A [i]calendar[/i] is a (finite) rectangular grid. A calendar is [i]valid[/i] if it satisfies the following conditions: (i) Each square of the calendar is colored white or red, and there are exactly 10 red squares. (ii) Suppose that there are $N$ columns of squares in the calendar. Then if we fill in the numbers $1,2,\ldots$ from the top row to the bottom row, and within each row from left to right, there do not exist $N$ consecutive numbers such that the squares they are in are all white. (iii) Suppose that there are $M$ rows of squares in the calendar. Then if we fill in the numbers $1,2,\ldots$ from the left-most column to the right-most column, and within each column from bottom to top, there do not exist $M$ consecutive numbers such that the squares they are in are all white. In other words, if we rotate the calendar clockwise by $90^{\circ}$, the resulting calendar still satisfies (ii). How many different kinds of valid calendars are there? (Remark: During the actual exam, the contestants were confused about what counts as different calendars. So although this was not in the actual exam, I would like to specify that two calendars are considered different if they have different side lengths or if the $10$ red squares are at different locations.)

We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of $100$ cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions: 1. The winner only depends on the relative order of the $200$ cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner. 2. If we write the elements of both sets in increasing order as $A =\{ a_1 , a_2 , \ldots, a_{100} \}$ and $B= \{ b_1 , b_2 , \ldots , b_{100} \}$, and $a_i > b_i$ for all $i$, then $A$ beats $B$. 3. If three players draw three disjoint sets $A, B, C$ from the deck, $A$ beats $B$ and $B$ beats $C$ then $A$ also beats $C$. How many ways are there to define such a rule? Here, we consider two rules as different if there exist two sets $A$ and $B$ such that $A$ beats $B$ according to one rule, but $B$ beats $A$ according to the other. [i]Proposed by Ilya Bogdanov, Russia[/i]

In CMI, each person has atmost $3$ friends. A disease has infected exactly $2023$ peoplein CMI . Each day, a person gets infected if and only if atleast two of their friends were infected on the previous day. Once someone is infected, they can neither die nor be cured. Given that everyone in CMI eventually got infected, what is the maximum possible number of people in CMI?

Four integers are marked on a circle. On each step we simultaneously replace each number by the difference between this number and next number on the circle, moving in a clockwise direction; that is, the numbers $ a,b,c,d$ are replaced by $ a\minus{}b,b\minus{}c,c\minus{}d,d\minus{}a.$ Is it possible after 1996 such to have numbers $ a,b,c,d$ such the numbers $ |bc\minus{}ad|, |ac \minus{} bd|, |ab \minus{} cd|$ are primes?

On the table there are written numbers $673, 674, \cdots, 2018, 2019.$ Nibab chooses arbitrarily three numbers $a,b$ and $c$, erases them and writes the number $\frac{\min(a,b,c)}{3}$, then he continues in an analogous way. After Nibab performed this operation $673$ times, on the table remained a single number $k$. Prove that $k\in(0,1).$

There are $a+b$ bowls arranged in a row, numbered $1$ through $a+b$, where $a$ and $b$ are given positive integers. Initially, each of the first $a$ bowls contains an apple, and each of the last $b$ bowls contains a pear. A legal move consists of moving an apple from bowl $i$ to bowl $i+1$ and a pear from bowl $j$ to bowl $j-1$, provided that the difference $i-j$ is even. We permit multiple fruits in the same bowl at the same time. The goal is to end up with the first $b$ bowls each containing a pear and the last $a$ bowls each containing an apple. Show that this is possible if and only if the product $ab$ is even.

The [i]liar's guessing game[/i] is a game played between two players $A$ and $B$. The rules of the game depend on two positive integers $k$ and $n$ which are known to both players. At the start of the game $A$ chooses integers $x$ and $N$ with $1 \le x \le N.$ Player $A$ keeps $x$ secret, and truthfully tells $N$ to player $B$. Player $B$ now tries to obtain information about $x$ by asking player $A$ questions as follows: each question consists of $B$ specifying an arbitrary set $S$ of positive integers (possibly one specified in some previous question), and asking $A$ whether $x$ belongs to $S$. Player $B$ may ask as many questions as he wishes. After each question, player $A$ must immediately answer it with [i]yes[/i] or [i]no[/i], but is allowed to lie as many times as she wants; the only restriction is that, among any $k+1$ consecutive answers, at least one answer must be truthful. After $B$ has asked as many questions as he wants, he must specify a set $X$ of at most $n$ positive integers. If $x$ belongs to $X$, then $B$ wins; otherwise, he loses. Prove that: 1. If $n \ge 2^k,$ then $B$ can guarantee a win. 2. For all sufficiently large $k$, there exists an integer $n \ge (1.99)^k$ such that $B$ cannot guarantee a win. [i]Proposed by David Arthur, Canada[/i]

Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is [i]clean[/i] if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.

