For nonzero real numbers $x$ and $y$, define $x\circ y = \tfrac{xy}{x+y}$. Compute \[2^1\circ \left(2^2\circ \left(2^3\circ\cdots\circ\left(2^{2016}\circ 2^{2017}\right)\right)\right).\]

Let $ n \ge 2$ and $x_1$, $x_2$, $...$, $x_n$ be real numbers from the segment $[1,\sqrt2]$. Prove that holds the inequality $$\frac{\sqrt{x_1^2-1}}{x_2}+\frac{\sqrt{x_2^2-1}}{x_3}+...+\frac{\sqrt{x_n^2-1}}{x_1} \le \frac{\sqrt2}{2} n.$$

Viorel participates in a mathematics competition with 50 problems. For each problem he answers correctly, he earns 4 points, and for each problem he answers incorrectly, he loses 1 point. If Viorel answered every problem and has 65 points, how many problems did he solve correctly?

Each term of an infinite sequence of natural numbers is obtained from the previous term by adding to it one of its nonzero digits. Prove that this sequence contains an even number.

If $\zeta$ denote the Riemann Zeta Function, and $s>1$ then $$\sum \limits_{k=1}^{\infty} \frac{1}{1+k^s}\geq \frac{\zeta (s)}{1+\zeta (s)}$$

Prove that every integer $ k$ greater than 1 has a multiple that is less than $ k^4$ and can be written in the decimal system with at most four different digits.

Show that $\varphi(2n)\mid n!$ for all positive integer $n$.

In the square $ABCD$ the points $E$ and $F$ are marked on the sides $AB$ and $BC$ respectively, in such a way that $EB = 2AE$ and $BF = FC$. Let $G$ be the intersection between $DF$ and $EC$. If the side of the square equals $10$, what is the distance from point $G$ to side $AB$?

Sketch the curve $$y= \frac{x}{1+x^6 (\sin x)^{2}},$$ and show that $$ \int_{0}^{\infty} \frac{x}{1+x^6 (\sin x)^{2}}\; dx$$ exists.

Vaysha has a board with $999$ consecutive numbers written and $999$ labels of the form [i]"This number is [b]not [/b]divisible by $i$"[/i], for $i \in \{ 2,3, \dots ,1000 \} $. She places each label next to a number on the board, so that each number has exactly one label. For each true statement on the stickers, Vaysha gets a piece of candy. How many pieces of candy can Vaysha guarantee to win, regardless of the numbers written on the board, if she plays optimally?

A finite sequence of decimal digits from $\{0,1,\cdots, 9\}$ is said to be [i]common[/i] if for each sufficiently large positive integer $n$, there exists a positive integer $m$ such that the expansion of $n$ in base $m$ ends with this sequence of digits. For example, $0$ is common because for any large $n$, the expansion of $n$ in base $n$ is $10$, whereas $00$ is not common because for any squarefree $n$, the expansion of $n$ in any base cannot end with $00$. Determine all common sequences. [i]Proposed by Wong Jer Ren[/i]

Consider three circles in the plane $\Gamma_1,\Gamma_2,\Gamma_3$ of radii $R$ passing through a point $O$, and denote by $\mathfrak D$ the set of points of the plane which belong to at least two of these circles. Find the position of $\Gamma_1,\Gamma_2,\Gamma_3$ for which the area of $\mathfrak D$ is the minimum possible. Justify your answer.

Let $ABC$ be a triangle whose angles $\alpha=\angle CAB$ and $\beta=\angle CBA$ are greater than $45^{\circ}$. Above the side $AB$ a right isosceles triangle $ABR$ is constructed with $AB$ as the hypotenuse, such that $R$ is inside the triangle $ABC$. Analogously we construct above the sides $BC$ and $AC$ the right isosceles triangles $CBP$ and $ACQ$, right at $P$ and in $Q$, but with these outside the triangle $ABC$. Prove that $CQRP$ is a parallelogram.

Ana and Beatriz take turns in a game that starts with a square of side $1$ drawn on an infinite grid. Each turn consists of drawing a square that does not overlap with the rectangle already drawn, in such a way that one of its sides is a (complete) side of the figure already drawn. A player wins if she completes a rectangle whose area is a multiple of $5$. If Ana goes first, does either player have a winning strategy?

