A. K. Tolpygo

The white rook is on the b2 square of an $8 \times 8$ chessboard, and the black rook is on the c4 square. The players take turns moving their rooks, with the whites starting. It is forbidden to place one's rook under attack by the other rook, or on a square that has already been visited by any rook. The player who cannot make a move loses. Which player can ensure a win, regardless of how the other plays? (On a move, a rook can slide horizontally or vertically by any number of squares, and it is considered to have visited only the starting and ending squares of this move.)
Let's divide the verticals into pairs, placing verticals b and c in one pair. We will also divide the horizontals into pairs, placing the 2nd and 4th in one pair. Then the cells will also be divided into pairs: a paired cell lies at the intersection of a paired horizontal and a paired vertical. Rooks on paired cells do not attack each other.

Strategy for the second player: respond to any move of the first by moving to the paired cell. This is possible: if the first player moved a rook from one cell to another, then these cells lie in the same row. But then the cells paired with them lie in a paired row, and a rook move between them is also possible. Thus, after each move of the second player, all unvisited cells will consist of the specified pairs, and the second player will always land on an unvisited cell. This means that the first player will eventually be unable to make a move and will lose.

## Answer

The second player.

## [ Decimal number system ] Problem 78188 Topics: [ $\quad$ Puzzles $\quad]$ [Arithmetic. Mental arithmetic and etc.]

Difficulty: 4

Classes: $7,8,9$

What integer must 999999999 be multiplied by to get a number consisting of all ones?

## Solution

Write the number 999999999 as 1000000000 - 1. Let $A$ be the desired number and $\bar{A}$ its decimal representation. Clearly, $999999999 \cdot A = 1000000000 A - A = \overline{A 000000000} - A$. This number must consist of all ones. Thus,

$$
A = 1000000000 A - 111 \ldots 11
$$

or

$$
A = \overline{A 000000000} - 11 \ldots 1 .\left(()^{*}\right)
$$

Since the last digits of the minuend are known, we can perform the subtraction ("in column"), thereby determining the last digits of the number $A$ on the left side of equation (*). By appending these digits to the left of the zeros in the right part, we can continue the subtraction, determining new digits, and so on, until this is possible (i.e., until the newly obtained digits of the number $A$ become equal to 1). Let's show how this process works using a shorter number 99 as an example:

$$
99 A = 100 A - A = 111 \ldots 11
$$

so

$$
A = 100 A - 111 \ldots 11 = \overline{A 00} - 111 \ldots 1
$$

Here is the sequential calculation of the digits:
$-\frac{\because 111}{89}$
$\cdots 8900$
778900
$\frac{111111}{667789}$

and so on. In our case, as it is not difficult to verify, the number obtained is

![](https://cdn.mathpix.com/cropped/2024_05_06_99ef25effe2e32f29eccg-02.jpg?height=126&width=788&top_left_y=1003&top_left_x=635)

The number $A_{1}$, obviously, is the smallest with the required property, since each digit of the number $A_{1}$ was necessarily obtained from the subsequent ones. But this number is not the only one, as our process can be continued by appending nine zeros to the left of $A_{1}$ ( $111 \ldots 11 - 111 \ldots 11 = 000 \ldots 00$ ) and resuming the subtraction. Clearly, any number $A_{\mathrm{n}}$ with the required property has the form:

$$
A_{\mathrm{n}} = A_{1} \underbrace{000 \ldots 00}_{9} A_{1} \underbrace{000 \ldots 00}_{9} A_{1} \ldots 0 A_{1} \underbrace{000 \ldots 00}_{9} A_{1}
$$

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