Školjka

%V0 A pile of $n$ pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a final configuration. For each $n$, show that, no matter what choices are made at each stage, the final configuration obtained is unique. Describe that configuration in terms of $n$. IMO ShortList 2001, combinatorics problem 7, alternative

%V0 Among a group of 120 people, some pairs are friends. A weak quartet is a set of four people containing exactly one pair of friends. What is the maximum possible number of weak quartets ?

%V0 3. $A$ and $B$ play a game, given an integer $N$, $A$ writes down $1$ first, then every player sees the last number written and if it is $n$ then in his turn he writes $n+1$ or $2n$, but his number cannot be bigger than $N$. The player who writes $N$ wins!, For wich values of $N$ does $B$ wins?

%V0 For an ${n\times n}$ matrix $A$, let $X_{i}$ be the set of entries in row $i$, and $Y_{j}$ the set of entries in column $j$, ${1\leq i,j\leq n}$. We say that $A$ is golden if ${X_{1},\dots ,X_{n},Y_{1},\dots ,Y_{n}}$ are distinct sets. Find the least integer $n$ such that there exists a ${2004\times 2004}$ golden matrix with entries in the set ${\{1,2,\dots ,n\}}$.

%V0 For $n\ge 2$, let $S_1$, $S_2$, $\ldots$, $S_{2^n}$ be $2^n$ subsets of $A = \{1, 2, 3, \ldots, 2^{n + 1}\}$ that satisfy the following property: There do not exist indices $a$ and $b$ with $a < b$ and elements $x$, $y$, $z\in A$ with $x < y < z$ and $y$, $z\in S_a$, and $x$, $z\in S_b$. Prove that at least one of the sets $S_1$, $S_2$, $\ldots$, $S_{2^n}$ contains no more than $4n$ elements. Proposed by Gerhard Woeginger, Netherlands

%V0 For any integer $n\geq 2$, we compute the integer $h(n)$ by applying the following procedure to its decimal representation. Let $r$ be the rightmost digit of $n$. If $r=0$, then the decimal representation of $h(n)$ results from the decimal representation of $n$ by removing this rightmost digit $0$.If $1\leq r \leq 9$ we split the decimal representation of $n$ into a maximal right part $R$ that solely consists of digits not less than $r$ and into a left part $L$ that either is empty or ends with a digit strictly smaller than $r$. Then the decimal representation of $h(n)$ consists of the decimal representation of $L$, followed by two copies of the decimal representation of $R-1$. For instance, for the number $17,151,345,543$, we will have $L=17,151$, $R=345,543$ and $h(n)=17,151,345,542,345,542$.Prove that, starting with an arbitrary integer $n\geq 2$, iterated application of $h$ produces the integer $1$ after finitely many steps. Proposed by Gerhard Woeginger, Austria