Let $S_n$ be the set of points $(x/2,y/2)\in\mathbb{R}^2$ such that $x$ and $y$ are odd integers and $|x|\leq y\leq 2n$. Let $T_n$ be the number of graphs $G$ with vertex set in $S_n$ satisfying the following conditions: [list] [*]G has no cycles. [*]If two points share an edge, then the distance between them is $1$. [*]For any path $P = (a,\dots,b)$ in $G$, the smallest $y$-coordinate among the points in $P$ is either that of $a$ or that of $b$. However, multiple points may share this $y$-coordinate. [/list] Find the $100$th-smallest positive integer $n$ such that the units digit of $T_{3n}$ is $4$.

A tiling of the plane with polygons consists of placing the polygons in the plane so that interiors of polygons do not overlap, each vertex of one polygon coincides with a vertex of another polygon, and no point of the plane is left uncovered. A unit polygon is a polygon with all sides of length one. It is quite easy to tile the plane with infinitely many unit squares. Likewise, it is easy to tile the plane with infinitely many unit equilateral triangles. (a) Prove that there is a tiling of the plane with infinitely many unit squares and infinitely many unit equilateral triangles in the same tiling. (b) Prove that it is impossible to find a tiling of the plane with infinitely many unit squares and finitely many (and at least one) unit equilateral triangles in the same tiling.

Three balloon vendors each offer two types of balloons - one offers red & blue, one offers blue & yellow, and one offers yellow & red. I like each vendor the same, so I must buy $7$ balloons from each. How many different possible triples $(x,y,z)$ are there such that I could buy $x$ blue, $y$ yellow, and $z$ red balloons?

Let $n \geq 1$ be an integer. What is the maximum number of disjoint pairs of elements of the set $\{ 1,2,\ldots , n \}$ such that the sums of the different pairs are different integers not exceeding $n$?

Let $n$ be a positive integer, and let $F$ be a family of subsets of $\{1, 2, ... , 2^n\}$ such that for any non-empty $ A\in F$ there exists $B \in F$ so that $|A| = |B| + 1$ and $B \subset A$. Suppose that F contains all $(2^n - 1)$-element subsets of $\{1, 2, ... , 2^n\}$ Determine the minimal possible value of $|F|$.

Given a convex pentagon $ABCDE$, prove that if an arbitrary point $M$ inside the pentagon is connected by lines with all the pentagon’s vertices, then either one or three or five of these lines cross the sides of the pentagon opposite the vertices they pass. Note: In reality, we need to exclude the points of the diagonals, because that in this case the drawn lines can pass not through the internal points of the sides, but through the vertices. But if the drawn diagonals are not considered or counted twice (because they are drawn from two vertices), then the statement remains true.

From the $9 \times 9$ chessboard, all $16$ unit squares whose row numbers and column numbers are both even have been removed. Disect the punctured board into rectangular pieces, with as few of them being unit squares as possible.

The sequence $a_1,a_2,\dots$ of integers satisfies the conditions: (i) $1\le a_j\le2015$ for all $j\ge1$, (ii) $k+a_k\neq \ell+a_\ell$ for all $1\le k<\ell$. Prove that there exist two positive integers $b$ and $N$ for which\[\left\vert\sum_{j=m+1}^n(a_j-b)\right\vert\le1007^2\]for all integers $m$ and $n$ such that $n>m\ge N$. [i]Proposed by Ivan Guo and Ross Atkins, Australia[/i]

A [i]windmill[/i] is a closed line segment of unit length with a distinguished endpoint, the [i]pivot[/i]. Let $S$ be a finite set of $n$ points such that the distance between any two points of $S$ is greater than $c$. A configuration of $n$ windmills is [i]admissible[/i] if no two windmills intersect and each point of $S$ is used exactly once as a pivot. An admissible configuration of windmills is initially given to Geoff in the plane. In one operation Geoff can rotate any windmill around its pivot, either clockwise or counterclockwise and by any amount, as long as no two windmills intersect during the process. Show that Geoff can reach any other admissible configuration in finitely many operations, where (i) $c = \sqrt 3$, (ii) $c = \sqrt 2$. [i]Proposed by Michael Ren[/i]

Let $S$ be the set of colorings of a $100 \times 100$ grid where each square is colored black or white and no $2\times2$ subgrid is colored like a chessboard. A random such coloring is chosen: what is the probability there is a path of black squares going from the top row to the bottom row where any two consecutive squares in the path are adjacent? [i]Proposed by Evan Chang (squareman), USA [/i]

For some positive integer $n$, there exists $n$ different positive integers $a_1, a_2, ..., a_n$ such that $(1)$ $a_1=1, a_n=2000$ $(2)$ $\forall i\in \mathbb{Z}$ $s.t.$ $2\le i\le n, a_i -a_{i-1}\in \{-3,5\}$ Determine the maximum value of n.

