We are given an infinite row of cells extending infinitely in both directions. Some cells contain one or more stones. The total number of stones is finite. At each move, the player performs one of the following three operations: [b]1. [/b]Take three stones from some cell, and add one stone to the cells located one cell to the left and one cell to the right, each skipping one cell in between. [b]2. [/b]Take two stones from some cell, and add one stone to the cell one cell to the left, skipping one cell and one stone to the adjacent cell to the right. [b]3.[/b] Take one stone from each of two adjacent cells, and add one stone to the cell to the right of these two cells. The process ends when no moves are possible. Prove that the process always terminates and the final distribution of stones does not depend on the choices of moves made by the player. [img]https://i.imgur.com/IjcIDOa.png[/img] [i]Proposed by Luka Tsulaia, Georgia[/i]

Given an $m \times n$ table consisting of $mn$ unit cells. Alice and Bob play the following game: Alice goes first and the one who moves colors one of the empty cells with one of the given three colors. Alice wins if there is a figure, such as the ones below, having three different colors. Otherwise Bob is the winner. Determine the winner for all cases of $m$ and $n$ where $m, n \ge 3$. Proposed by [i]Toghrul Abbasov, Azerbaijan[/i]

On a table, there are $11$ piles of ten stones each. Pete and Basil play the following game. In turns they take $1, 2$ or $3$ stones at a time: Pete takes stones from any single pile while Basil takes stones from different piles but no more than one from each. Pete moves first. The player who cannot move, loses. Which of the players, Pete or Basil, has a winning strategy?

The archipelago Barrantes - $n$ is a group of islands connected by bridges as follows: there are a main island (Humberto), in the first step I place an island below Humberto and one above from Humberto and I connect these 2 islands to Humberto. I put $2$ islands to the left of these $2$ new islands and I connect them with a bridge to the island that they have on their right. In the second step I take the last $2$ islands and I apply the same process that I applied to Humberto. In the third step I apply the same process to the $4$ new islands. We repeat this step n times we reflect the archipelago that we have on a vertical line to the right of Humberto. We connect Humberto with his reflection and so we have the archipelago Barrantes -$n$. However, the archipelago Barrantes -$n$ exists on a small planet cylindrical, so that the islands to the left of the archipelago are in fact the islands that are connected to the islands on the right. The figure shows the Barrantes archipelago -$2$, The islands at the edges are still numbered to show how the archipelago connects around the cylindrical world, the island numbered $1$ on the left is the same as the island numbered $1$ on the right. [img]https://cdn.artofproblemsolving.com/attachments/e/c/803d95ce742c2739729fdb4d74af59d4d0652f.png[/img] One day two bands of pirates arrive at the archipelago Barrantes - $n$: The pirates Black Beard and the Straw Hat Pirates. Blackbeard proposes a game to Straw Hat: The first player conquers an island, the next player must conquer an island connected to the island that was conquered in the previous turn (clearly not conquered on a previous shift). The one who cannot conquer any island in his turn loses. Straw Hat decides to give the first turn to Blackbeard. Prove that Straw Hat has a winning strategy for every $n$.

An ordered quadruple of numbers is called [i]ten-esque[/i] if it is composed of 4 nonnegative integers whose sum is equal to $10$. Ana chooses a ten-esque quadruple $(a_1, a_2, a_3, a_4)$ and Banana tries to guess it. At each stage Banana offers a ten-esque quadtruple $(x_1,x_2,x_3,x_4)$ and Ana tells her the value of \[|a_1-x_1|+|a_2-x_2|+|a_3-x_3|+|a_4-x_4|\] How many guesses are needed for Banana to figure out the quadruple Ana chose?

Among the $n$ inhabitants of an island, every two are either friends or enemies. Some day, the chief of the island orders that each inhabitant (including himself) makes and wears a necklace consisting of marbles, in such a way that the necklaces of two friends have at least one marble of the same type and that the necklaces of two enemies differ at all marbles. (A necklace may also be marbleless). Show that the chief’s order can be achieved by using $\left\lfloor\frac{n^2}4\right\rfloor$ different types of stones, but not necessarily by using fewer types.

Let $n$ be a positive integer and $M = \{1, 2, 3, 4, 5, 6\}$. Two persons $A$ and $B$ play in the following Way: $A$ writes down a digit from $M$, $B$ appends a digit from $M$, and so it becomes alternately one digit from $M$ is appended until the $2n$-digit decimal representation of a number has been created. If this number is divisible by $9$, $B$ wins, otherwise $A$ wins. For which $n$ can $A$ and for which $n$ can $B$ force the win?

