Found problems: 304
Consider $2009$ cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two player, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of $50$ consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins.
(a) Does the game necessarily end?
(b) Does there exist a winning strategy for the starting player?
[i]Proposed by Michael Albert, Richard Guy, New Zealand[/i]
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.
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 magician and his assistant are performing the following trick. There is a row of $13$ empty closed boxes. The magician leaves the room, and a person from the audience hides a coin in each of two boxes of his choice, so that the assistant knows which boxes contain coins. The magician returns and the assistant is allowed to open one box that does not contain a coin. Next, the magician selects four boxes, which are then simultaneously opened. The goal of the magician is to open both boxes that contain coins. Devise a method that will allow the magician and his assistant to always successfully perform the trick.
(Igor Zhizhilkin)
[url=https://artofproblemsolving.com/community/c6h1801447p11962869]junior version posted here[/url]
We have two piles with $2000$ and $2017$ coins respectively.
Ann and Bob take alternate turns making the following moves:
The player whose turn is to move picks a pile with at least two coins, removes from that pile $t$ coins for some $2\le t \le 4$, and adds to the other pile $1$ coin. The players can choose a different $t$ at each turn, and the player who cannot make a move loses.
If Ann plays first determine which player has a winning strategy.
The main building of ETH Zurich is a rectangle divided into unit squares. Every side of a square is a wall, with certain walls having doors. The outer wall of the main building has no doors. A number of participants of the SMO have gathered in the main building lost. You can only move from one square to another through doors. We have indicates that there is a walkable path between every two squares of the main building.
Cyril wants the participants to find each other again by having everyone on the same square leads. To do this, he can give them the following instructions via walkie-talkie: North, East, South or West. After each instruction, each participant simultaneously attempts a square in that direction to go. If there is no door in the corresponding wall, he remains standing.
Show that Cyril can reach his goal after a finite number of directions, no matter which one square the participants at the beginning.
[hide=original wording]Das Hauptgebäude der ETH Zürich ist ein in Einheitsquadrate unterteiltes Rechteck. Jede Seite eines Quadrates ist eine Wand, wobei gewisse Wände Türen haben. Die Aussenwand des Hauptgebäudes hat keine Türen. Eine Anzahl von Teilnehmern der SMO hat sich im Hauptgebäude verirrt. Sie können sich nur durch Türen von einem Quadrat zum anderen bewegen. Wir nehmen an, dass zwischen je zwei Quadraten des Hauptgebäudes ein begehbarer Weg existiert.
Cyril möchte erreichen, dass sich die Teilnehmer wieder nden, indem er alle auf dasselbe Quadrat führt. Dazu kann er ihnen per Walkie-Talkie folgende Anweisungen geben: Nord, Ost, Süd oder West. Nach jeder Anweisung versucht jeder Teilnehmer gleichzeitig, ein Quadrat in diese Richtung zu gehen. Falls in der entsprechenden Wand keine Türe ist, bleibt er stehen.
Zeige, dass Cyril sein Ziel nach endlich vielen Anweisungen erreichen kann, egal auf welchen Quadraten sich die Teilnehmer am Anfang benden. [/hide]
Sixty-four dice with the numbers ”one” to ”six” are placed on one table and formed into a square with eight horizontal and eight vertical rows of cubes pushed together. By rotating the dice, while maintaining their place, we want to finally have all sixty-four dice the "one" points upwards. Each dice however, may not be turned individually, but only every eight dice in a horizontal or vertical row together by $90^o$ to the longitudinal axis of this row may turn. Prove that it is always possible to solve the dice by repeatedly applying the permitted type of rotation to the required end position.
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]
The "Sea battle" game.
a) You are trying to find the $4$-field ship -- a rectangle $1x4$, situated on the $7x7$ playing board. You are allowed to ask a question, whether it occupies the particular field or not. How many questions is it necessary to ask to find that ship surely?
b) The same question, but the ship is a connected (i.e. its fields have common sides) set of $4$ fields.
There is $207$ boxes on the table which numbered $1,2, \dots , 207$ respectively. Firstly Aslı puts a red ball in each of the $100$ boxes that she chooses and puts a white ball in each of the remaining ones. After that Zehra, writes a pair $(i,j)$ on the blackboard such that $1\leq i \leq j \leq 207$. Finally, Aslı tells Zehra that for every pair; whether the color of the balls which is inside the box which numbered by these numbers are the same or not. Find the least possible value of $N$ such that Zehra can guarantee finding all colors that has been painted to balls in each of the boxes with writing $N$ pairs on the blackboard.
Agustín and Lucas, by turns, each time mark a box that has not yet been marked on a $101\times 101$ grid board. Augustine starts the game. You cannot check a box that already has two checked boxes in its row or column. The one who can't make his move loses. Decide which of the two players has a winning strategy.
