A dance with 2018 couples takes place in Havana. For the dance, 2018 distinct points labeled $0, 1,\ldots, 2017$ are marked in a circumference and each couple is placed on a different point. For $i\geq1$, let $s_i=i\ (\textrm{mod}\ 2018)$ and $r_i=2i\ (\textrm{mod}\ 2018)$. The dance begins at minute $0$. On the $i$-th minute, the couple at point $s_i$ (if there's any) moves to point $r_i$, the couple on point $r_i$ (if there's any) drops out, and the dance continues with the remaining couples. The dance ends after $2018^2$ minutes. Determine how many couples remain at the end. Note: If $r_i=s_i$, the couple on $s_i$ stays there and does not drop out.

There is a board of $2008\times 2008$ and $2008$ pieces, one in each row and each column of the board. It is allowed to do one of the following movements: a) Take two steps to the right and $10$ up. b) Take two steps to the right and $6$ steps down. c) Take two steps to the left and $6$ steps up. d) Take two steps to the left and $10$ steps down. If the path down cannot be completed, it is skipped to the upper part along the same column and the route continues normally, similarly in the other directions. In each play you will move a checker using any of the allowed operations. Would it be possible that at some point, after a finite number of played, the pieces are located forming a square of side $44$ in the upper left corner of the board and the remaining $72$ are in the last row in the first $72$ boxes?

Let $X$ be a set containing $n$ elements. Find the number of ordered triples $(A,B, C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C$.

Let $n$ be a positive integer. Let $P_n=\{2^n,2^{n-1}\cdot 3, 2^{n-2}\cdot 3^2, \dots, 3^n \}.$ For each subset $X$ of $P_n$, we write $S_X$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \leq y \leq 3^{n+1}-2^{n+1}.$ Prove that there is a subset $Y$ of $P_n$ such that $0 \leq y-S_Y < 2^n$

Assume that the following generating function equation is correct, prove the following statement: $\Pi_{i=1}^{\infty} (1+x^{3i})\Pi_{j=1}^{\infty} (1-x^{6j+3})=1$ Statement: The number of partitions of $n$ to numbers not of the form $6k+1$ or $6k-1$ is equal to the number of partitions of $n$ in which each summand appears at least twice. (10 points) [i]Proposed by Morteza Saghafian[/i]

Each of $A$, $B$, $C$, $D$, $E$, and $F$ knows a piece of gossip. They communicate by telephone via a central switchboard, which can connect only two of them at a time. During a conversation, each side tells the other everything he or she knows at that point. Determine the minimum number of calls for everyone to know all six pieces of gossip.

In a board of $2000\times2001$ squares with integer coordinates $(x,y)$, $0\leq{x}\leq1999$ and $0\leq{y}\leq2000$. A ship in the table moves in the following way: before a move, the ship is in position $(x,y)$ and has a velocity of $(h,v)$ where $x,y,h,v$ are integers. The ship chooses new velocity $(h^\prime,v^\prime)$ such that $h^\prime-h,v^\prime-v\in\{-1,0,1\}$. The new position of the ship will be $(x^\prime,y^\prime)$ where $x^\prime$ is the remainder of the division of $x+h^\prime$ by $2000$ and $y^\prime$ is the remainder of the division of $y+v^\prime$ by $2001$. There are two ships on the board: The Martian ship and the Human trying to capture it. Initially each ship is in a different square and has velocity $(0,0)$. The Human is the first to move; thereafter they continue moving alternatively. Is there a strategy for the Human to capture the Martian, independent of the initial positions and the Martian’s moves? [i]Note[/i]: The Human catches the Martian ship by reaching the same position as the Martian ship after the same move.