A set $S$ is called [i]neighbouring [/i] if it has the following two properties: a) $S$ has exactly four elements b) for every element $x$ of $S$, at least one of the numbers $x - 1$ or $x+1$ belongs to $S$. Find the number of all [i]neighbouring [/i] subsets of the set $\{1,2,... ,n\}$.

The vertices of a regular polygon with $N$ sides are marked on the blackboard. Ana and Beto play alternately, Ana begins. Each player, in turn, must do the following: $\bullet$ join two vertices with a segment, without cutting another already marked segment; or $\bullet$ delete a vertex that does not belong to any marked segment. The player who cannot take any action on his turn loses the game. Determine which of the two players can guarantee victory: a) if $N=28$ b) if $N=29$

Given positive integer $n \geq 2$. Now Cindy fills each cell of the $n*n$ grid with a positive integer not greater than $n$ such that the numbers in each row are in a non-decreasing order (from left to right) and numbers in each column is also in a non-decreasing order (from top to bottom). We call two adjacant cells form a $domino$ , if they are filled with the same number. Now Cindy wants the number of $domino$s as small as possible. Find the smallest number of $dominos$ Cindy can reach. (Here, two cells are called adjacant if they share one common side)

There are $n \geq 1$ notebooks, numbered from $1$ to $n$, stacked in a pile. Zahar repeats the following operation: he randomly chooses a notebook whose number $k$ does not correspond to its location in this stack, counting from top to bottom, and returns it to the $k$th position, counting from the top, without changing the location of the other notebooks. If there is no such notebook, he stops. Is it guaranteed that Zahar will arrange all the notebooks in ascending order of numbers in a finite number of operations? [i]Proposed by Zahar Naumets[/i]

For a sequence with some ones and zeros, we count the number of continuous runs of equal digits in it. (For example the sequence $011001010$ has $7$ continuous runs: $0,11,00,1,0,1,0$.) Find the sum of the number of all continuous runs for all possible sequences with $2019$ ones and $2019$ zeros.

Merlin's castle has 100 rooms and 1000 corridors. Each corridor links some two rooms. Each pair of rooms is linked by one corridor at most. Merlin has given out the plan of the castle to the wise men and declared the rules of the challenge. The wise men need to scatter across the rooms in a manner they wish. Each minute Merlin will choose a corridor and one of the wise men will have to pass along it from one of the rooms at its ends to the other one. Merlin wins when in both rooms on the ends of the chosen corridor there are no wise men. Let us call a number $m$ the magic number of the castle if $m$ wise men can pre-agree before the challenge and act in such a way that Merlin never wins, $m$ being the minimal possible number. What are the possible values of the magic number of the castle? (Merlin and all the wise men always know the location of all the wise men). Timofey Vasilyev

Let $m,n$ be naturals satisfying $n \geq m \geq 2$ and let $S$ be a set consisting of $n$ naturals. Prove that $S$ has at least $2^{n-m+1}$ distinct subsets, each whose sum is divisible by $m$. (The zero set counts as a subset).

Let $P(A)$ be the arithmetic-means of all elements of set $A = \{ a_1, a_2, \ldots, a_n \}$, namely $P(A) = \frac{1}{n} \sum^{n}_{i=1}a_i$. We denote $B$ "balanced subset" of $A$, if $B$ is a non-empty subset of $A$ and $P(B) = P(A)$. Let set $M = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$. Find the number of all "balanced subset" of $M$.

