The expression $1\oplus2\oplus3\oplus4\oplus5\oplus6\oplus7\oplus8\oplus9$ is written on a blackboard. Bill and Peter play the following game. They replace $\oplus$ by $+$ or $\cdot$, making their moves in turn, and one of them can use only $+$, while the other one can use only $\cdot$. At the beginning, Bill selects the sign he will use, and he tries to make the result an even number. Peter tries to make the result an odd number. Prove that Peter can always win. [hide=Original Wording]The expression $1*2*3*4*5*6*7*8*9$ is written on a blackboard. Bill and Peter play the following game. They replace $*$ by $+$ or $\cdot$, making their moves in turn, and one of them can use only $+$, while the other one can use only $\cdot$. At the beginning Bill selects the sign he will use, and he tries to make the result an even number. Peter tries to make the result an odd number. Prove that Peter can always win.[/hide]

In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned 1/2 point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

Two players, Agon and Besa, choose a number from the set $\{1,2,3,4,5,6,7,8\}$, in turns, until no number is left. Then, each player sums all the numbers that he has chosen. We say that a player wins if the sum of his chosen numbers is a prime and the sum of the numbers that his opponent has chosen is composite. In the contrary, the game ends in a draw. Agon starts first. Does there exist a winning strategy for any of the players?

There are 2019 cards in a box. Each card has a number written on one of its sides and a letter on the other side. Amy and Ben play the following game: in the beginning Amy takes all the cards, places them on a line and then she flips as many cards as she wishes. Each time Ben touches a card he has to flip it and its neighboring cards. Ben is allowed to have as many as 2019 touches. Ben wins if all the cards are on the numbers' side, otherwise Amy wins. Determine who has a winning strategy.

Peter and Basil play the following game on a horizontal table $1\times{2019}$. Initially Peter chooses $n$ positive integers and writes them on a board. After that Basil puts a coin in one of the cells. Then at each move, Peter announces a number s among the numbers written on the board, and Basil needs to shift the coin by $s$ cells, if it is possible: either to the left, or to the right, by his decision. In case it is not possible to shift the coin by $s$ cells neither to the left, nor to the right, the coin stays in the current cell. Find the least $n$ such that Peter can play so that the coin will visit all the cells, regardless of the way Basil plays.

A grasshopper is sitting at an integer point in the Euclidean plane. Each second it jumps to another integer point in such a way that the jump vector is constant. A hunter that knows neither the starting point of the grasshopper nor the jump vector (but knows that the jump vector for each second is constant) wants to catch the grasshopper. Each second the hunter can choose one integer point in the plane and, if the grasshopper is there, he catches it. Can the hunter always catch the grasshopper in a finite amount of time?

General Tilly and the Duke of Wallenstein play "Divide and rule!" (Divide et impera!). To this end, they arrange $N$ tin soldiers in $M$ companies and command them by turns. Both of them must give a command and execute it in their turn. Only two commands are possible: The command "[i]Divide![/i]" chooses one company and divides it into two companies, where the commander is free to choose their size, the only condition being that both companies must contain at least one tin soldier. On the other hand, the command "[i]Rule![/i]" removes exactly one tin soldier from each company. The game is lost if in your turn you can't give a command without losing a company. Wallenstein starts to command. a) Can he force Tilly to lose if they start with $7$ companies of $7$ tin soldiers each? b) Who loses if they start with $M \ge 1$ companies consisting of $n_1 \ge 1, n_2 \ge 1, \dotsc, n_M \ge 1$ $(n_1+n_2+\dotsc+n_M=N)$ tin soldiers?

A 2-player game is played on $n\geq 3$ points, where no 3 points are collinear. Each move consists of selecting 2 of the points and drawing a new line segment connecting them. The first player to draw a line segment that creates an odd cycle loses. (An odd cycle must have all its vertices among the $n$ points from the start, so the vertices of the cycle cannot be the intersections of the lines drawn.) Find all $n$ such that the player to move first wins.

Peter and Basil play the following game on a horizontal table $1\times{2019}$. Initially Peter chooses $n$ positive integers and writes them on a board. After that Basil puts a coin in one of the cells. Then at each move, Peter announces a number s among the numbers written on the board, and Basil needs to shift the coin by $s$ cells, if it is possible: either to the left, or to the right, by his decision. In case it is not possible to shift the coin by $s$ cells neither to the left, nor to the right, the coin stays in the current cell. Find the least $n$ such that Peter can play so that the coin will visit all the cells, regardless of the way Basil plays.

