[Algorithm Theory (Miscellaneous).]

Given two fuse cords, each of which burns for exactly one minute if lit from one end (but may burn unevenly).

How can you use these cords to measure 45 seconds? (You can light the cord from either of the two ends.)

#
One wick can be lit at the moment when the other one is completely burned out.

## Solution

We will light the first wick from both ends simultaneously, and the second wick from one end. The wick lit from both ends will burn twice as fast, that is, in 30 seconds. At the moment the first wick is completely burned, we light the second wick from the second end. If we hadn't done this, the second wick would have burned for another 30 seconds. Therefore, after it was lit from the second end, it had 15 seconds left to burn. In total, from the initial moment to the moment the second wick is completely burned, 30 + 15 = 45 seconds have passed.

The game board is a horizontal strip of size \(1 \times 100\) cells. A chip is placed in the leftmost cell. Two players take turns moving the chip to the right, and in one move, it is allowed to move the chip to the right by a distance of 1 to 10 cells. The player who cannot make a move (i.e., the chip is in the rightmost cell before their turn) loses. Who will win with correct play?

## Hint

The second player can play in such a way that after their move, the number of free cells to the right of the chip is a multiple of 11.

## Solution

The winning strategy for the second player is as follows: after each move of the first player by \(k\) cells to the right, move the chip by \(11-k\) cells. Thus, after a pair of moves by the first and second players, the chip will move 11 cells to the right. Initially, there were 99 empty cells to the right of the chip. Therefore, after 9 pairs of moves by the first and second players, the chip will be in the rightmost cell.

## Answer

The second player.