Problems

A daisy has a) 12 petals; b) 11 petals. Consider the game with two players where: in one turn a player is allowed to remove either exactly one petal or two petals which are next to each other. The loser is the one who cannot make a turn. How should the second player act, in cases a) and b), in order to win the game regardless of the moves of the first player?

In the first pile there are 100 sweets and in the second there are 200. Consider the game with two players where: in one turn a player can take any amount of sweets from one of the piles. The winner is the one who takes the last sweet. Which player would win by using the correct strategy?

A cat tries to catch a mouse in labyrinths A, B, and C. The cat walks first, beginning with the node marked with the letter “K”. Then the mouse (from the node “M”) moves, then again the cat moves, etc. From any node the cat and mouse go to any adjacent node. If at some point the cat and mouse are in the same node, then the cat eats the mouse.

Can the cat catch the mouse in each of the cases A, B, C?

Two play a game on a chessboard \(8 \times 8\). The player who makes the first move puts a knight on the board. Then they take turns moving it (according to the usual rules), whilst you can not put the knight on a cell which he already visited. The loser is one who has nowhere to go. Who wins with the right strategy – the first player or his partner?

a) The vertices (corners) in a regular polygon with 10 sides are colored black and white in an alternating fashion (i.e. one vertex is black, the next is white, etc). Two people play the following game. Each player in turn draws a line connecting two vertices of the same color. These lines must not have common vertices (i.e. must not begin or end on the same dot as another line) with the lines already drawn. The winner of the game is the player who made the final move. Which player, the first or the second, would win if the right strategy is used?

b) The same problem, but for a regular polygon with 12 sides.

There is a chocolate bar with five longitudinal and eight transverse grooves, along which it can be broken (in total into \(9 * 6 = 54\) squares). Two players take part, in turns. A player in his turn breaks off the chocolate bar a strip of width 1 and eats it. Another player who plays in his turn does the same with the part that is left, etc. The one who breaks a strip of width 2 into two strips of width 1 eats one of them, and the other is eaten by his partner. Prove that the first player can act in such a way that he will get at least 6 more chocolate squares than the second player.

Two people are playing. The first player writes out numbers from left to right, randomly alternating between 0 and 1, until there are 2021 numbers in total. Each time after the first one writes out the next digit, the second switches two numbers from the already written row (when only one digit is written, the second misses its move). Is the second player always able to ensure that, after his last move, the arrangement of the numbers is symmetrical relative to the middle number?

On a table there are 2022 cards with the numbers 1, 2, 3, ..., 2022. Two players take one card in turn. After all the cards are taken, the winner is the one who has a greater last digit of the sum of the numbers on the cards taken. Find out which of the players can always win regardless of the opponent’s strategy, and also explain how he should go about playing.

The March Hare and the Dormouse also decided to play a game. They made two piles of matches on the table. The first pile contains 7 matches, and the second one 8. The March Hare set the rules: the players divide a pile into two piles in turns, i.e. the first player divides one of the piles into two, then the second player divides one of the piles on the table into two, then the first player divides one of the piles into two and so on. The loser is the one who cannot not find a pile to divide. The March Hare starts the game. Can the March Hare play in such a way that he always wins?

A board \(7\times7\) is coloured in chessboard fashion in such a way that all the corners are black. The Queen orders the Hatter to colour the board white but sets the rule: in one go it is allowed to repaint only two adjacent cells into opposite colours. The Hatter tries to explain that this is impossible. Can you help the Hatter to present his arguments?