With $2018$ points, a network composed of hexagons is formed as a sample the figure: [asy] unitsize(1 cm); int i; path hex = dir(30)--(0,1)--dir(150)--dir(210)--(0,-1)--dir(330)--cycle; draw(hex); draw(shift((sqrt(3),0))*(hex)); draw(shift((2*sqrt(3),0))*(hex)); draw(shift((4*sqrt(3),0))*(hex)); draw(shift((5*sqrt(3),0))*(hex)); dot((3*sqrt(3) - 0.3,0)); dot((3*sqrt(3),0)); dot((3*sqrt(3) + 0.3,0)); dot(dir(150)); dot(dir(210)); for (i = 0; i <= 5; ++i) { if (i != 3) { dot((0,1) + i*(sqrt(3),0)); dot(dir(30) + i*(sqrt(3),0)); dot(dir(330) + i*(sqrt(3),0)); dot((0,-1) + i*(sqrt(3),0)); } } dot(dir(150) + 4*(sqrt(3),0)); dot(dir(210) + 4*(sqrt(3),0)); [/asy] A bee and a fly play the following game: initially the bee chooses one of the $2018$ dots and paints it red, then the fly chooses one of the $2017$ unpainted dots and paint it blue. Then the bee chooses an unpainted point and paints it red and then the fly chooses an unpainted one and paints it blue and so they alternate. If at the end of the game there is an equilateral triangle with red vertices, the bee wins, otherwise Otherwise the fly wins. Determine which of the two insects has a winning strategy.

Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. [i]Proposed by Amine Natik, Morocco[/i]

There are $n$ coins in the pile. Two players play a game by alternately performing a move. A move consists of taking $5,7$ or $11$ coins away from the pile. The player unable to perform a move loses the game. Which player - the one playing first or second - has the winning strategy if: (a) $n=2001$; (b) $n=5000$?

Two players play a game on four piles of pebbles labeled with the numbers $1,2,3,4$ respectively. The players take turns in an alternating fashion. On his or her turn, a player selects integers $m$ and $n$ with $1\leq m<n\leq 4$, removes $m$ pebbles from pile $n$, and places one pebble in each of the piles $n-1,n-2,\dots,n-m$. A player loses the game if he or she cannot make a legal move. For each starting position, determine the player with a winning strategy.

Alberto wants to organize a poker game with his friends this evening. Bruno and Barbara together go to gym once in three evenings, whereas Carla, Corrado, Dario and Davide are busy once in two evenings (not necessarily the same day). Moreover, Dario is not willing to play with Davide, since they have a quarrel over a girl. A poker game requires at least four persons (including Alberto). What is the probability that the game will be played?

Peter buys a lottery ticket on which he enters an $n$-digit number, none of the digits being $0$. On the draw date, the lottery administrators will reveal an $n\times n$ table, each cell containing one of the digits from $1$ to $9$. A ticket wins a prize if it does not match any row or column of this table, read in either direction. Peter wants to bribe the administrators to reveal the digits on some cells chosen by Peter, so that Peter can guarantee to have a winning ticket. What is the minimum number of digits Peter has to know?

Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n + 1$ squares in a row, numbered $0$ to $n$ from left to right. Initially, $n$ stones are put into square $0$, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of these stones and moves it to the right by at most $k$ squares (the stone should say within the board). Sisyphus' aim is to move all $n$ stones to square $n$. Prove that Sisyphus cannot reach the aim in less than \[ \left \lceil \frac{n}{1} \right \rceil + \left \lceil \frac{n}{2} \right \rceil + \left \lceil \frac{n}{3} \right \rceil + \dots + \left \lceil \frac{n}{n} \right \rceil \] turns. (As usual, $\lceil x \rceil$ stands for the least integer not smaller than $x$. )

Oleksii and Solomiya play the following game on a square $6n\times 6n$, where $n$ is a positive integer. Oleksii in his turn places a piece of type $F$, consisting of three cells, on the board. Solomia, in turn, after each move of Oleksii, places the numbers $0, 1, 2$ in the cells of the figure that Oleksii has just placed, using each of the numbers exactly once. If two of Oleksii's pieces intersect at any moment (have a common square), he immediately loses. Once the square is completely filled with numbers, the game stops. In this case, if the sum of the numbers in each row and each column is divisible by $3$, Solomiya wins, and otherwise Oleksii wins. Who can win this game if the figure of type $F$ is: a) a rectangle ; b) a corner of three cells? [i]Proposed by Oleksii Masalitin[/i]

Two players play a game with a pile of $N$ coins on a table. On a player's turn, if there are $n$ coins, the player can take at most $n/2+1$ coins, and must take at least one coin. The player who grabs the last coin wins. For how many values of $N$ between $1$ and $100$ (inclusive) does the first player have a winning strategy?

