In a company of seven people, any six can sit at a round table so that every two neighbors turn out to be acquaintances. Prove that the whole company can be seated at a round table so that every two neighbors turn out to be acquaintances.

There are $2018$ peoples. We call the group of people as "club" if all members of same "club" are all friends, but not friends with a nonmember of "club". Prove, that we can divide peoples for $90$ rooms, such that no one room has all members of some "club".

Let $n \geq 3$ be a positive integer. A [i]chipped $n$-board[/i] is a $2 \times n$ checkerboard with the bottom left square removed. Lino wants to tile a chipped $n$-board and is allowed to use the following types of tiles: [list] [*] Type 1: any $1 \times k$ board where $1 \leq k \leq n$ [*] Type 2: any chipped $k$-board where $1 \leq k \leq n$ that must cover the left-most tile of the $2 \times n$ checkerboard. [/list] Two tilings $T_1$ and $T_2$ are considered the same if there is a set of consecutive Type 1 tiles in both rows of $T_1$ that can be vertically swapped to obtain the tiling $T_2$. For example, the following three tilings of a chipped $7$-board are the same: [img]http://i.imgur.com/8QaSgc0.png[/img] For any positive integer $n$ and any positive integer $1 \leq m \leq 2n - 1$, let $c_{m,n}$ be the number of distinct tilings of a chipped $n$-board using exactly $m$ tiles (any combination of tile types may be used), and define the polynomial $$P_n(x) = \sum^{2n-1}_{m=1} c_{m,n}x^m.$$ Find, with justification, polynomials $f(x)$ and $g(x)$ such that $$P_n(x) = f(x)P_{n-1}(x) + g(x)P_{n-2}(x)$$ for all $n \geq 3$.

We have $ n \geq 2$ lamps $ L_{1}, . . . ,L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp $ L_{i}$ and its neighbours (only one neighbour for $ i \equal{} 1$ or $ i \equal{} n$, two neighbours for other $ i$) are in the same state, then $ L_{i}$ is switched off; – otherwise, $ L_{i}$ is switched on. Initially all the lamps are off except the leftmost one which is on. $ (a)$ Prove that there are infinitely many integers $ n$ for which all the lamps will eventually be off. $ (b)$ Prove that there are infinitely many integers $ n$ for which the lamps will never be all off.

$5.$Let x be a real number with $0<x<1$ and let $0.c_1c_2c_3...$ be the decimal expansion of x.Denote by $B(x)$ the set of all subsequences of $c_1c_2c_3$ that consist of 6 consecutive digits. For instance , $B(\frac{1}{22})={045454,454545,545454}$ Find the minimum number of elements of $B(x)$ as $x$ varies among all irrational numbers with $0<x<1$

$ 30$ students participated in the mathematical Olympiad. Each of them was given $ 8$ problems to solve. Jury estimated their work with the following rule: 1) Each problem was worth $ k$ points, if it wasn't solved by exactly $ k$ students; 2) Each student received the maximum possible points in each problem or got $ 0$ in it; Lasha got the least number of points. What's the maximal number of points he could have? Remark: 1) means that if the problem was solved by exactly $ k$ students, than each of them got $ 30 \minus{} k$ points in it.

In the adjacent figure, can the numbers $1,2,3, 4,..., 18$ be placed, one on each line segment, such that the sum of the numbers on the three line segments meeting at each point is divisible by $3$?

The numbers $1, 2, 3, . . . , 101$ are written in a row in some order. Prove that it is always possible to erase $90 $ of the numbers so that the remaining $11$ numbers remain arranged in either increasing or decreasing order.

An ant is walking on a sidewalk and discovers $12$ sidewalk panels with leaves inscribed in them, as shown below. Find the number of ways in which the ant can traverse from point $A$ to point $B$ if it can only move [list] [*] up, down, or right (along the border of a sidewalk panel), or [*] up-right (along one of two margin halves of a leaf) [/list] and cannot visit any border or margin half more than once (an example path is highlighted in red). [asy] unitsize(1cm); int r = 4; int c = 5; for (int i = 0; i < r; ++i) { for (int j = 0; j < c; ++j) { pair A = (j,i); } } for (int i = 0; i < r; ++i) { for (int j = 0; j < c; ++j) { if (j != c-1) { draw((j,i)--(j+1,i)); } if (i != r-1) { draw((j,i)--(j,i+1)); } } } for (int i = 1; i < r+1; ++i) { for (int j = 0; j < c-2; ++j) { fill(arc((i,j),1,90,180)--cycle,deepgreen); fill(arc((i-1,j+1),1,270,360)--cycle,deepgreen); draw((i-1,j)--(i,j+1), heavygreen+linewidth(0.5)); draw((i-2/3,j+1/3)--(i-2/3,j+1/3+0.1), heavygreen); draw((i-1/3,j+2/3)--(i-1/3,j+2/3+0.1), heavygreen); draw((i-2/3,j+1/3)--(i-2/3+0.1,j+1/3), heavygreen); draw((i-1/3,j+2/3)--(i-1/3+0.1,j+2/3), heavygreen); draw(arc((i,j),1,90,180)); draw(arc((i-1,j+1),1,270,360)); } } draw((0,3)--(0,1), red+linewidth(1.5)); draw((0,3)--(0,1), red+linewidth(1.5)); draw(arc((1,1),1,90,180), red+linewidth(1.5)); draw((1,2)--(1,1)--(2,1), red+linewidth(1.5)); draw(arc((2,2),1,270,360), red+linewidth(1.5)); draw(arc((4,2),1,90,180), red+linewidth(1.5)); draw((4,3)--(4,0), red+linewidth(1.5)); dot((0,3)); dot((4,0)); label("$A$", (0,3), NW); label("$B$", (4,0), SE); [/asy]