Amy and Bec play the following game. Initially, there are three piles, each containing $2020$ stones. The players take turns to make a move, with Amy going first. Each move consists of choosing one of the piles available, removing the unchosen pile(s) from the game, and then dividing the chosen pile into $2$ or $3$ non-empty piles. A player loses the game if he/she is unable to make a move. Prove that Bec can always win the game, no matter how Amy plays.

Andrey and Sasha play the game, making moves alternate. On his turn, Andrey marks on the plane an arbitrary point that has not yet been marked. After that, Sasha colors this point in one of two colors: white and black. Sasha wins if after his move it is impossible to draw a line such that all white points lie in one half-plane, while all black points lie in another half-plane with respect to this line. [b]a)[/b] Prove that Andrey can make moves in such a way that Sasha will never win. [b]b)[/b] Suppose that Andrey can mark only integer points on the Cartesian plane. Can Sasha guarantee himself a win regardless of Andrey's moves? [i](N. Naradzetski)[/i]

A number line with the integers $1$ through $20,$ from left to right, is drawn. Ten coins are placed along this number line, with one coin at each odd number on the line. A legal move consists of moving one coin from its current position to a position of strictly greater value on the number line that is not already occupied by another coin. How many ways can we perform two legal moves in sequence, starting from the initial position of the coins (different two-move sequences that result in the same position are considered distinct)?

Alice and Bob play a game. There are $100$ gold coins, $100$ silver coins, and $100$ bronze coins. Players take turns to take at least one coin, but they cannot take two or more coins of same kind at once. Alice goes first. The player who cannot take any coin loses. Who has a winning strategy?

Two players play a game as follows: there $n > 1$ rounds and $d \geq 1$ is fixed. In the first round A picks a positive integer $m_1$, then B picks a positive integer $n_1 \not = m_1$. In round $k$ (for $k = 2, \ldots , n$), A picks an integer $m_k$ such that $m_{k-1} < m_k \leq m_{k-1} + d$. Then B picks an integer $n_k$ such that $n_{k-1} < n_k \leq n_{k-1} + d$. A gets $\gcd(m_k,n_{k-1})$ points and B gets $\gcd(m_k,n_k)$ points. After $n$ rounds, A wins if he has at least as many points as B, otherwise he loses. For each $(n, d)$ which player has a winning strategy?

Two players play a game as follows. The first player chooses two non-zero integers A and B. The second player forms a quadratic with A, B and 1998 as coefficients (in any order). The first player wins iff the equation has two distinct rational roots. Show that the first player can always win.

A [i]Nim-style game[/i] is defined as follows. Two positive integers $k$ and $n$ are specified, along with a finite set $S$ of $k$-tuples of integers (not necessarily positive). At the start of the game, the $k$-tuple $(n, 0, 0, ..., 0)$ is written on the blackboard. A legal move consists of erasing the tuple $(a_1,a_2,...,a_k)$ which is written on the blackboard and replacing it with $(a_1+b_1, a_2+b_2, ..., a_k+b_k)$, where $(b_1, b_2, ..., b_k)$ is an element of the set $S$. Two players take turns making legal moves, and the first to write a negative integer loses. In the event that neither player is ever forced to write a negative integer, the game is a draw. Prove that there is a choice of $k$ and $S$ with the following property: the first player has a winning strategy if $n$ is a power of 2, and otherwise the second player has a winning strategy. [i]Proposed by Linus Hamilton[/i]

Peter wrote $100$ distinct integers on a board. Basil needs to fill the cells of a table $100\times{100}$ with integers so that the sum in each rectangle $1\times{3}$ (either vertical, or horizontal) is equal to one of the numbers written on the board. Find the greatest $n$ such that, regardless of numbers written by Peter, Basil can fill the table so that it would contain each of numbers $(1,2,...,n)$ at least once (and possibly some other integers).

