Let $n$ be a given positive integer. Ana and Bob play the following game: Ana chooses a polynomial $p$ of degree $n$ with integer coefficients. In each round, Bob can choose a finite set $S$ of positive integers, and Ana responds with a list containing the values of the polynomial $p$ evaluated at the elements of $S$ with multiplicity (sorted in increasing order). Determine, in terms of $n$, the smallest positive integer $k$ such that Bob can always determine the polynomial $p$ in at most $k$ rounds. [i]Proposed by: Andrei Chirita, Cambridge[/i]

Let $M$ be the point of intersection of the diagonals, and $K$ be the point of intersection of the bisectors of the angles $B$ and $C$ of the convex quadrilateral $ABCD$. Prove that points $A, B, M, K$ lie on the same circle if the following relation holds: $|AB|=|BC|=|CD|$

Prove that $2014$ divides $53n^{55}- 57n^{53} + 4n$ for all integer $n$.

Cyclic quadrilateral $ABCD$ has positive integer side lengths $AB$, $BC$, $CA$, $AD$. It is known that $AD=2005$, $\angle{ABC}=\angle{ADC} = 90^o$, and $\max \{ AB,BC,CD \} < 2005$. Determine the maximum and minimum possible values for the perimeter of $ABCD$.

Consider a convex quadrilateral $ABCD$. Pairs of its opposite sides are continued until they intersect: $BA$ and $CD$ at the point $P$, $BC$ and $AD$ at the point $Q$. Let $K$ be the intersection point of the exterior bisectors of the angles $A$ and $C$ of the quadrilateral, $L$ be the intersection point of the exterior bisectors of the angles $B$ and $D$ of the quadrilateral, and $M$ be the intersection point of the exterior bisectors of the angles $P$ and $Q$ (the exterior bisector of an angle $X$ is the line passing through X and perpendicular to its ordinary bisector). Prove that the points $K$, $L$ and $M$ lie on a straight line. (S Markelov)

$11$ people are sitting around a circle table, orderly (means that the distance between two adjacent persons is equal to others) and $11$ cards with numbers $1$ to $11$ are given to them. Some may have no card and some may have more than $1$ card. In each round, one [and only one] can give one of his cards with number $ i $ to his adjacent person if after and before the round, the locations of the cards with numbers $ i-1,i,i+1 $ don’t make an acute-angled triangle. (Card with number $0$ means the card with number $11$ and card with number $12$ means the card with number $1$!) Suppose that the cards are given to the persons regularly clockwise. (Mean that the number of the cards in the clockwise direction is increasing.) Prove that the cards can’t be gathered at one person.

[b]a)[/b] Prove that the sum of all the elements of a finite union of sets of elements of finite cyclic subgroups of the group of complex numbers, is an integer number. [b]b)[/b] Show that there are finite union of sets of elements of finite cyclic subgroups of the group of complex numbers such that the sum of all its elements is equal to any given integer. [i]Paltin Ionescu[/i]

Given that $5^{2018}$ has $1411$ digits and starts with $3$ (the leftmost non-zero digit is $3$), for how many integers $1\leq n\leq2017$ does $5^n$ start with $1$? [i]2018 CCA Math Bonanza Tiebreaker Round #3[/i]

Find the number of integer solutions of $||x| - 2023| < 2020$.

Let $AD$ be a bisector of triangle $ABC$. Points $M$ and $N$ are projections of $B$ and $C$ respectively to $AD$. The circle with diameter $MN$ intersects $BC$ at points $X$ and $Y$. Prove that $\angle BAX = \angle CAY$.

Find all pairs of non-negative integers $(x,y)$ such that \[\sqrt{x+y}-\sqrt{x}-\sqrt{y}+2=0.\]

Gleb picked positive integers $N$ and $a$ ($a < N$). He wrote the number $a$ on a blackboard. Then each turn he did the following: he took the last number on the blackboard, divided the number $N$ by this last number with remainder and wrote the remainder onto the board. When he wrote the number $0$ onto the board, he stopped. Could he pick $N$ and $a$ such that the sum of the numbers on the blackboard would become greater than $100N$ ? Ivan Mitrofanov

Let $A(XYZ)$ be the area of the triangle $XYZ$. A non-regular convex polygon $P_1 P_2 \ldots P_n$ is called [i]guayaco[/i] if exists a point $O$ in its interior such that \[A(P_1OP_2) = A(P_2OP_3) = \cdots = A(P_nOP_1).\] Show that, for every integer $n \ge 3$, a guayaco polygon of $n$ sides exists.

If $ a,b,c>0 $ and $ ab+bc+ca+abc=4, $ then $ \sqrt{ab} +\sqrt{bc} +\sqrt{ca} \le 3\le a+b+c. $

Let $n$ be a fixed positive odd integer. Take $m+2$ [b]distinct[/b] points $P_0,P_1,\ldots ,P_{m+1}$ (where $m$ is a non-negative integer) on the coordinate plane in such a way that the following three conditions are satisfied: 1) $P_0=(0,1),P_{m+1}=(n+1,n)$, and for each integer $i,1\le i\le m$, both $x$- and $y$- coordinates of $P_i$ are integers lying in between $1$ and $n$ ($1$ and $n$ inclusive). 2) For each integer $i,0\le i\le m$, $P_iP_{i+1}$ is parallel to the $x$-axis if $i$ is even, and is parallel to the $y$-axis if $i$ is odd. 3) For each pair $i,j$ with $0\le i<j\le m$, line segments $P_iP_{i+1}$ and $P_jP_{j+1}$ share at most $1$ point. Determine the maximum possible value that $m$ can take.

Is it possible to find a set of $100$ (or $200$) points on the boundary of a cube such that this set remains fixed under all rotations which leave the cube fixed ?