In a circumference there are $99$ natural numbers. If $a$ and $b$ are two consecutive numbers in the circle, then they must satisfies one of the following conditions: $a-b=1, a-b=2$ or $\frac{a}{b}=2$. Prove that, in the circle exists a number multiple of $3$.

Let $ABCD$ a trapezium with $AB$ parallel to $CD, \Omega$ the circumcircle of $ABCD$ and $A_1,B_1$ points on segments $AC$ and $BC$ respectively, such that $DA_1B_1C$ is a cyclic cuadrilateral. Let $A_2$ and $B_2$ the symmetric points of $A_1$ and $B_1$ with respect of the midpoint of $AC$ and $BC$, respectively. Prove that points $A, B, A_2, B_2$ are concyclic.

Let $n>1$ be an integer. Find all non-constant real polynomials $P(x)$ satisfying , for any real $x$ , the identy \[P(x)P(x^2)P(x^3)\cdots P(x^n)=P(x^{\frac{n(n+1)}{2}})\]

The diagonals of a convex quadrilateral $ABCD$ intersect in $E$. Let $S_1, S_2, S_3$ and $S_4$ the areas of the triangles $AEB, BEC, CED, DEA$ respectively. Prove that, if exists real numbers $w, x, y$ and $z$ such that $$S_1=x+y+xy, S_2=y+z+yz, S_3=w+z+wz, S_4=w+x+wx,$$ then $E$ is the midpoint of $AC$ or $E$ is the midpoint of $BD$.

Martin and Chayo have an bag with $2016$ chocolates each one. Both empty his bag on a table making a pile of chocolates. They decide to make a competence to see who gets the chocolates, as follows: A movement consist that a player take two chocolates of his pile, keep a chocolate in his bag and put the other chocolate in the pile of the other player, in his turn the player needs to make at least one movement and he can repeat as many times as he wish before passing his turn. Lost the player that can not make at least one movement in his turn. If Martin starts the game, who can ensure the victory and keep all the chocolates?

Let $M$ the midpoint of $AC$ of an acutangle triangle $ABC$ with $AB>BC$. Let $\Omega$ the circumcircle of $ABC$. Let $P$ the intersection of the tangents to $\Omega$ in point $A$ and $C$ and $S$ the intersection of $BP$ and $AC$. Let $AD$ the altitude of triangle $ABP$ with $D$ in $BP$ and $\omega$ the circumcircle of triangle $CSD$. Let $K$ and $C$ the intersections of $\omega$ and $\Omega (K\neq C)$. Prove that $\angle CKM=90^\circ$.