[b]p1.[/b] Let $n$ be the number so that $1 - 2 + 3 - 4 + ... - (n - 1) + n = 2012$. What is $4^{2012}$ (mod $n$)? [b]p2. [/b]Consider three unit squares placed side by side. Label the top left vertex $P$ and the bottom four vertices $A,B,C,D$ respectively. Find $\angle PBA + \angle PCA + \angle PDA$. [b]p3.[/b] Given $f(x) = \frac{3}{x-1}$ , then express $\frac{9(x^2-2x+1)}{x^2-8x+16}$ entirely in terms of $f(x)$. In other words, $x$ should not be in your answer, only $f(x)$. [b]p4.[/b] Right triangle with right angle $B$ and integer side lengths has $BD$ as the altitude. $E$ and $F$ are the incenters of triangles $ADB$ and $BDC$ respectively. Line $EF$ is extended and intersects $BC$ at $G$, and $AB$ at $H$. If $AB = 15$ and $BC = 8$, find the area of triangle $BGH$. [b]p5.[/b] Let $a_1, a_2, ..., a_n$ be a sequence of real numbers. Call a $k$-inversion $(0 < k\le n)$ of a sequence to be indices $i_1, i_2, .. , i_k$ such that $i_1 < i_2 < .. < i_k$ but $a_{i1} > a_{i2} > ...> a_{ik}$ . Calculate the expected number of $6$-inversions in a random permutation of the set $\{1, 2, ... , 10\}$. [b]p6.[/b] Chell is given a strip of squares labeled $1, .. , 6$ all placed side to side. For each $k \in {1, ..., 6}$, she then chooses one square at random in $\{1, ..., k\}$ and places a Weighted Storage Cube there. After she has placed all $6$ cubes, she computes her score as follows: For each square, she takes the number of cubes in the pile and then takes the square (i.e. if there were 3 cubes in a square, her score for that square would be $9$). Her overall score is the sum of the scores of each square. What is the expected value of her score? PS. You had better use hide for answers.

Consider $4n$ points in the plane such that no three of them are collinear ($n\geq 1$). Show that the set of centroids of all the triangles formed by any three of these points contains at least $4n$ elements. [i]Radu Bumbăcea[/i]

On the board are written $n$ natural numbers, $n\in \mathbb{N}$. In one move it is possible to choose two equal written numbers and increase one by $1$ and decrease the other by $1$. Prove that in this the game cannot be played more than $\frac{n^3}{6}$ moves.

There are $100$ cities in Matland. Every road in Matland connects two cities, does not pass through any other city and does not form crossroads with other roads (although roads can go through tunnels one after the other). Driving in Matlandia by road, it is possible to get from any city to any other. Prove that that it is possible to repair some of the roads of Matlandia so that from an odd number of repaired roads would go in each city.

Integers not divisible by $2012$ are arranged on the arcs of an oriented graph. We call the [i]weight of a vertex [/i]the difference between the sum of the numbers on the arcs coming into it and the sum of the numbers on the arcs going away from it. It is known that the weight of each vertex is divisible by $2012$. Prove that non-zero integers with absolute values not exceeding $2012$ can be arranged on the arcs of this graph, so that the weight of each vertex is zero. [i]Proposed by W. Tutte[/i]

All vertices of a hexagon, whose sides may intersect at points other than the vertices, lie on a circle. (a) Draw a hexagon such that it has the largest possible number of points of self-intersection. (b) Prove that this number is indeed maximum. (NB Vassiliev)

We are given a natural number $k$. Let us consider the following game on an infinite onedimensional board. At the start of the game, we distrubute $n$ coins on the fields of the given board (one field can have multiple coins on itself). After that, we have two choices for the following moves: $(i)$ We choose two nonempty fields next to each other, and we transfer all the coins from one of the fields to the other. $(ii)$ We choose a field with at least $2$ coins on it, and we transfer one coin from the chosen field to the $k-\mathrm{th}$ field on the left , and one coin from the chosen field to the $k-\mathrm{th}$ field on the right. $\mathbf{(a)}$ If $n\leq k+1$, prove that we can play only finitely many moves. $\mathbf{(b)}$ For which values of $k$ we can choose a natural number $n$ and distribute $n$ coins on the given board such that we can play infinitely many moves.

Prove that for every positive integer $t$ there is a unique permutation $a_0, a_1, \ldots , a_{t-1}$ of $0, 1, \ldots , t-1$ such that, for every $0 \leq i \leq t-1$, the binomial coefficient $\binom{t+i}{2a_i}$ is odd and $2a_i \neq t+i$.

