There are $n \geq 2$ coins numbered from $1$ to $n$. These coins are placed around a circle, not necessarily in order. In each turn, if we are on the coin numbered $i$, we will jump to the one $i$ places from it, always in a clockwise order, beginning with coin number 1. For an example, see the figure below. Find all values of $n$ for which there exists an arrangement of the coins in which every coin will be visited.

Consider the isosceles right triangle $ABC$ with $\angle BAC = 90^\circ$. Let $\ell$ be the line passing through $B$ and the midpoint of side $AC$. Let $\Gamma$ be the circumference with diameter $AB$. The line $\ell$ and the circumference $\Gamma$ meet at point $P$, different from $B$. Show that the circumference passing through $A,\ C$ and $P$ is tangent to line $BC$ at $C$.

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R}\longrightarrow \mathbb{R}$ so that the equality $$f(x+yf(x+y)) +xf(x)= f(xf(x+y+1))+y^2$$is true for any real numbers $x,y$.

Lucía multiplies some positive one-digit numbers (not necessarily distinct) and obtains a number $n$ greater than 10. Then, she multiplies all the digits of $n$ and obtains an odd number. Find all possible values of the units digit of $n$. $\textit{Proposed by Pablo Serrano, Ecuador}$

Celeste has an unlimited amount of each type of $n$ types of candy, numerated type 1, type 2, ... type n. Initially she takes $m>0$ candy pieces and places them in a row on a table. Then, she chooses one of the following operations (if available) and executes it: $1.$ She eats a candy of type $k$, and in its position in the row she places one candy type $k-1$ followed by one candy type $k+1$ (we consider type $n+1$ to be type 1, and type 0 to be type $n$). $2.$ She chooses two consecutive candies which are the same type, and eats them. Find all positive integers $n$ for which Celeste can leave the table empty for any value of $m$ and any configuration of candies on the table. $\textit{Proposed by Federico Bach and Santiago Rodriguez, Colombia}$

Let $ABC$ be a triangle with incenter $I$, and $A$-excenter $\Gamma$. Let $A_1,B_1,C_1$ be the points of tangency of $\Gamma$ with $BC,AC$ and $AB$, respectively. Suppose $IA_1, IB_1$ and $IC_1$ intersect $\Gamma$ for the second time at points $A_2,B_2,C_2$, respectively. $M$ is the midpoint of segment $AA_1$. If the intersection of $A_1B_1$ and $A_2B_2$ is $X$, and the intersection of $A_1C_1$ and $A_2C_2$ is $Y$, prove that $MX=MY$.