A regular hexagon in the plane is called sweet if its area is equal to $1$. Is it possible to place $2000000$ sweet hexagons in the plane such that the union of their interiors is a convex polygon of area at least $1900000$? Remark: A subset $S$ of the plane is called convex if for every pair of points in $S$, every point on the straight line segment that joins the pair of points also belongs to $S$. The hexagons may overlap.

Let $ n \geq 2$ be a positive integer and $ \lambda$ a positive real number. Initially there are $ n$ fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points $ A$ and $ B$, with $ A$ to the left of $ B$, and letting the flea from $ A$ jump over the flea from $ B$ to the point $ C$ so that $ \frac {BC}{AB} \equal{} \lambda$. Determine all values of $ \lambda$ such that, for any point $ M$ on the line and for any initial position of the $ n$ fleas, there exists a sequence of moves that will take them all to the position right of $ M$.

There are $6$ pieces of cheese of different weights. For any two pieces we can identify the heavier piece. Given that it is possible to divide them into two groups of equal weights with three pieces in each. Give the explicit way to find these groups by performing two weightings on a regular balance.

Let $n$ be a positive integer. A Steiner tree associated with a finite set $S$ of points in the Euclidean $n$-space is a finite collection $T$ of straight-line segments in that space such that any two points in $S$ are joined by a unique path in $T$ , and its length is the sum of the segment lengths. Show that there exists a Steiner tree of length $1+(2^{n-1}-1)\sqrt{3}$ associated with the vertex set of a unit $n$-cube.

We consider permutations $f$ of the set $\mathbb{N}$ of non-negative integers, i.e. bijective maps $f$ from $\mathbb{N}$ to $\mathbb{N}$, with the following additional properties: \[f(f(x)) = x \quad \mbox{and}\quad \left| f(x)-x\right| \leqslant 3\quad\mbox{for all } x \in\mathbb{N}\mbox{.}\] Further, for all integers $n > 42$, \[\left.M(n)=\frac{1}{n+1}\sum_{j=0}^n \left|f(j)-j\right|<2,011\mbox{.}\right.\] Show that there are infinitely many natural numbers $K$ such that $f$ maps the set \[\left\{ n\mid 0\leqslant n\leqslant K\right\}\] onto itself.

Let $a_1,a_2, \dots a_n$ be positive real numbers. Define $b_1,b_2, \dots b_n$ as follows. \begin{align*} b_1&=a_1 \\ b_2&=max(a_1,a_2)\\ b_i&=max(b_{i-1},b_{i-2}+a_i) \text{ for } i=3,4 \dots n \end{align*} Also define $c_1,c_2 \dots c_n$ as follows. \begin{align*} c_n&=a_n \\ c_{n-1}&=max(a_n,a_{n-1})\\ c_i&=max(c_{i+1},c_{i+2}+a_i) \text{ for } i=n-2,n-3 \dots 1 \end{align*} Prove that $b_n=c_1$.\\

Numbers $m$ and $n$ are given positive integers. There are $mn$ people in a party, standing in the shape of an $m\times n$ grid. Some of these people are police officers and the rest are the guests. Some of the guests may be criminals. The goal is to determine whether there is a criminal between the guests or not.\\ Two people are considered \textit{adjacent} if they have a common side. Any police officer can see their adjacent people and for every one of them, know that they're criminal or not. On the other hand, any criminal will threaten exactly one of their adjacent people (which is likely an officer!) to murder. A threatened officer will be too scared, that they deny the existence of any criminal between their adjacent people.\\ Find the least possible number of officers such that they can take position in the party, in a way that the goal is achievable. (Note that the number of criminals is unknown and it is possible to have zero criminals.) [i]Proposed by Abolfazl Asadi[/i]

In the country of [i]Sugarland[/i], there are $13$ students in the IMO team selection camp. $6$ team selection tests were taken and the results have came out. Assume that no students have the same score on the same test.To select the IMO team, the national committee of math Olympiad have decided to choose a permutation of these $6$ tests and starting from the first test, the person with the highest score between the remaining students will become a member of the team.The committee is having a session to choose the permutation. Is it possible that all $13$ students have a chance of being a team member? [i]Proposed by Morteza Saghafian[/i]