You are responsible for arranging a banquet for an agency. In the agency, some pairs of agents are enemies. A group of agents are called [i]avengers[/i], if and only if the number of agents in the group is odd and at least $3$, and it is possible to arrange all of them around a round table so that every two neighbors are enemies. You figure out a way to assign all agents to $11$ tables so that any two agents on the same tables are not enemies, and that’s the minimum number of tables you can get. Prove that there are at least $2^{10}-11$ avengers in the agency. This problem is adapted from 2015 IMO Shortlist C7.

Given is a convex $2024$-gon $A_1A_2\ldots A_{2024}$ and $1000$ points inside it, so that no three points are collinear. Some pairs of the points are connected with segments so that the interior of the polygon is divided into triangles. Every point is assigned one number among $\{1, -1, 2, - 2\}$, so that the sum of the numbers written in $A_i$ and $A_{i+1012}$ is zero for all $i=1,2, \ldots, 1012$. Prove that there is a triangle, such that the sum of the numbers in some two of its vertices is zero. [hide=Remark on source of 11.3] It appears as Estonia TST 2004/5, so it will not be posted.

The vertices of a regular polygon with $N$ sides are marked on the blackboard. Ana and Beto play alternately, Ana begins. Each player, in turn, must do the following: $\bullet$ join two vertices with a segment, without cutting another already marked segment; or $\bullet$ delete a vertex that does not belong to any marked segment. The player who cannot take any action on his turn loses the game. Determine which of the two players can guarantee victory: a) if $N=28$ b) if $N=29$

For each finite set $ U$ of nonzero vectors in the plane we define $ l(U)$ to be the length of the vector that is the sum of all vectors in $ U.$ Given a finite set $ V$ of nonzero vectors in the plane, a subset $ B$ of $ V$ is said to be maximal if $ l(B)$ is greater than or equal to $ l(A)$ for each nonempty subset $ A$ of $ V.$ (a) Construct sets of 4 and 5 vectors that have 8 and 10 maximal subsets respectively. (b) Show that, for any set $ V$ consisting of $ n \geq 1$ vectors the number of maximal subsets is less than or equal to $ 2n.$

Prove that for arbitary positive integer $ n\geq 4$, there exists a permutation of the subsets that contain at least two elements of the set $ G_{n} \equal{} \{1,2,3,\cdots,n\}$: $ P_{1},P_{2},\cdots,P_{2^n \minus{} n \minus{} 1}$ such that $ |P_{i}\cap P_{i \plus{} 1}| \equal{} 2,i \equal{} 1,2,\cdots,2^n \minus{} n \minus{} 2.$

Find the number of ways to color $n \times m$ board with white and black colors such that any $2 \times 2$ square contains the same number of black and white cells.

Given any six points in the plane, no three collinear, show that we can always find three which form an obtuse-angled triangle with one angle at least $120^o$.

we named a $n*n$ table $selfish$ if we number the row and column with $0,1,2,3,...,n-1$.(from left to right an from up to down) for every {$ i,j\in{0,1,2,...,n-1}$} the number of cell $(i,j)$ is equal to the number of number $i$ in the row $j$. for example we have such table for $n=5$ 1 0 3 3 4 1 3 2 1 1 0 1 0 1 0 2 1 0 0 0 1 0 0 0 0 prove that for $n>5$ there is no $selfish$ table

A $10 \times 10$ table consists of $100$ unit cells. A [i]block[/i] is a $2 \times 2$ square consisting of $4$ unit cells of the table. A set $C$ of $n$ blocks covers the table (i.e. each cell of the table is covered by some block of $C$ ) but no $n -1$ blocks of $C$ cover the table. Find the largest possible value of $n$.

There are $n$ cities in a country, one of which is the capital. An airline operates bi-directional flights between some pairs of cities such that one can reach any city from any other city. The airline wants to close down some (possibly zero) number of flights, such that the number of flights needed to reach any particular city from the capital does not increase. Suppose that there are an odd number of ways that the airline can do this. Prove that the set of cities can be divided into two groups, such that there is no flight between two cities of the same group. [i]Proposed by Pranjal Srivastava[/i]