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]

[b]3.[/b] Let $f(x)= x^n +a_1 x^(n-1)+ \dots + a_n$ ($n\geq 1$) be an irreducible polynomial over the field $K$. Show that every non-zero matrix commuting with the matrix $ \begin{bmatrix} 0 & 1 & 0 & \dots & 0 & 0 \\ 0 & 0 & 1 & \dots & 0 & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & 0 & 1 \\ -a_n & -a_{n-1} & -a_{n-2} & \dots & -a_2 & -a_1 \end{bmatrix} $ is invertible. [b](A. 4)[/b]

The numbers $1$ through $4^{n}$ are written on a board. In each step, Pedro erases two numbers $a$ and $b$ from the board, and writes instead the number $\frac{ab}{\sqrt{2a^2+2b^2}}$. Pedro repeats this procedure until only one number remains. Prove that this number is less than $\frac{1}{n}$, no matter what numbers Pedro chose in each step.

A hexagonal table is given, as the one on the drawing, which has $~$ $2012$ $~$ columns. There are $~$ $2012$ $~$ hexagons in each of the odd columns, and there are $~$ $2013$ $~$ hexagons in each of the even columns. The number $~$ $i$ $~$ is written in each hexagon from the $~$ $i$-th column. Changing the numbers in the table is allowed in the following way: We arbitrarily select three adjacent hexagons, we rotate the numbers, and if the rotation is clockwise then the three numbers decrease by one, and if we rotate them counterclockwise the three numbers increase by one (see the drawing below). What's the maximum number of zeros that can be obtained in the table by using the above-defined steps.

On a square board divided into $15 \times 15$ little squares there are $15$ rooks that do not attack each other. Then each rook makes one move like that of a knight. Prove that after this is done a pair of rooks will necessarily attack each other.

Let $n>1$ be a positive integer. Each cell of an $n\times n$ table contains an integer. Suppose that the following conditions are satisfied: [list=1] [*] Each number in the table is congruent to $1$ modulo $n$. [*] The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^2$. [/list] Let $R_i$ be the product of the numbers in the $i^{\text{th}}$ row, and $C_j$ be the product of the number in the $j^{\text{th}}$ column. Prove that the sums $R_1+\hdots R_n$ and $C_1+\hdots C_n$ are congruent modulo $n^4$.

Suppose that $a,b,c,d$ are positive real numbers satisfying $(a+c)(b+d)=ac+bd$. Find the smallest possible value of $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}.$$ [i]Israel[/i]

Let $(x_1, y_1, z_1), (x_2, y_2, z_2), \cdots, (x_{19}, y_{19}, z_{19})$ be integers. Prove that there exist pairwise distinct subscripts $i, j, k$ such that $x_i+x_j+x_k$, $y_i+y_j+y_k$, $z_i+z_j+z_k$ are all multiples of $3$.

An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding $2m$ to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.

Let $S$ be the set of $10$-tuples of non-negative integers that have sum $2019$. For any tuple in $S$, if one of the numbers in the tuple is $\geq 9$, then we can subtract $9$ from it, and add $1$ to the remaining numbers in the tuple. Call thus one operation. If for $A,B\in S$ we can get from $A$ to $B$ in finitely many operations, then denote $A\rightarrow B$. (1) Find the smallest integer $k$, such that if the minimum number in $A,B\in S$ respectively are both $\geq k$, then $A\rightarrow B$ implies $B\rightarrow A$. (2) For the $k$ obtained in (1), how many tuples can we pick from $S$, such that any two of these tuples $A,B$ that are distinct, $A\not\rightarrow B$.

Alan has three pins that form a right triangle with legs $1$ and $4$ at first. Every move, he can pick any one of the pins, pick any new point $\mathcal{P}$ on the opposite side, and move the pin to its $\textit{reflection}$ across $\mathcal{P}$. After a series of moves, can the pins eventually form a right triangle with legs $2$ and $3$? [center][img width=75]https://cdn.artofproblemsolving.com/attachments/e/0/f50c28102c8cefd1fd1f4c327fd3f24f12748d.png[/img][/center] [i]Jason Lee[/i]

Let $r>1$ be a rational number. Alice plays a solitaire game on a number line. Initially there is a red bead at $0$ and a blue bead at $1$. In a move, Alice chooses one of the beads and an integer $k \in \mathbb{Z}$. If the chosen bead is at $x$, and the other bead is at $y$, then the bead at $x$ is moved to the point $x'$ satisfying $x'-y=r^k(x-y)$. Find all $r$ for which Alice can move the red bead to $1$ in at most $2021$ moves.

Ten million fireflies are glowing in $\mathbb{R}^3$ at midnight. Some of the fireflies are friends, and friendship is always mutual. Every second, one firefly moves to a new position so that its distance from each one of its friends is the same as it was before moving. This is the only way that the fireflies ever change their positions. No two fireflies may ever occupy the same point. Initially, no two fireflies, friends or not, are more than a meter away. Following some finite number of seconds, all fireflies find themselves at least ten million meters away from their original positions. Given this information, find the greatest possible number of friendships between the fireflies. [i]Nikolai Beluhov[/i]

Let $ X$ be a set of 10,000 integers, none of them is divisible by 47. Prove that there exists a 2007-element subset $ Y$ of $ X$ such that $ a \minus{} b \plus{} c \minus{} d \plus{} e$ is not divisible by 47 for any $ a,b,c,d,e \in Y.$ [i]Author: Gerhard Wöginger, Netherlands[/i]

Let $H$ be a complex Hilbert space. Let $T:H\to H$ be a bounded linear operator such that $|(Tx,x)|\le\lVert x\rVert^2$ for each $x\in H$. Assume that $\mu\in\mathbb C$, $|\mu|=1$, is an eigenvalue with the corresponding eigenspace $E=\{\phi\in H:T\phi=\mu\phi\}$. Prove that the orthogonal complement $E^\perp=\{x\in H:\forall\phi\in E:(x,\phi)=0\}$ of $E$ is $T$-invariant, i.e., $T(E^\perp)\subseteq E^\perp$.