Školjka

%V0 Consider $2009$ cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two player, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of $50$ consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. (a) Does the game necessarily end? (b) Does there exist a winning strategy for the starting player? Proposed by Michael Albert, Richard Guy, New Zealand

%V0 Let $n \in \mathbb N$ and $A_n$ set of all permutations $(a_1, \ldots, a_n)$ of the set $\{1, 2, \ldots , n\}$ for which $$k|2(a_1 + \cdots+ a_k), \text{ for all } 1 \leq k \leq n.$$ Find the number of elements of the set $A_n.$

%V0 This ISL 2005 problem has not been used in any TST I know. A pity, since it is a nice problem, but in its shortlist formulation, it is absolutely incomprehensible. Here is a mathematical restatement of the problem: Let $k$ be a nonnegative integer. A forest consists of rooted (i. e. oriented) trees. Each vertex of the forest is either a leaf or has two successors. A vertex $v$ is called an extended successor of a vertex $u$ if there is a chain of vertices $u_{0}=u$, $u_{1}$, $u_{2}$, ..., $u_{t-1}$, $u_{t}=v$ with $t>0$ such that the vertex $u_{i+1}$ is a successor of the vertex $u_{i}$ for every integer $i$ with $0\leq i\leq t-1$. A vertex is called dynastic if it has two successors and each of these successors has at least $k$ extended successors. Prove that if the forest has $n$ vertices, then there are at most $\frac{n}{k+2}$ dynastic vertices.

%V0 There are $10001$ students at an university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold: i.) Each pair of students are in exactly one club. ii.) For each student and each society, the student is in exactly one club of the society. iii.) Each club has an odd number of students. In addition, a club with ${2m+1}$ students ($m$ is a positive integer) is in exactly $m$ societies. Find all possible values of $k$. RemarkIn IMOTC 2005, it is given that $m=2005$.

%V0 Let $n \geq 2, n \in \mathbb{N}$ and $A_0 = (a_{01},a_{02}, \ldots, a_{0n})$ be any $n-$tuple of natural numbers, such that $0 \leq a_{0i} \leq i-1,$ for $i = 1, \ldots, n.$ $n-$tuples $A_1= (a_{11},a_{12}, \ldots, a_{1n}), A_2 = (a_{21},a_{22}, \ldots, a_{2n}), \ldots$ are defined by: $a_{i+1,j} = Card \{a_{i,l}| 1 \leq l \leq j-1, a_{i,l} \geq a_{i,j}\},$ for $i \in \mathbb{N}$ and $j = 1, \ldots, n.$ Prove that there exists $k \in \mathbb{N},$ such that $A_{k+2} = A_{k}.$

%V0 a) Show that the set $\mathbb{Q}^{ + }$ of all positive rationals can be partitioned into three disjoint subsets. $A,B,C$ satisfying the following conditions: $$BA = B; B^2 = C; BC = A;$$ where $HK$ stands for the set $\{hk: h \in H, k \in K\}$ for any two subsets $H, K$ of $\mathbb{Q}^{ + }$ and $H^2$ stands for $HH.$ b) Show that all positive rational cubes are in $A$ for such a partition of $\mathbb{Q}^{ + }.$ c) Find such a partition $\mathbb{Q}^{ + } = A \cup B \cup C$ with the property that for no positive integer $n \leq 34,$ both $n$ and $n + 1$ are in $A,$ that is, $$\text{min} \{n \in \mathbb{N}: n \in A, n + 1 \in A \} > 34.$$