Anna and Orjan play the following game: they start with a positive integer $n>1$, Anna writes it as the sum of two other positive integers, $n = n_1+n_2$. Orjan deletes one of them, $n_1$ or $n_2$. If the remaining number is larger than $1$, the process is repeated, i.e. Anna writes it as the sum of two positive integers, $ n_3+n_4$, Orjan deletes one of them etc. The game ends when the last number is $1$. Orjan is the winner if there are two equal numbers among the numbers he has deleted, otherwise Anna wins. Who is winning the game if n = 2008 and they both play optimally?
Let $n > 1$ be a fixed integer. Alberto and Barbara play the following game:
(i) Alberto chooses a positive integer,
(ii) Barbara chooses an integer greater than $1$ which is a multiple or submultiple of the number Alberto chose (including itself),
(iii) Alberto increases or decreases the Barbara’s number by $1$.
Steps (ii) and (iii) are alternatively repeated. Barbara wins if she succeeds to reach the number $n$ in at most $50$ moves. For which values of $n$ can she win, no matter how Alberto plays?
Two players Arnaldo and Betania play alternately, with Arnaldo being the first to play. Initially there are two piles of stones containing $x$ and $y$ stones respectively. In each play, it is possible to perform one of the following operations:
1. Choose two non-empty piles and take one stone from each pile.
2. Choose a pile with an odd amount of stones, take one of their stones and, if possible, split into two piles with the same amount of stones.
The player who cannot perform either of operations 1 and 2 loses.
Determine who has the winning strategy based on $x$ and $y$.
Alma and Bertha play the following game. There are $100$ round, $200$ triangular and $200$ square pieces on a table. In each move a player must remove two pieces, but it cannot be a triangle and a square. Alma starts, and one loses if one is unable to move or if there are no pieces left when it is one’s turn. Which player has a winning strategy?
The King gives the following task to his two wizards. The First Wizard should choose $7$ distinct positive integers with total sum $100$ and secretly submit them to the King. To the Second Wizard he should tell only the fourth largest number. The Second Wizard must figure out all the chosen numbers. Can the wizards succeed for sure? The wizards cannot discuss their strategy beforehand.
(Mikhail Evdokimov)
The numbers from $1$ to $500$ are written on the board. Two players $A$ and $B$ erase alternately one number at a time, and $A$ deletes the first number. If the sum of the last two number on the board is divisible by $3$, $B$ wins, otherwise $A$ wins. Which player can lay out a strategy that ensures this player's victory?
Two players $A$ and $B$ play alternatively in a convex polygon with $n \geq 5$ sides. In each turn, the corresponding player has to draw a diagonal that does not cut inside the polygon previously drawn diagonals. A player loses if after his turn, one quadrilateral is formed such that its two diagonals are not drawn. $A$ starts the game.
For each positive integer $n$, find a winning strategy for one of the players.
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?
(Game) At the beginning of the game the organisers place $4$ piles of paper disks onto the table. The player who is in turn takes away a pile, then divides one of the remaining piles into two nonempty piles. Whoever is unable to move, loses.
[i]Defeat the organisers in this game twice in a row! A starting position will be given and then you can decide whether you want to go first or second.[/i]
We are given a set $P$ of points and a set $L$ of straight lines. At the beginning there are 4 points, no three of which are collinear, and $L=\emptyset $. Two players are taking turns adding one or two lines to $L$, where each of these lines has to pass through at least two of the points in $P$. After that all intersection points of the lines in $L$ are added to $P$, if they are not already part of it. A player wins, if after his turn there are three collinear points from $P$, which lie on a line that isn’t from $L$. Find who of the two players has a winning strategy.
The numbers $3$, $4$ ,$...$, $2019$ are written on a blackboard. David and Edgardo play alternately, starting with David. On their turn, each player must erase a number from the board and write two positive integers whose sum is
equal to the number just deleted. The winner is the one who makes all the numbers on the board equal. Determine who has a winning strategy and describe it.
The following expression $x^{30} + *x^{29} +...+ *x+8 = 0$ is written on a blackboard. Two players $A$ and $B$ play the following game. $A$ starts the game. He replaces all the asterisks by the natural numbers from $1$ to $30$ (using each of them exactly once). Then player $B$ replace some of" $+$ "by ” $-$ "(by his own choice). The goal of $A$ is to get the equation having a real root greater than $10$, while the goal of $B$ is to get the equation having a real root less that or equal to $10$. If both of the players achieve their goals or nobody of them achieves his goal, then the result of the game is a draw. Otherwise, the player achieving his goal is a winner.
Who of the players wins if both of them play to win?
(I.Bliznets)
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]
Twenty cubes of the same size and appearance are made of either aluminum or of heavier duralumin. How can one find the number of duralumin cubes using not more than $11$ weighings on a balance without weights? (We assume that all cubes can be made of aluminum, but not all of duralumin.)