown in the figure. Then he will not win if Vasya's cells are the gray-marked ones in the same figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_bf0205b4cbbe4b58a1adg-25.jpg?height=346&width=352&top_left_y=1205&top_left_x=859)

## Answer

In two moves.