Find a subset of $\{1,2, ...,2022\}$ with maximum number of elements such that it does not have two elements $a$ and $b$ such that $a = b + d$ for some divisor $d$ of $b$.

In a classroom there are $20$ rows of $22$ desks each $(22$ desks have noone in front of them). The $440$ contestants of a certain regional math contest are going to sit at the desks. Before the exam, the organizers left some amount of sweets on each desk, which amount can be any positive integer. When the students go into the room, just before sitting down they look at the desk behind them, the one on the left and the one diagonally opposite to the right of theirs, thus seeing how many sweets each one has; if there is no desk in any of these directions, they simply ignore that position. Then they sit and watch their own sweets. A student gets angry if any of the desks he saw has more than one candy more than his. The organizers managed to distribute the sweets in such a way that no student gets angry. Prove that there are $8$ students with the same amount of sweets.

In my isosceles triangle $\vartriangle ABC$ with $AB = CA$, we draw $D$ the midpoint of $BC$. Let $E$ be a point on $AC$ such that $\angle CDE = 60^o$ and $M$ the midpoint of $DE$. Prove that $\angle AME = \angle BMD$.

Prove that in all triangles $\vartriangle ABC$ with $\angle A = 2 \angle B$ it holds that, if $D$ is the foot of the perpendicular from $C$ to the perpendicular bisector of $AB$, $\frac{AC}{DC}$ is constant for any value of $\angle B$.

Determine all positive integers $n$ such that $\lfloor \sqrt{n} \rfloor - 1$ divides $n + 1$ and $\lfloor \sqrt{n} \rfloor +2$ divides $ n + 4$.

There is a $2021 \times 2023$ board that has a white piece in the central square, on which Mich and Moka are going to play in turns. First Mich places a green token on any free space so that it is not in the same row or column as the white token, then Moka places a red token on any free space so that it is not in the same row or column as the white token. white or green. From now on, Mich will place green tokens and Moka will place red tokens alternately according to the following rules: $\bullet$ For the placed piece there must be another piece of the same color in its row or column, such that there is no other piece between both pieces. $\bullet$ If there is at least one box that meets the previous rule, then it is mandatory to place a token. When a token is placed, it changes all the tokens that are on squares adjacent to it to the same color. The game ends when one of the players can no longer place tiles. If when the game ends the board has more green tiles then Mich wins, and if it has more red tiles then Moka wins. Determine if either player has a winning strategy.