Physicists at Princeton are trying to analyze atom entanglement using the following experiment. Originally there is one atom in the space and it starts splitting according to the following procedure. If after $n$ minutes there are atoms $a_1, \dots, a_N$, in the following minute every atom $a_i$ splits into four new atoms, $a_i^{(1)},a_i^{(2)},a_i^{(3)},a_i^{(4)}$. Atoms $a_i^{(j)}$ and $a_k^{(j)}$ are entangled if and only the atoms $a_i$ and $a_k$ were entangled after $n$ minutes. Moreover, atoms $a_i^{(j)}$ and $a_k^{(j+1)}$ are entangled for all $1 \le i$, $k \le N$ and $j = 1$, $2$, $3$. Therefore, after one minute there is $4$ atoms, after two minutes there are $16$ atoms and so on. Physicists are now interested in the number of unordered quadruplets of atoms $\{b_1, b_2, b_3, b_4\}$ among which there is an odd number of entanglements. What is the number of such quadruplets after $3$ minutes? [i]Remark[/i]. Note that atom entanglement is not transitive. In other words, if atoms $a_i$, $a_j$ are entangled and if $a_j$, $a_k$ are entangled, this does not necessarily mean that $a_i$ and $a_k$ are entangled.

The infinite sequence $a_0,a _1, a_2, \dots$ of (not necessarily distinct) integers has the following properties: $0\le a_i \le i$ for all integers $i\ge 0$, and \[\binom{k}{a_0} + \binom{k}{a_1} + \dots + \binom{k}{a_k} = 2^k\] for all integers $k\ge 0$. Prove that all integers $N\ge 0$ occur in the sequence (that is, for all $N\ge 0$, there exists $i\ge 0$ with $a_i=N$).

Let $\sigma$ be a random permutation of $\{0, 1, \ldots, 6\}$. Let $L(\sigma)$ be the length of the longest initial monotonic consecutive subsequence of $\sigma$ not containing $0$; for example, \[L(\underline{2,3,4},6,5,1,0) = 3,\ L(\underline{3,2},4,5,6,1,0) = 2,\ L(0,1,2,3,4,5,6) = 0.\] If the expected value of $L(\sigma)$ can be written as $\frac mn$, where $m$ and $n$ are relatively prime positive integers, then find $m + n$.

For each integer $n\ge 1,$ compute the smallest possible value of \[\sum_{k=1}^{n}\left\lfloor\frac{a_k}{k}\right\rfloor\] over all permutations $(a_1,\dots,a_n)$ of $\{1,\dots,n\}.$ [i]Proposed by Shahjalal Shohag, Bangladesh[/i]

Let $m,n\in \mathbb{Z}_{\ge 1}$ and a rectangular board $m\times n$ sliced by parallel lines to the rectangle's sides into $mn$ unit squares. At moment $t=0$, there is an ant inside every square, positioned exactly in its centre, such that it is oriented towards one of the rectangle's sides. Every second, all the ants move exactly a unit following their current orientation; however, if two ants meet at the centre of a unit square, both of them turn $180^o$ around (the turn happens instantly, without any loss of time) and the next second they continue their motion following their new orientation. If two ants meet at the midpoint of a side of a unit square, they just continue moving, without changing their orientation.\\ \\ Prove that, after finitely many seconds, some ant must fall off the table.\\ \\ [i](Oliver Hayman)[/i]

Lavidópolis is a city with 2024 neighborhoods. Lavi Dopes was elected mayor, and since he saw that there were no roads in the city, he asked Gil Bento, the monster engineer, to design the city's roads according to the following rules: 1. Any two neighborhoods are connected by at most one two-way road; 2. For any two neighborhoods, there is exactly one route from one neighborhood to another, which may pass through some intermediate neighborhoods, but never passes through the same neighborhood more than once. Mayor Lavi Dopes wants to try for re-election, but since he knows nothing about the city and only shows up during campaign times (he spent all this time stealing... I mean, thinking about math problems), he wants to find a pair of neighborhoods such that the number of roads that are part of the route connecting them is maximized among all pairs of neighborhoods. To do this, he starts asking Gil Bento various questions, all in the following manner: he chooses two of the 2024 neighborhoods, say A and B, and asks: "Given neighborhoods A and B, how many roads are part of the route connecting A to B?" Knowing that Gil Bento always answers correctly to each question, determine the minimum number of questions that Lavi Dopes needs to ask to achieve his goal, regardless of how Gil Bento has designed the roads of Lavidópolis.