Find the smallest positive integer $k$ such that, for any subset $A$ of $S=\{1,2,\ldots,2012\}$ with $|A|=k$, there exist three elements $x,y,z$ in $A$ such that $x=a+b$, $y=b+c$, $z=c+a$, where $a,b,c$ are in $S$ and are distinct integers. [i]Proposed by Huawei Zhu[/i]

A sequence of real numbers is called a [i]Fibonacci [/i] sequence if $$t_{n+2} = t_{n+1} + t_n$$ for $n= 1,2,3,. .$ . Two Fibonacci sequences are said to be [i]essentially different[/i] if the terms of one sequence cannot be obtained by multiplying the terms of the other by a constant. For example, the Fibonacci sequences $1,2,3,5,8,...$ and $1,3,4,7,11,...$ are essentially different, but the sequences $1,2,3,5,8,...$ and $2,4,6,10,16,...$ are not. (a) Prove that there exist real numbers $p$ and $q$ such that the sequences $1,p,p^2,p^3,...$ and $1,q,q^2,q^3,...$ are essentially different Fibonacci sequences. (b) Let $a_1,a_2,a_3,...$ and $b_1,b_2,b_3,...$ be essentially different Fibonacci sequences. Prove that for every Fibonacci sequence $t_1,t_2,t_3,...$, there exists exactly one number $\alpha$ and exactly one number $\beta$, such that: $$t_n = \alpha a_n + \beta b_n$$ for $n = 1,2,3,...$ (c) $t_1,t_2,t_3,...$, is the Fibonacci sequence with $t_1 = 1$ and $t_2= 2$. Express $t_n$ in terms of $n$.

A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$, with the sides parallel to those of the box, then exactly $40$ percent of the volume of the box is occupied. Determine the possible dimensions of the box.

Your friend sitting to your left (or right?) is unable to solve any of the eight problems on his or her Combinatorics $B$ test, and decides to guess random answers to each of them. To your astonishment, your friend manages to get two of the answers correct. Assuming your friend has equal probability of guessing each of the questions correctly, what is the average possible value of your friend’s score? Recall that each question is worth the point value shown at the beginning of each question.

Let $n$ be a given positive integer. Luca, the lazy flea sits on one of the vertices of a regular $2n$-gon. For each jump, Luca picks an axis of symmetry of the polygon, and reflects herself on the chosen axis of symmetry. Let $P(n)$ denote the number of different ways Luca can make $2n$ jumps such that she returns to her original position in the end, and does not pick the same axis twice. (It is possible that Luca's jump does not change her position, however, it still counts as a jump.) [b]a)[/b] Find the value of $P(n)$ if $n$ is odd. [b]b)[/b] Prove that if $n$ is even, then \[P(n)=(n-1)!\cdot n!\cdot \sum_{d\mid n}\left(\varphi\left(\frac{n}d\right)\binom{2d}{d}\right).\] [i]Proposed by Péter Csikvári and Kartal Nagy, Budapest[/i]

Let $x$ and $y$ be positive integers we have a table $x\times y$ where $(x + 1)(y + 1)$ points are red(the points are the vertices of the squares). Initially there is one ant in each red point, in a moment the ants walk by the lines of the table with same speed, each turn that an ant arrive in a red point the ant moves $90º$ to some direction. Determine all values of $x$ and $y$ where is possible that the ants move indefinitely where can't be in any moment two ants in the same red point.

There is a central train station in point $O$, which is connected to other train stations $A_1, A_2, \ldots, A_8$ with tracks. There is also a track between stations $A_i$ and $A_{i+1}$ for each $i$ from $1$ to $8$ (here $A_9 = A_1$). The length of each track $A_iA_{i+1}$ is equal to $1$, and the length of each track $OA_i$ is equal to $2$, for each $i$ from $1$ to $8$. There are also $8$ trains $B_1, B_2, \ldots, B_8$, with speeds $1, 2, \ldots, 8$ correspondently. Trains can move only by the tracks above, in both directions. No time is wasted on changing directions. If two or more trains meet at some point, they will move together from now on, with the speed equal to that of the fastest of them. Is it possible to arrange trains into stations $A_1, A_2, \ldots, A_8$ (each station has to contain one train initially), and to organize their movement in such a way, that all trains arrive at $O$ in time $t < \frac{1}{2}$? [i](Proposed by Bogdan Rublov)[/i]

A certain frog that was placed on a vertex of a convex polygon chose to jump to another vertex, either clockwise skipping one vertex, either counterclockwise skipping two vertexes, and repeated the procedure. If the number of jumps that the frog made is equal to the number of sides of the polygon, the frog has passed through all its vertexes and ended up on the initial vertex, what´s the set formed by all the possible values that this number can take? [i]Andrei Eckstein[/i]