Chekanov Yu.V.

A square board is divided by seven lines parallel to one side of the board and seven lines parallel to the other side of the board into 64 rectangular cells, which are painted in white and black in a checkerboard pattern. The distances between adjacent lines are not necessarily the same, so the cells can be of different sizes. However, it is known that the ratio of the area of each white cell to the area of any black cell is no more than 2. Find the greatest possible ratio of the total area of the white cells to the total area of the black cells.
Let the top-left cell be white. All odd columns are shifted to the left, and then all odd rows are shifted up. Now all the white cells are gathered in two rectangles in the top-left and bottom-right corners, and the ratio of the area of each white rectangle to the area of each black one is no more than 2. If the width of the left white rectangle is greater than $1 / 2$, then by increasing its height, we increase the total white area. Therefore, the maximum ratio of the white area to the black area is achieved when the top-left rectangle is a square with a side of $2 / 3$, and it is equal to $(4 / 9 + 1 / 9):(2 \cdot 2 / 9) = 5 / 4$.

## Answer

## [ Prime numbers and their properties ] Problem 111788 Topics: [ [Divisibility of numbers. General properties] [The extremal principle (miscellaneous). $\quad]$ Author: Senderov V.A. Do there exist such prime numbers $p_{1}, p_{2}, \ldots, p_{2007}$, that $p_{1}^{2}-1$ is divisible by $p_{2}$, $p_{2}^{2}-1$ is divisible by $p_{3}$, ..., $p_{2007}^{2}-1$ is divisible by $p_{1}$?

## Solution

Suppose such numbers exist. Let one of the numbers be 2, say, $p_{1}=2$. Then sequentially we get $p_{2}=3, p_{3}=2$,

$p_{4}=3, p_{5}=2, \ldots, p_{2007}=2, p_{1}=3$. Contradiction.

Suppose all numbers are odd primes, and $p_{1}$ is the largest of them. Let $p_{2007}=q \leq p_{1}$. Then $q^{2}-1=2 \cdot 2 \cdot q^{-} 1 / 2 \cdot q^{+1} / 2$, where all factors are integers and do not exceed $q^{++1} / 2 < q \leq p_{1}$. But this contradicts the fact that $q^{2}-1$ is divisible by $p_{1}$.

## Answer

No, they do not exist.