Given is $2022\times 2022$ cells table. We can select $4$ cells, such that they make the figure $L$ (rotations, symmetric still count) (left one) and put a ball in each of them, or select $4$ cell which makes up the right figure (rotations, symmetric still count) and get one ball from each of them. For which $k$ is it possible in a given moment to be exactly $k$ points in each of the cells

A natural number is written on the blackboard. In each step, we erase the units digit and add four times the erased digit to the remaining number, and write the result on the blackboard instead of the initial number. Starting with the number $13^{2006}$, is it possible to obtain the number $2006^{13}$ by repeating this step finitely many times?

It is given $5$ numbers $1$, $3$, $5$, $7$, $9$. We get the new $5$ numbers such that we take arbitrary $4$ numbers(out of current $5$ numbers) $a$, $b$, $c$ and $d$ and replace them with $\frac{a+b+c-d}{2}$, $\frac{a+b-c+d}{2}$, $\frac{a-b+c+d}{2}$ and $\frac{-a+b+c+d}{2}$. Can we, with repeated iterations, get numbers: $a)$ $0$, $2$, $4$, $6$ and $8$ $b)$ $3$, $4$, $5$, $6$ and $7$

The only pieces on an $8\times8$ chessboard are three rooks. Each moves along a row or a column without running to or jumping over another rook. The white rook starts at the bottom left corner, the black rook starts in the square directly above the white rook, and the red rook starts in the square directly to the right of the white rook. The white rook is to finish at the top right corner, the black rook in the square directly to the left of the white rook, and the red rook in the square directly below the white rook. At all times, each rook must be either in the same row or the same column as another rook. Is it possible to get the rooks to their destinations?

Consider a quartet of positive numbers $(a,b,c,d)$. In one step, we transform it to $(ab,bc,cd,da)$. Prove that you can never obtain the initial set if neither of $a,b,c,d$ is $1$.

Given a positive integer $k$ and other two integers $b > w > 1.$ There are two strings of pearls, a string of $b$ black pearls and a string of $w$ white pearls. The length of a string is the number of pearls on it. One cuts these strings in some steps by the following rules. In each step: [b](i)[/b] The strings are ordered by their lengths in a non-increasing order. If there are some strings of equal lengths, then the white ones precede the black ones. Then $k$ first ones (if they consist of more than one pearl) are chosen; if there are less than $k$ strings longer than 1, then one chooses all of them. [b](ii)[/b] Next, one cuts each chosen string into two parts differing in length by at most one. (For instance, if there are strings of $5, 4, 4, 2$ black pearls, strings of $8, 4, 3$ white pearls and $k = 4,$ then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts $(4,4), (3,2), (2,2)$ and $(2,2)$ respectively.) The process stops immediately after the step when a first isolated white pearl appears. Prove that at this stage, there will still exist a string of at least two black pearls. [i]Proposed by Bill Sands, Thao Do, Canada[/i]

Let $ A$, $ B$, $ C$, $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$, $ X$, $ Y$, $ Z$, $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ and $ P$ be pairwise distinct points in space such that $ A^{\prime} \in BC;\ B^{\prime}\in CA;\ C^{\prime}\in AB;\ X^{\prime}\in YZ;\ Y^{\prime}\in ZX;\ Z^{\prime}\in XY;$ $ P \in AX;\ P\in BY;\ P\in CZ;\ P\in A^{\prime}X^{\prime};\ P\in B^{\prime}Y^{\prime};\ P\in C^{\prime}Z^{\prime}$. Prove that $ \frac {BA^{\prime}}{A^{\prime}C}\cdot\frac {CB^{\prime}}{B^{\prime}A}\cdot\frac {AC^{\prime}}{C^{\prime}B} \equal{} \frac {YX^{\prime}}{X^{\prime}Z}\cdot\frac {ZY^{\prime}}{Y^{\prime}X}\cdot\frac {XZ^{\prime}}{Z^{\prime}Y}$.

Let $ R $ be the circumradius of a triangle $ ABC. $ The points $ B,C, $ lie on a circle of radius $ \rho $ that intersects $ AB,AC $ at $ E,D, $ respectively. $ \rho' $ is the circumradius of $ ADE. $ Show that there exists a triangle with sides $ R,\rho ,\rho' , $ and having an angle whose value doesn't depend on $ \rho . $ [i]Laurențiu Panaitopol[/i]

Prove that each finite set of integers can be arranged without intersection.

For a finite set $ X$ of positive integers, let $ \Sigma(X) \equal{} \sum_{x \in X} \arctan \frac{1}{x}.$ Given a finite set $ S$ of positive integers for which $ \Sigma(S) < \frac{\pi}{2},$ show that there exists at least one finite set $ T$ of positive integers for which $ S \subset T$ and $ \Sigma(S) \equal{} \frac{\pi}{2}.$ [i]Kevin Buzzard, United Kingdom[/i]

Let $n > 3$ be a positive integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their immediate neighbors on the right and left. Determine all initial distributions of the coins from which it is possible that, after a finite number of steps, each child has exactly one coin.