Suppose that $V$ is a finite dimensional vector space over the real numbers equipped with an inner product and $S:V\times V \longrightarrow \mathbb R$ is a skew symmetric function that is linear for each variable when others are kept fixed. Prove there exists a linear transformation $T:V \longrightarrow V$ such that $\forall u,v \in V: S(u,v)=<u,T(v)>$. We know that there always exists $v\in V$ such that $W=<v,T(v)>$ is invariant under $T$. (it means $T(W)\subseteq W$). Prove that if $W$ is invariant under $T$ then the following subspace is also invariant under $T$: $W^{\perp}=\{v\in V:\forall u\in W <v,u>=0\}$. Prove that if dimension of $V$ is more than $3$, then there exist a two dimensional subspace $W$ of $V$ such that the volume defined on it by function $S$ is zero!!!! (This is the way that we can define a two dimensional volume for each subspace $V$. This can be done for volumes of higher dimensions.)

Consider a $2018 \times 2019$ board with integers in each unit square. Two unit squares are said to be neighbours if they share a common edge. In each turn, you choose some unit squares. Then for each chosen unit square the average of all its neighbours is calculated. Finally, after these calculations are done, the number in each chosen unit square is replaced by the corresponding average. Is it always possible to make the numbers in all squares become the same after finitely many turns?

The integers $ 1,2,\dots,20$ are written on the blackboard. Consider the following operation as one step: [i]choose two integers $ a$ and $ b$ such that $ a\minus{}b \ge 2$ and replace them with $ a\minus{}1$ and $ b\plus{}1$[/i]. Please, determine the maximum number of steps that can be done. [i]Yudi Satria, Jakarta[/i]

Consider a circle of radius $4$ with center $O_1$, a circle of radius $2$ with center $O_2$ that lies on the circumference of circle $O_1$, and a circle of radius $1$ with center $O_3$ that lies on the circumference of circle $O_2$. The centers of the circle are collinear in the order $O_1$, $O_2$, $O_3$. Let $A$ be a point of intersection of circles $O_1$ and $O_2$ and $B$ be a point of intersection of circles $O_2$ and $O_3$ such that $A$ and $B$ lie on the same semicircle of $O_2$. Compute the length of $AB$.

A grid consists of all points of the form $(m, n)$ where $m$ and $n$ are integers with $|m|\le 2019,|n| \le 2019$ and $|m| +|n| < 4038$. We call the points $(m,n)$ of the grid with either $|m| = 2019$ or $|n| = 2019$ the [i]boundary points[/i]. The four lines $x = \pm 2019$ and $y= \pm 2019$ are called [i]boundary lines[/i]. Two points in the grid are called [i]neighbours [/i] if the distance between them is equal to $1$. Anna and Bob play a game on this grid. Anna starts with a token at the point $(0,0)$. They take turns, with Bob playing first. 1) On each of his turns. Bob [i]deletes [/i] at most two boundary points on each boundary line. 2) On each of her turns. Anna makes exactly three [i]steps[/i] , where a [i]step [/i] consists of moving her token from its current point to any neighbouring point, which has not been deleted. As soon as Anna places her token on some boundary point which has not been deleted, the game is over and Anna wins. Does Anna have a winning strategy? [i]Proposed by Demetres Christofides, Cyprus[/i]

Let $G$ be a finite group, and let $H_1, H_2 \subset G$ be two subgroups. Suppose that for any representation of $G$ on a finite-dimensional complex vector space $V$, one has that \[\text{dim} V^{H_1}=\text{dim} V^{H_2},\] where $V^{H_i}$ is the subspace of $H_i$-invariant vectors in $V$ ($i=1,2$). Prove that \[Z(G) \cap H_1=Z(G) \cap H_2.\] Here $Z(G)$ denotes the center of $G$.

Recall that the [i]Fibonacci numbers[/i] are defined recursively by the equation $F_n = F_{n - 1} + F_{n - 2}$ for every integer $n \ge 2$, with initial values $F_0 = 0$ and $F_1 = 1$. Let $k$ be a positive integer. Say that an integer is $k$-[i]summable[/i] if it is the sum of $k$ Fibonacci numbers (not necessarily distinct). (a) Prove that every positive integer less than $F_{2k + 3} - 1$ is $k$-summable. (b) Prove that $F_{2k + 3} - 1$ is not $k$-summable.