Seven different odd primes are given. Is it possible that for any two of them, the difference of their eight powers to be divisible by all the remaining ones ? Proposed by F. Petrov, K. Sukhov

We call a sequence of integers a Fibonacci-type sequence if it is infinite in both ways and $a_{n}=a_{n-1}+a_{n-2}$ for any $n\in\mathbb{Z}$. How many Fibonacci-type sequences can we find, with the property that in these sequences there are two consecutive terms, strictly positive, and less or equal than $N$ ? (two sequences are considered to be the same if they differ only by shifting of indices) Proposed by I. Pevzner

A line $d$ is given in the plane. Let $B\in d$ and $A$ another point, not on $d$, and such that $AB$ is not perpendicular on $d$. Let $\omega$ be a variable circle touching $d$ at $B$ and letting $A$ outside, and $X$ and $Y$ the points on $\omega$ such that $AX$ and $AY$ are tangent to the circle. Prove that the line $XY$ passes through a fixed point. Proposed by F. Bakharev

Find all functions $f: (0,\infty)\rightarrow(0,\infty)$ with the following properties: $f(x+1)=f(x)+1$ and $f\left(\frac{1}{f(x)}\right)=\frac{1}{x}$. Proposed by P. Volkmann

There are 100 boxers, each of them having different strengths, who participate in a tournament. Any of them fights each other only once. Several boxers form a plot. In one of their matches, they hide in their glove a horse shoe. If in a fight, only one of the boxers has a horse shoe hidden, he wins the fight; otherwise, the stronger boxer wins. It is known that there are three boxers who obtained (strictly) more wins than the strongest three boxers. What is the minimum number of plotters ? Proposed by N. Kalinin

Click for solution Let's assume we have $p$ plotters and $u$ usual boxers, ie boxers who are not the three strongest and who do not participate in the plot. If $p$ is minimum, the plotters should not use horseshoes on normal boxers, because it does not matter how many games they win. So we will assume that all plotters are stronger than usual boxers. In that case, the three strongest boxers and the plotters each get $u$ wins. Let boxer $s_{1}$ be the strongest boxer, $s_{2}$ the second strongest one, etc.(note that in that case the plotters are boxes $s_{i}$, $4\le i\le p+3$) Then boxer $s_{1}$ beat boxer $s_{2}$ and $s_{3}$ and got $2$ wins among the three strongets boxers; boxer $s_{2}$ got one win, boxer $s_{3}$ got $0$ wins. Among the $p$ boxers, boxer $s_{4}$ got $p-1$ wins, boxer $s_{5}$ got $p-2$ wins, boxer $s_{6}$ got $p-3$ wins. We want minimum $p$ so the boxers $s_{4},s_{5},s_{6}$ should be the ones getting more wins than boxers $s_{1},s_{2},s_{3}$. Each boxer $s_{4},s_{5},s_{6}$ should use a horseshoe on one of $s_{1},s_{2}, s_{3}$(to make $p$ a minimum) and so will get an extra win, therefore the total number of wins $s_{4}$ will get is $u+(p-1)+1$, $s_{5}$ will have $u+(p-2)+1$ wins, $s_{6}$ will obtain $u+(p-3)+1$ wins. To minimize the number of wins of $s_{1},s_{2},s_{3}$ each plotter should use a horseshoe on them. To minimmize te maximum number of wins that $s_{1},s_{2},s_{3}$ will get, we need $\frac{p}{3}+1$ plotters using horseshoe on $s_{1}$, $\frac{p}{3}$ plotters using horseshoe on $s_{2}$, $\frac{p}{3}-1$ plotters using horseshoe on $s_{3}$. Then the total number of wins each of $s_{1},s_{2},s_{3}$ will get is $u+1+\frac{2p}{3}$. Now, we need the minimum number of wins obtained by $s_{4},s_{5},s_{6}$ to be greater than maximum number of wins obtained by $s_{1},s_{2},s_{3}$, si have: $u+(p-3)+1>u+1+\frac{2p}{3}$ $\frac{p}{3}>3$. $p>9$. Taking $p=10$, we cannot reach the desired result:need at least $5$ plotters use horseshoe on $s_{1}$, at least $4$ to use horseshoe on $s_{2}$, at least $3$ to use horseshoe on $s_{3}$. Taking $p=11$, we cannot reach the desired result:need at least $5$ plotters use horseshoe on $s_{1}$, at least $4$ to use horseshoe on $s_{2}$, at least $3$ to use horseshoe on $s_{3}$. Finally, $p=12$ works: $5$ plotters use horseshoe on $s_{1}$, $4$ use horseshoe on $s_{2}$, $3$ use horseshoe on $s_{3}$. So, answer is $12$.

Let $ABC$ be a triangle, $G$ it`s centroid, $H$ it`s orthocenter, and $M$ the midpoint of the arc $\widehat{AC}$ (not containing $B$). It is known that $MG=R$, where $R$ is the radius of the circumcircle. Prove that $BG\geq BH$. Proposed by F. Bakharev

From a $n\times (n-1)$ rectangle divided into unit squares, we cut the corner, which consists of the first row and the first column. (that is, the corner has $2n-2$ unit squares). For the following, when we say corner we reffer to the above definition, along with rotations and symmetry. Consider an infinite lattice of unit squares. We will color the squares with $k$ colors, such that for any corner, the squares in that corner are coloured differently (that means that there are no squares coloured with the same colour). Find out the minimum of $k$. Proposed by S. Berlov

For a positive integer, we define it's set of exponents the unordered list of all the exponents of the primes, in it`s decomposition. For example, $18=2\cdot 3^{2}$ has it`s set of exponents $1,2$ and $300=2^{2}\cdot 3\cdot 5^{2}$ has it`s set of exponents $1,2,2$. There are given two arithmetical progressions $\big(a_{n}\big)_{n}$ and $\big(b_{n}\big)_{n}$, such that for any positive integer $n$, $a_{n}$ and $b_{n}$ have the same set of exponents. Prove that the progressions are proportional (that is, there is $k$ such that $a_{n}=kb_{n}$ for any $n$). Proposed by A. Golovanov