Let $n > 1$ be a positive integer. A 2-dimensional grid, infinite in all directions, is given. Each 1 by 1 square in a given $n$ by $n$ square has a counter on it. A [i]move[/i] consists of taking $n$ adjacent counters in a row or column and sliding them each by one space along that row or column. A [i]returning sequence[/i] is a finite sequence of moves such that all counters again fill the original $n$ by $n$ square at the end of the sequence. [list] [*] Assume that all counters are distinguishable except two, which are indistinguishable from each other. Prove that any distinguishable arrangement of counters in the $n$ by $n$ square can be reached by a returning sequence. [*] Assume all counters are distinguishable. Prove that there is no returning sequence that switches two counters and returns the rest to their original positions.[/list] [i]Mitchell Lee and Benjamin Gunby.[/i]

Let $n$ be a positive integer. There are $n$ colors available. Each of the integers from $1$ to $1000$ must be painted with one of the $n$ colors such that any two different numbers, if one divides the other, are painted in different colors. Determine the smallest value of $n$ for which this is possible.

Given an $n \times n \times n$ grid of unit cubes, a cube is [i]good[/i] if it is a sub-cube of the grid and has side length at least two. If a good cube contains another good cube and their faces do not intersect, the first good cube is said to [i]properly[/i] contain the second. What is the size of the largest possible set of good cubes such that no cube in the set properly contains another cube in the set?

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 $A$ be a finite set of positive numbers , $B=\{\frac{a+b}{c+d} |a,b,c,d \in A \}$. Show that: $\left | B \right | \ge 2\left | A \right |^2-1 $, where $|X| $ be the number of elements of the finite set $X$. (High School Affiliated to Nanjing Normal University )

A brick has the shape of a cube of size $2$ with one corner unit cube removed. Given a cube of side $2^{n}$ divided into unit cubes from which an arbitrary unit cube is removed, show that the remaining figure can be built using the described bricks.

Mugdho and Dipto play a game on a numbered row of $n \geq 5$ squares. At the beginning, a pebble is put on the first square and then the players make consecutive moves; Mugdho starts. During a move a player is allowed to choose one of the following: [list] [*] move the pebble one square rightward [*] move the pebble four squares rightward [*] move the pebble two squares leftward [/list] All of the possible moves are only allowed if the pebble stays within the borders of the square row. The player who moves the pebble to the last square (a. k. a $n$-th) wins. Determine for which values of $n$ each of the players has a winning strategy.

There are three positive integers written on a blackboard every minute. You can pick two written numbers $a$ and $b$ and replace them with $a \cdot b$ and $|a-b|$. Prove that it is always possible to make two of the numbers zero.

You plan to organize your birthday party, which will be attended either by exactly $m$ persons or by exactly $n$ persons (you are not sure at the moment). You have a big birthday cake and you want to divide it into several parts (not necessarily equal), so that you are able to distribute the whole cake among the people attending the party with everybody getting cake of equal mass (however, one may get one big slice, while others several small slices - the sizes of slices may differ). What is the minimal number of parts you need to divide the cake, so that it is possible, regardless of the number of guests.

One eastern country was ruled by an old Shah. The population of the country consisted of inhabitants and satraps. Each resident had his own place of residence (place of registration). Satraps moved around the country and carried out the decrees of the Shah. One day the Shah issued a decree containing the following points: 1) Some residents are bandits. 2) Every bandit must be destroyed. 3) Together with the bandit, all those residents who are located closer to the bandit than the Shah (in other words, than the location of the Shah’s palace) must be destroyed. Finding out which of the residents was a bandit was entrusted to the Shah's adviser, known for his connections with one hostile state. Prove that: a) if the country in question is on a plane, then the adviser has the opportunity to declare no more than six inhabitants bandits in such a way that all inhabitants of the country must be destroyed in accordance with the decrees; b) if the country is located on a sphere, then you can get by with five bandits.

Let $A$ be the set of $n$-digit integers whose digits are all from $\{ 1, 2, 3, 4, 5 \}$. $B$ is subset of $A$ such that it contains digit $5$, and there is no digit $3$ in front of digit $5$ (i.e. for $n = 2$, $35$ is not allowed, but $53$ is allowed). How many elements does set $B$ have?

