Nicholas and Peter are dividing $(2n+1)$ nuts. Each wants to get more. Three ways for that were suggested. (Each consist of three stages.) First two stages are common. 1 stage: Peter divides nuts onto $2$ heaps, each contain not less than $2$ nuts. 2 stage: Nicholas divides both heaps onto $2$ heaps, each contain not less than $1$ nut. 3 stage: 1 way: Nicholas takes the biggest and the least heaps. 2 way: Nicholas takes two middle size heaps. 3 way: Nicholas takes either the biggest and the least heaps or two middle size heaps, but gives one nut to the Peter for the right of choice. Find the most and the least profitable method for the Nicholas.

On each of the 2006 cards a natural number is written. Cards are placed arbitrarily in a row. 2 players take in turns a card from any end of the row until all the cards are taken. After that each player calculates sum of the numbers written of his cards. If the sum of the first player is not less then the sum of the second one then the first player wins. Show that there's a winning strategy for the first player.

The numbers $1, 2, ..., 2020$ and $2021$ are written on a blackboard. The following operation is executed: Two numbers are chosen, both are erased and replaced by the absolute value of their difference. This operation is repeated until there is only one number left on the blackboard. (a) Show that $2021$ can be the final number on the blackboard. (b) Show that $2020$ cannot be the final number on the blackboard. (Karl Czakler)

There are two bowls on the table, in one there are $p$, in the other $q$ stones ($p, q \in N*$ ). Two players $A$ and $B$ take turns playing, starting with $A$. Who's turn: $\bullet$ takes a stone from one of the bowls $\bullet$or removes one stone from each bowl $\bullet$ or puts a stone from one of the bowls into the other. Whoever takes the last stone wins. Under what conditions can $A$ and under what conditions can $B$ force the win? The answer must be justified.

Pedro and Tiago are playing a game with a deck of n cards, numbered from $1$ to $n$. Starting with Pedro, they choose cards alternately, and receive the number of points indicated by the cards. However, whenever the player chooses the card with the highest number among those remaining in the deck, he is forced to pass his next turn, not choosing any card. When the deck runs out, the player with the most points wins. Knowing that Tiago can at least draw, regardless of Pedro's moves, how many cards are in the deck? Indicates all possibilities,

Susana and Brenda play a game writing polynomials on the board. Susana starts and they play taking turns. 1) On the preparatory turn (turn 0), Susana choose a positive integer $n_0$ and writes the polynomial $P_0(x)=n_0$. 2) On turn 1, Brenda choose a positive integer $n_1$, different from $n_0$, and either writes the polynomial $$P_1(x)=n_1x+P_0(x) \textup{ or } P_1(x)=n_1x-P_0(x)$$ 3) In general, on turn $k$, the respective player chooses an integer $n_k$, different from $n_0, n_1, \ldots, n_{k-1}$, and either writes the polynomial $$P_k(x)=n_kx^k+P_{k-1}(x) \textup{ or } P_k(x)=n_kx^k-P_{k-1}(x)$$ The first player to write a polynomial with at least one whole whole number root wins. Find and describe a winning strategy.

Let $N\geq 4$ be a fixed positive integer. Two players, $A$ and $B$ are forming an ordered set $\{x_1,x_2,...\},$ adding elements alternatively. $A$ chooses $x_1$ to be $1$ or $-1,$ then $B$ chooses $x_2$ to be $2$ or $-2,$ then $A$ chooses $x_3$ to be $3$ or $-3,$ and so on. (at the $k^{th}$ step, the chosen number must always be $k$ or $-k$) The winner is the first player to make the sequence sum up to a multiple of $N.$ Depending on $N,$ find out, with proof, which player has a winning strategy.

Two players in turn put two or three coins into their own hats (before the game starts, the hats are empty). Each time, after the second player duplicated the move of the first player, they exchange hats. The player wins, if after his move his hat contains one hundred or more coins. Which player has a winning strategy?

a) Given a convex hexagon $ABCDEF$, which has a center of symmetry. Prove that the perimeter of triangle $ACE$ is greater than half the perimeter of hexagon $ABCDEF$. b) Given a convex $(2n)$-gon $P$ having a center of symmetry, its vertices are colored alternately red and blue. Let $Q$ be an $n$-gon with red vertices. Is it possible to say that the perimeter of $Q$ is certainly greater than half the perimeter $P$? Solve the problem for $n = 4$ and $n = 5$.

The vertices of the regular $n$-gon are marked. Two players play the following game: they, in turn, select a vertex and connect it by a segment to either the adjacent vertex or the center of the $n$-gon. The winner is a player if after his move it is possible to get any vertex from any other vertex moving along segments. For each integer $n\geqslant 3$ determine who has a winning strategy.

Juan and Pedro play alternately on the given grid. Each one in turn traces $1$ to $5$ routes different from the ones outlined above, that join $A$ with $B$, moving only to the right and upwards on the grid lines. Juan starts playing. The one who traces a route that passes through $C$ or $D$ loses. Prove that one of them can win regardless of how the other plays. [img]https://cdn.artofproblemsolving.com/attachments/2/7/6a24ca9c4c1c710bd41e44bfcab3d3b61b6d4f.png[/img]

Two players are playing a turn based game on a $n \times n$ chessboard. At the beginning, only the bottom left corner of the chessboard contains a piece. At each turn, the player moves the piece to either the square just above, or the square just right, or the diagonal square just right-top. If a player cannot make a move, he loses the game. The game is played once on each $6\times 7$, $6 \times 8$, $7 \times 7$, $7 \times 8$, and $8 \times 8$ chessboard. In how many of them, can the first player guarantee to win? $ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \text{None} $

In two wooden boxes, there are $1994$ and $2024$ marbles, respectively. Spiro and Cvetko play the following game: alternately, each player takes a turn and removes some marbles from one of the boxes, so that the number of removed marbles in that turn is a divisor of the current number of marbles in the other box. The winner of the game is the one after whose turn both boxes are empty. Spiro takes the first turn. Which of the players has a winning strategy?

Player Zero and Player One play a game on a $n \times n$ board ($n \ge 1$). The columns of this $n \times n$ board are numbered $1,2,4,\dots,2^{n-1}$. Turn my turn, the players put their own number in one of the free cells (thus Player Zero puts a $0$ and Player One puts a $1$). Player Zero begins. When the board is filled, the game ends and each row yields a (reverse binary) number obtained by adding the values of the columns with a $1$ in that row. For instance, when $n=4$, a row with $0101$ yields the number $0 \cdot1+1 \cdot 2+0 \cdot 4+1 \cdot 8=10$. a) For which natural numbers $n$ can Player One always ensure that at least one of the row numbers is divisible by $4$? b) For which natural numbers $n$ can Player One always ensure that at least one of the row numbers is divisible by $3$?

There are $50$ coins in a row; each coin has a value. Two people are playing a game alternating moves. In one move a player can take either the leftmost or the rightmost coin. Who can always accumulate coins whose total value is at least the value of the coins of the opponent?

For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with non-overlapping trominos? When it is possible, what is the minimum number of trominos needed?