Two players play the following game. At the outset there are two piles, containing $10,000$ and $20,000$ tokens,respectively . A move consists of removing any positive number of tokens from a single pile $or$ removing $x>0$ tokens from one pile and $y>0$ tokens from the other , where $x+y$ is divisible by $2015$. The player who can not make a move loses. Which player has a winning strategy

A game is played on a $n \times n$ chessboard. In the beginning Bars the cat occupies any cell according to his choice. The $d$ sparrows land on certain cells according to their choice (several sparrows may land in the same cell). Bars and the sparrows play in turns. In each turn of Bars, he moves to a cell adjacent by a side or a vertex (like a king in chess). In each turn of the sparrows, precisely one of the sparrows flies from its current cell to any other cell of his choice. The goal of Bars is to get to a cell containing a sparrow. Can Bars achieve his goal a) if $d=\lfloor \frac{3\cdot n^2}{25}\rfloor$, assuming $n$ is large enough? b) if $d=\lfloor \frac{3\cdot n^2}{19}\rfloor$, assuming $n$ is large enough? c) if $d=\lfloor \frac{3\cdot n^2}{14}\rfloor$, assuming $n$ is large enough?

There are 60 empty boxes $B_1,\ldots,B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game. In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps: (a) Bob chooses an integer $k$ with $1\leq k\leq 59$ and splits the boxes into the two groups $B_1,\ldots,B_k$ and $B_{k+1},\ldots,B_{60}$. (b) Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group. Bob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning. [i]Czech Republic[/i]

Let it \(k\) be a fixed positive integer. Alberto and Beralto play the following game: Given an initial number \(N_0\) and starting with Alberto, they alternately do the following operation: change the number \(n\) for a number \(m\) such that \(m < n\) and \(m\) and \(n\) differ, in its base-2 representation, in exactly \(l\) consecutive digits for some \(l\) such that \(1 \leq l \leq k\). If someone can't play, he loses. We say a non-negative integer \(t\) is a [i]winner[/i] number when the gamer who receives the number \(t\) has a winning strategy, that is, he can choose the next numbers in order to guarrantee his own victory, regardless the options of the other player. Else, we call it [i]loser[/i]. Prove that, for every positive integer \(N\), the total of non-negative loser integers lesser than \(2^N\) is \(2^{N-\lfloor \frac{log(min\{N,k\})}{log 2} \rfloor}\)

Navi and Ozna are playing a game where Ozna starts first and the two take turn making moves. A positive integer is written on the waord. A move is to (i) subtract any positive integer at most 2015 from it or (ii) given that the integer on the board is divisible by $2014$, divide by $2014$. The first person to make the integer $0$ wins. To make Navi's condition worse, Ozna gets to pick integers $a$ and $b$, $a\ge 2015$ such that all numbers of the form $an+b$ will not be the starting integer, where $n$ is any positive integer. Find the minimum number of starting integer where Navi wins.

There are $64$ booths around a circular table and on each one there is a chip. The chips and the corresponding booths are numbered $1$ to $64$ in this order. At the center of the table there are $1996$ light bulbs which are all turned off. Every minute the chips move simultaneously in a circular way (following the numbering sense) as follows: chip $1$ moves one booth, chip $2$ moves two booths, etc., so that more than one chip can be in the same booth. At any minute, for each chip sharing a booth with chip $1$ a bulb is lit. Where is chip $1$ on the first minute in which all bulbs are lit?

Mugdho and Dipto play a game on a numbered row of $n \geq 5$ squares. At the beginning, a pebble is put on the first square and then the players make consecutive moves; Mugdho starts. During a move a player is allowed to choose one of the following: [list] [*] move the pebble one square rightward [*] move the pebble four squares rightward [*] move the pebble two squares leftward [/list] All of the possible moves are only allowed if the pebble stays within the borders of the square row. The player who moves the pebble to the last square (a. k. a $n$-th) wins. Determine for which values of $n$ each of the players has a winning strategy.

Two students $A$ and $B$ play a game on a $20 \text{ x } 20$ chessboard. It is known that two squares are said to be [i]adjacent[/i] if the two squares have a common side. At the beginning, there is a chess piece in a certain square of the chessboard. Given that $A$ will be the first one to move the chess piece, $A$ and $B$ will alternately move this chess piece to an adjacent square. Also, the common side of any pair of adjacent squares can only be passed once. If the opponent cannot move anymore, then he will be declared the winner (to clarify since the wording wasn’t that good, you lose if you can’t move). Who among $A$ and $B$ has a winning strategy? Justify your claim.