Let $S$ be a set, $|S|=35$. A set $F$ of mappings from $S$ to itself is called to be satisfying property $P(k)$, if for any $x,y\in S$, there exist $f_1, \cdots, f_k \in F$ (not necessarily different), such that $f_k(f_{k-1}(\cdots (f_1(x))))=f_k(f_{k-1}(\cdots (f_1(y))))$. Find the least positive integer $m$, such that if $F$ satisfies property $P(2019)$, then it also satisfies property $P(m)$.

How many ways can you roll three 20-sided dice such that the sum of the three rolls is exactly $42$? Here the order of the rolls matter. [i](Note that a 20-sided die is is very much like a regular 6-sided die other than the fact that it has $20$ faces instead of $6$)[/i]

A convex polygon $\mathcal{P}$ with a center of symmetry $O{}$ is drawn in the plane. Prove that it is possible to place a rhombus in $\mathcal{P}$ whose image following a homothety of factor two centered at $O$ contains $\mathcal{P}$. [i]Proposed by I. Bogdanov, S. Gerdzhikov and N. Nikolov[/i]

Given $25$ different positive numbers. Prove that you can choose two of them such, that none of the other numbers equals neither to the sum nor to the difference between the chosen numbers.

A black unit equilateral triangle is drawn on the plane. How can we place nine tiles, each a unit equilateral triangle, on the plane so that they do not overlap, and each tile covers at least one interior point of the black triangle? (Folklore)

A number is called [i]trilegal[/i] if its digits belong to the set \(\{1, 2, 3\}\) and if it is divisible by \(99\). How many trilegal numbers with \(10\) digits are there?

$n$ distinct integers are given, all of which are bigger than $-a$, where $a$ is a positive integer. It turned out that the amount of odd numbers among them is equal to the biggest even number, and the amount of even numbers is equal to the biggest odd numbers a) Find the least possible value of $n$ for all $a$ b) For each $a \geq 2$ find the maximum possible value of $n$

Each square in an $n \times n$ table is painted black or white. The routes where two rows meet two columns, called a quartet if the remaining squares are the same color. (a) What is the largest possible number of black squares in a $4 \times 4$ table without quartets? (b) Is it possible to paint a $5 \times 5$ table so that it has no quartets?

In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height $0$. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn: [list] [*] The gardener chooses a square in the garden. Each tree on that square and all the surrounding squares (of which there are at most eight) then becomes one unit taller. [*] The lumberjack then chooses four different squares on the board. Each tree of positive height on those squares then becomes one unit shorter. [/list] We say that a tree is [i]majestic[/i] if its height is at least $10^6$. Determine the largest $K$ such that the gardener can ensure there are eventually $K$ majestic trees on the board, no matter how the lumberjack plays.

Prair has a box with some combination of red and green balls. If she randomly draws two balls out of the box (without replacement), the probability of drawing two balls of the same color is equal to the probability of drawing two balls of different colors! How many possible values between $200$ and $1000$ are there for the total number of balls in the box? [i]Proposed by Andrew Wu[/i]

There are $n$ people standing on a circular track. We want to perform a number of [i]moves[/i] so that we end up with a situation where the distance between every two neighbours is the same. The [i]move[/i] that is allowed consists in selecting two people and asking one of them to walk a distance $d$ on the circular track clockwise, and asking the other to walk the same distance on the track anticlockwise. The two people selected and the quantity $d$ can vary from move to move. Prove that it is possible to reach the desired situation (where the distance between every two neighbours is the same) after at most $n-1$ moves.

Given three automates that deal with the cards with the pairs of natural numbers. The first, having got the card with ($a,b)$, produces new card with $(a+1,b+1)$, the second, having got the card with $(a,b)$, produces new card with $(a/2,b/2)$, if both $a$ and $b$ are even and nothing in the opposite case; the third, having got the pair of cards with $(a,b)$ and $(b,c)$ produces new card with $(a,c)$. All the automates return the initial cards also. Suppose there was $(5,19)$ card initially. Is it possible to obtain a) $(1,50)$? b) $(1,100)$? c) Suppose there was $(a,b)$ card initially $(a<b)$. We want to obtain $(1,n)$ card. For what $n$ is it possible?

Put the numbers $1,2,3,...,n$ on the circumference of a circle in such a way that the difference between neighboring numbers is at most $2$. Prove that there is just one solution (if regard is paid only to the order in which the numbers are arranged).

The familiar $3$-dimensional cube has $6$ $2$-dimensional faces, $12$ $1$-dimensional edges, and $8$ $0$-dimensional vertices. Find the number of $9$-dimensional sub-subfaces in a $12$-dimensional cube.