Let $f_n\ (n=1,\ 2,\ \cdots)$ be a linear transformation expressed by a matrix $\left( \begin{array}{cc} 1-n & 1 \\ -n(n+1) & n+2 \end{array} \right)$ on the $xy$ plane. Answer the following questions: (1) Prove that there exists 2 lines passing through the origin $O(0,\ 0)$ such that all points of the lines are mapped to the same lines, then find the equation of the lines. (2) Find the area $S_n$ of the figure enclosed by the lines obtained in (1) and the curve $y=x^2$. (3) Find $\sum_{n=1}^{\infty} \frac{1}{S_n-\frac 16}.$ [i]2011 Tokyo Institute of Technlogy entrance exam, Problem 1[/i]

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$.

Consider a table with one real number in each cell. In one step, one may switch the sign of the numbers in one row or one column simultaneously. Prove that one can obtain a table with non-negative sums in each row and each column.

On the board written numbers from 1 to 25 . Bob can pick any three of them say $a,b,c$ and replace by $a^3+b^3+c^3$ . Prove that last number on the board can not be $2013^3$.

Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]

The points $(1,1),(2,3),(4,5)$ and $(999,111)$ are marked in the coordinate system. We continue to mark points in the following way : [list] [*]If points $(a,b)$ are marked then $(b,a)$ and $(a-b,a+b)$ can be marked [*]If points $(a,b)$ and $(c,d)$ are marked then so can be $(ad+bc, 4ac-4bd)$. [/list] Can we, after some finite number of these steps, mark a point belonging to the line $y=2x$.

The operations below can be applied on any expression of the form \(ax^2+bx+c\). $(\text{I})$ If \(c \neq 0\), replace \(a\) by \(4a-\frac{3}{c}\) and \(c\) by \(\frac{c}{4}\). $(\text{II})$ If \(a \neq 0\), replace \(a\) by \(-\frac{a}{2}\) and \(c\) by \(-2c+\frac{3}{a}\). $(\text{III}_t)$ Replace \(x\) by \(x-t\), where \(t\) is an integer. (Different values of \(t\) can be used.) Is it possible to transform \(x^2-x-6\) into each of the following by applying some sequence of the above operations? $(\text{a})$ \(5x^2+5x-1\) $(\text{b})$ \(x^2+6x+2\)

Consider a $ 7\times 7$ numbers table $ a_{ij} \equal{} (i^2 \plus{} j)(i \plus{} j^2), 1\le i,j\le 7.$ When we add arbitrarily each term of an arithmetical progression consisting of $ 7$ integers to corresponding to term of certain row (or column) in turn, call it an operation. Determine whether such that each row of numbers table is an arithmetical progression, after a finite number of operations.

The circles $ C_{1}$ and $ C_{2}$ touch externally at $ M$ and the radius of $ C_{2}$ is larger than that of $ C_{1}$. $ A$ is any point on $ C_{2}$ which does not lie on the line joining the centers of the circles. $ B$ and $ C$ are points on $ C_{1}$ such that $ AB$ and $ AC$ are tangent to $ C_{1}$. The lines $ BM$, $ CM$ intersect $ C_{2}$ again at $ E$, $ F$ respectively. $ D$ is the intersection of the tangent at $ A$ and the line $ EF$. Show that the locus of $ D$ as $ A$ varies is a straight line.

The vertices of a regular $n$-gon are initially marked with one of the signs $+$ or $-$ . A [i]move[/i] consists in choosing three consecutive vertices and changing the signs from the vertices , from $+$ to $-$ and from $-$ to $+$. [b]a)[/b] Prove that if $n=2015$ then for any initial configuration of signs , there exists a sequence of [i]moves[/i] such that we'll arrive at a configuration with only $+$ signs. [b]b)[/b] Prove that if $n=2016$ , then there exists an initial configuration of signs such that no matter how we make the [i]moves[/i] we'll never arrive at a configuration with only $+$ signs.

On an infinite chessboard, a solitaire game is played as follows: at the start, we have $n^2$ pieces occupying a square of side $n.$ The only allowed move is to jump over an occupied square to an unoccupied one, and the piece which has been jumped over is removed. For which $n$ can the game end with only one piece remaining on the board?

Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x>y$ and $x$ is to the left of $y$, and replaces the pair $(x,y)$ by either $(y+1,x)$ or $(x-1,x)$. Prove that she can perform only finitely many such iterations. [i]Proposed by Warut Suksompong, Thailand[/i]

Prove that $ \forall x\in \mathbb{R} $ , $ \cos ^7x+\cos ^7(x+\frac {2\pi}{3})+\cos ^7(x+\frac {4\pi}{3})=\frac {63}{64}\cos 3x $

On the board written numbers from 1 to 25 . Bob can pick any three of them say $a,b,c$ and replace by $a^3+b^3+c^3$ . Prove that last number on the board can not be $2013^3$.