A hundred tourists arrive to a hotel at night. They know that in the hotel there are single rooms numbered as $1, 2, \ldots , n$, and among them $k{}$ (the tourists do not know which) are under repair, the other rooms are free. The tourists, one after another, check the rooms in any order (maybe different for different tourists), and the first room not under repair is taken by the tourist. The tourists don’t know whether a room is occupied until they check it. However it is forbidden to check an occupied room, and the tourists may coordinate their strategy beforehand to avoid this situation. For each $k{}$ find the smallest $n{}$ for which the tourists may select their rooms for sure. [i]Fyodor Ivlev[/i]

We have a set of $343$ closed jars, each containing blue, yellow and red marbles with the number of marbles from each color being at least $1$ and at most $7$. No two jars have exactly the same contents. Initially all jars are with the caps up. To flip a jar will mean to change its position from cap-up to cap-down or vice versa. It is allowed to choose a triple of positive integers $(b; y; r) \in \{1; 2; ...; 7\}^3$ and flip all the jars whose number of blue, yellow and red marbles differ by not more than $1$ from $b, y, r$, respectively. After $n$ moves all the jars turned out to be with the caps down. Find the number of all possible values of $n$, if $n \le 2021$.

There are $2019$ students sitting around circular table. Initially each of them have one candy. Teacher is allowed to pick one student, who has at least one can candy, and this student can decide, whether he gives his candy to his neighbour on the right or on the left. Prove that no matter what students teacher picks during the process, students can always ensure that any point of time no student has more than $2$ candies.

Consider the segment $[0; 1]$. At each step we may split one of the available segments into two new segments and write the product of lengths of these two new segments onto a blackboard. Prove that the sum of the numbers on the blackboard never will exceed $1/2$. [i]Mikhail Lukin[/i]

A hundred tourists arrive to a hotel at night. They know that in the hotel there are single rooms numbered as $1, 2, \ldots , n$, and among them $k{}$ (the tourists do not know which) are under repair, the other rooms are free. The tourists, one after another, check the rooms in any order (maybe different for different tourists), and the first room not under repair is taken by the tourist. The tourists don’t know whether a room is occupied until they check it. However it is forbidden to check an occupied room, and the tourists may coordinate their strategy beforehand to avoid this situation. For each $k{}$ find the smallest $n{}$ for which the tourists may select their rooms for sure. [i]Fyodor Ivlev[/i]

On a table there are $100$ red and $k$ white buckets for which all of them are initially empty. In each move, a red and a white bucket is selected and an equal amount of water is added to both of them. After some number of moves, there is no empty bucket and for every pair of buckets that are selected together at least once during the moves, the amount of water in these buckets is the same. Find all the possible values of $k$.

Initially on the blackboard eight zeros are written. In one step, it is allowed to choose numbers $a,b,c,d$, erase them and replace them with the numbers $a+1$, $b+2$, $c+3$, $d+3$. Determine: a) the minimum number of steps required to achieve $8$ consecutive integers on the board b) whether it is possible to achieve that sum of the numbers is $2021$ c) whether it is possible to achieve that product of the numbers is $2145$

On a board are written some positive reals (not necessarily distinct). For every two numbers in the frame $a$ and $b$ distinct such that $$\frac{1}{2}<\frac{a}{b}<2,$$ an allowed operation is to delete $a$ and $b$ and write $2a-b$ and $2b-a$ in their place. Show that we can do the operation only a finite number of times.

Evan is a $n$-dimensional being that lives in a house formed by the points of $\mathbb{Z}_{\geq 0}^n$. His room is the set of points in which coordinates are all less than or equal to $2021$. Evan's room has been infested with bees, so he decides to flush them out through $\textit{captures}$. A $\textit{capture}$ can be performed by eliminating a bee from point $ (a_1, a_2, \ldots, a_n) $ and replacing it with $ n $ bees, one in each of the points: $$ (a_1 + 1, a_2 , \ldots, a_n), (a_1, a_2 + 1, \ldots, a_n), \ldots, (a_1, a_2, \ldots, a_n + 1) $$ However, two bees can never occupy the same point in the house. Determine, for every $ n $, the greatest value $ A (n) $ of bees such that, for some initial arrangement of these bees in Evan's room, he is able to accomplish his goal with a finite amount of $\textit{captures}$.

Let $n$ be a positive integer. Initially, a bishop is placed in each square of the top row of a $2^n \times 2^n$ chessboard; those bishops are numbered from $1$ to $2^n$ from left to right. A [i]jump[/i] is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row. Find the total number of permutations $\sigma$ of the numbers $1, 2, \ldots, 2^n$ with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order $\sigma(1), \sigma(2), \ldots, \sigma(2^n)$, from left to right. [i]Israel[/i]

Consider the segment $[0; 1]$. At each step we may split one of the available segments into two new segments and write the product of lengths of these two new segments onto a blackboard. Prove that the sum of the numbers on the blackboard never will exceed $1/2$. [i]Mikhail Lukin[/i]

A sequence of $2n + 1$ non-negative integers $a_1, a_2, ..., a_{2n + 1}$ is given. There's also a sequence of $2n + 1$ consecutive cells enumerated from $1$ to $2n + 1$ from left to right, such that initially the number $a_i$ is written on the $i$-th cell, for $i = 1, 2, ..., 2n + 1$. Starting from this initial position, we repeat the following sequence of steps, as long as it's possible: [i]Step 1[/i]: Add up the numbers written on all the cells, denote the sum as $s$. [i]Step 2[/i]: If $s$ is equal to $0$ or if it is larger than the current number of cells, the process terminates. Otherwise, remove the $s$-th cell, and shift shift all cells that are to the right of it one position to the left. Then go to Step 1. Example: $(1, 0, 1, \underline{2}, 0) \rightarrow (1, \underline{0}, 1, 0) \rightarrow (1, \underline{1}, 0) \rightarrow (\underline{1}, 0) \rightarrow (0)$. A sequence $a_1, a_2,. . . , a_{2n+1}$ of non-negative integers is called balanced, if at the end of this process there’s exactly one cell left, and it’s the cell that was initially enumerated by $(n + 1)$, i.e. the cell that was initially in the middle. Find the total number of balanced sequences as a function of $n$. [i]Proposed by Viktor Simjanoski, North Macedonia[/i]

At the start of the day, the four numbers $(a_0,b_0,c_0,d_0)$ were written on the board. Every minute, Danny replaces the four numbers written on the board with new ones according to the following rule: if the numbers written on the board are $(a,b,c,d)$, then Danny first calculates the numbers \begin{align*} a'&=a+4b+16c+64d\\ b'&=b+4c+16d+64a\\ c'&=c+4d+16a+64b\\ d'&=d+4a+16b+64c \end{align*} and replaces the numbers $(a,b,c,d)$ with the numbers $(a'd',c'd',c'b',b'a')$. For which initial quadruples $(a_0,b_0,c_0,d_0)$, will Danny write at some point a quadruple of numbers all of which are divisible by $5780^{5780}$?

A single player game has the following rules: initially, there are $10$ piles of stones with $1,2,...,10$ stones, respectively. A movement consists on making one of the following operations: [b]i)[/b] to choose $2$ piles, both of them with at least $2$ stones, combine them and then add $2$ stones to the new pile; [b]ii)[/b] to choose a pile with at least $4$ stones, remove $2$ stones from it, and then split it into two piles with amount of piles to be chosen by the player. The game continues until is not possible to make an operation. Show that the number of piles with one stone in the end of the game is always the same, no matter how the movements are made.