On a table are given $1999$ red candies and $6661$ yellow candies. The candies are indistinguishable due to the same packing. A gourmet applies the following procedure as long as it is possible: (i) He picks any of the remaining candies, notes its color, eats it and goes to (ii). (ii) He picks any of the remaining candies, and notes its color: if it is the same as the color of the last eaten candy, eats it and goes to (ii); otherwise returns it upon repacking and goes to (i). Prove that all the candies will be eaten and find the probability that the last eaten candy will be red.

Alice and Bob determine a number with $2018$ digits in the decimal system by choosing digits from left to right. Alice starts and then they each choose a digit in turn. They have to observe the rule that each digit must differ from the previously chosen digit modulo $3$. Since Bob will make the last move, he bets that he can make sure that the final number is divisible by $3$. Can Alice avoid that? [i](Proposed by Richard Henner)[/i]

Batman, Robin, and The Joker are in three of the vertex cells in a square $2025 \times 2025$ board, such that Batman and Robin are on the same diagonal (picture). In each round, first The Joker moves to an adjacent cell (having a common side), without exiting the board. Then in the same round Batman and Robin move to an adjacent cell. The Joker wins if he reaches the fourth "target" vertex cell (marked T). Batman and Robin win if they catch The Joker i.e. at least one of them is on the same cell as The Joker. If in each move all three can see where the others moved, who has a winning strategy, The Joker, or Batman and Robin? Explain the answer. [b]Comment.[/b] Batman and Robin decide their common strategy at the beginning. [img]https://i.imgur.com/PeLBQNt.png[/img]

Stekel and Prick play a game on an $ m \times n$ board, where $m$ and $n$ are positive are integers. They alternate turns, with Stekel starting. Spine bets on his turn, he always takes a pawn on a square where there is no pawn yet. Prick does his turn the same, but his pawn must always come into a square adjacent to the square that Spike just placed a pawn in on his previous turn. Prick wins like the whole board is full of pawns. Spike wins if Prik can no longer move a pawn on his turn, while there is still at least one empty square on the board. Determine for all pairs $(m, n)$ who has a winning strategy.

(a) An infinite sheet is divided into squares by two sets of parallel lines. Two players play the following game: the first player chooses a square and colours it red, the second player chooses a non-coloured square and colours it blue, the first player chooses a non-coloured square and colours it red, the second player chooses a non-coloured square and colours it blue, and so on. The goal of the first player is to colour four squares whose vertices form a square with sides parallel to the lines of the two parallel sets. The goal of the second player is to prevent him. Can the first player win? (b) What is the answer to this question if the second player is permitted to colour two squares at once? (DG Azov) PS. (a) for Juniors, (a),(b) for Seniors

Mojtaba and Hooman are playing a game. Initially Mojtaba draws $2018$ vectors with zero sum. Then in each turn, starting with Mojtaba, the player takes a vector and puts it on the plane. After the first move, the players must put their vector next to the previous vector (the beginning of the vector must lie on the end of the previous vector). At last, there will be a closed polygon. If this polygon is not self-intersecting, Mojtaba wins. Otherwise Hooman. Who has the winning strategy? [i]Proposed by Mahyar Sefidgaran, Jafar Namdar [/i]

There are two numbers on a board, $1/2009$ and $1/2008$. Alex and Ben play the following game. At each move, Alex names a number $x$ (of his choice), while Ben responds by increasing one of the numbers on the board (of his choice) by $x$. Alex wins if at some moment one of the numbers on the board becomes $1$. Can Alex win (no matter how Ben plays)?

On an infinite chessboard, a solitaire game is played as follows: at the start, we have $n^2$ pieces occupying a square of side $n.$ The only allowed move is to jump over an occupied square to an unoccupied one, and the piece which has been jumped over is removed. For which $n$ can the game end with only one piece remaining on the board?

Turbo the snail sits on a point on a circle with circumference $1$. Given an infinite sequence of positive real numbers $c_1, c_2, c_3, \dots$, Turbo successively crawls distances $c_1, c_2, c_3, \dots$ around the circle, each time choosing to crawl either clockwise or counterclockwise. Determine the largest constant $C > 0$ with the following property: for every sequence of positive real numbers $c_1, c_2, c_3, \dots$ with $c_i < C$ for all $i$, Turbo can (after studying the sequence) ensure that there is some point on the circle that it will never visit or crawl across.