Three legendary knights are fighting against a multiheaded dragon. Whenever the first knight attacks, he cuts off half of the current number of heads plus one more. Whenever the second knight attacks, he cuts off one third of the current number of heads plus two more. Whenever the third knight attacks, he cuts off one fourth of the current number of heads plus three more. They repeatedly attack in an arbitrary order so that at each step an integer number of heads is being cut off. If all the knights cannot attack as the number of heads would become non-integer, the dragon eats them. Will the knights be able to cut off all the dragon’s heads if it has $41!$ heads? Alexey Zaslavsky

(a) Prove that \[ [5x]\plus{}[5y] \ge [3x\plus{}y] \plus{} [3y\plus{}x],\] where $ x,y \ge 0$ and $ [u]$ denotes the greatest integer $ \le u$ (e.g., $ [\sqrt{2}]\equal{}1$). (b) Using (a) or otherwise, prove that \[ \frac{(5m)!(5n)!}{m!n!(3m\plus{}n)!(3n\plus{}m)!}\] is integral for all positive integral $ m$ and $ n$.

Let $ABCD$ be a convex quadrilateral such that the circle with diameter $AB$ touches the line $CD$. Prove that that the circle with diameter $CD$ touches the line $AB$ if and only if $BC$ and $AD$ are parallel.

Let $M(t)$ be measurable and locally bounded function, that is, \[M(t) \le C_{a,b}, \quad \forall 0 \le a \le t \le b<\infty\] with some constant $C_{a,b}$, from $[0,\infty)$ to $[0,\infty)$ such that \[M(t) \le 1+\int_0^t M(t-s)(1+t)^{-1}s^{-1/2} ds, \quad \forall t \ge 0.\] Show that \[M(t) \le 10+2\sqrt{5}, \quad \forall t \ge 0.\]

If $1<k_1<k_2<...<k_n$ and $a_1,a_2,...,a_n$ are integers such that for every integer $N,$ $k_i \mid N-a_i$ for some $1 \leq i \leq n,$ find the smallest possible value of $n.$

Let $ G$ be an infinite graph such that for any countably infinite vertex set $ A$ there is a vertex $ p$, not in $A$, joined to infinitely many elements of $ A$. Show that $ G$ has a countably infinite vertex set $ A$ such that $ G$ contains uncountably infinitely many vertices $ p$ joined to infinitely many elements of $ A$. [i]P. Erdos, A. Hajnal[/i]

The numbers $1,2,...,100$ are written in a board. We are allowed to choose any two numbers from the board $a,b$ to delete them and replace on the board the number $a+b-1$. What are the possible numbers u can get after $99$ consecutive operations of these?

$\definecolor{A}{RGB}{70,255,50}\color{A}\fbox{C6.}$ There are $n$ $\definecolor{A}{RGB}{0,0,255}\color{A}\text{boys}$ and $n$ $\definecolor{A}{RGB}{255,0,255}\color{A}\text{girls}$ in a club. Some of them are friends with each other. The $\definecolor{A}{RGB}{0,0,255}\color{A}\text{boys}$ want to get into a [i]relationship[/i], so some subset of them wants to ask some $\definecolor{A}{RGB}{255,0,255}\color{A}\text{girls}$ out for a trip. Because the $\definecolor{A}{RGB}{0,0,255}\color{A}\text{boys}$ are shy, for a nonempty set $B$ of $\definecolor{A}{RGB}{0,0,255}\color{A}\text{boys}$, they want to make sure that each of the girl they ask out is friend with one of the $\definecolor{A}{RGB}{0,0,255}\color{A}\text{boys}$ in $B$. If the number of $\definecolor{A}{RGB}{255,0,255}\color{A}\text{girls}$ they are able to ask out is smaller than the number of the $\definecolor{A}{RGB}{0,0,255}\color{A}\text{boys}$ in $B$, then the nonempty set $B$ of those $\definecolor{A}{RGB}{0,0,255}\color{A}\text{boys}$ is called a group of complete losers. Show that for any $0 \le k < 2n$, there exists an arrangement of the [i]friendships[/i] among those $2n$ people so that there are exactly $k$ groups of complete losers. [i]Proposed by [/i][b][color=#419DAB]ltf0501[/color][/b]. [color=#3D9186]#1737[/color]

Let $n$ be a positive integer, and consider a sequence $a_1 , a_2 , \dotsc , a_n $ of positive integers. Extend it periodically to an infinite sequence $a_1 , a_2 , \dotsc $ by defining $a_{n+i} = a_i $ for all $i \ge 1$. If \[a_1 \le a_2 \le \dots \le a_n \le a_1 +n \] and \[a_{a_i } \le n+i-1 \quad\text{for}\quad i=1,2,\dotsc, n, \] prove that \[a_1 + \dots +a_n \le n^2. \]

Find the number of ordered pairs $(x,y)$ of positive integers that satisfy $x\le 2y\le 60$ and $y\le 2x\le 60.$

Let $\alpha,\beta,\gamma$ be angles of a triangle. Determine all real triplets $x,y,z$ satisfying the system \begin{align*} x\cos\beta+\frac1z\cos\alpha &=1, \\ y\cos\gamma+\frac1x\cos\beta &=1, \\ z\cos\alpha+\frac1y\cos\gamma &=1. \end{align*}