A number is called nillless if it is integer and positive and contains no zeros. You can make a positive integer nillless by simply omitting the zeros. We denote this with square brackets, for example $[2050] = 25$ and $[13] = 13$. When we multiply, add, and subtract we indicate with square brackets when we omit the zeros. For example, $[[4 \cdot 5] + 7] = [[20] + 7] = [2 + 7] = [9] = 9$ and $[[5 + 5] + 9] = [[10] + 9] = [1 + 9] = [10] = 1$. The following is known about the two numbers $a$ and $b$: $\bullet$ $a$ and $b$ are nillless, $\bullet$ $1 < a < b < 100$, $\bullet$ $[[a \cdot b] - 1] = 1$. Which pairs $(a, b)$ satisfy these three requirements?

In a room there are $2023$ vases numbered from $1$ to $2023$. In each vase we want to put a note with a positive integer from $1$, $2$ $...$ , $2023$ on it. The numbers on the notes do not necessarily have to be distinct. The following should now apply to each vase. Look at the note inside the vase, find the (not necessarily different) vase with the number written on the note, and look at the note inside this vase. Then the average of the numbers on the two notes must be exactly equal to the number of the first selected vase. For example, if we put a note with the number $5$ in vase $13$, then vase $5$ should contain a note with the number $21$ on it: after all, the average of $5$ and $ 21$ is $13$. Determine all possible ways to provide each vase with a note.

Felix chooses a positive integer as the starting number and writes it on the board. He then repeats the next step: he replaces the number $n$ on the board by $\frac12n$ if $n$ is even and by $n^2 + 3$ if $n$ is odd. For how many choices of starting numbers below $2023$ will Felix never write a number of more than four digits on the board?

In acute-angled triangle $ABC$ with $|BC| < |BA|$, point $N$ is the midpoint of $AC$. The circle with diameter $AB$ intersects the bisector of $\angle B$ in two points: $B$ and $X$. Prove that $XN$ is parallel to $BC$.

A maths teacher has $10$ cards with the numbers $1$ to $10$ on them, one number per card. She places these cards in some order in a line next to each other on the table. The students come to the table, one at a time. The student whose turn it is goes once through the line of cards from left to right and removes every card she encounters that is (at that moment) the lowest card on the table. This continues till all cards are removed from the table. For example, if the line is in order $3$, $1$, $4$, $5$, $8,$ $6$, $9$, $10$, $2$, $7$ from left to right, the first student takes cards $1$ and $2$. Then the second student comes who, in our example, takes the cards $3$, $4$, $5$, $6$, and $7$. The third student then takes the cards $8$, $9$, and $10$. Let $A$ be the number of sequences of cards that the teacher can choose so that exactly nine students get a turn to pick cards. Let $B$ be the number of sequences of cards that the teacher can choose so that exactly two students get a turn to pick cards. Prove that $A = B$.