« Vrati se
na "cetvrtini" beskonacne kvadratne mreze ( tamo gdje je x \geqslant 0 i y  \geqslant 0 ), igra se sljedeca igra: u kvadraticima s koordinatama (0,0), (0,1), (0, 2), (1,0), (1,1), (2, 0) postavljen je po jedan zeton. dozvoljen potez je maknuti jedan zeton s polja (m, n), gdje su m, n \in \mathbb{N}_0 takvi da na poljima (m+1,n) i (m, n+1) ne stoji zeton, te staviti po jedan zeton na polje (m+1,n) i na polje (m, n+1). je li moguce konacnom primjenom ovih poteza doci u situaciju da na pocetnih sest polja ne stoji vise niti jedan zeton?

Slični zadaci

Dano je n+1 prirodnih brojeva čiji je zbroj 3n i oni su poredani u krug. Dokaži da postoji nekoliko uzastopnih brojeva (moguće i samo 1) čiji je zbroj jednak točno 2n .
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
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.
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.
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.
For n an odd positive integer, the unit squares of an n\times n chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an L-shape formed by three connected unit squares. For which values of n is it possible to cover all the black squares with non-overlapping trominos? When it is possible, what is the minimum number of trominos needed?