We have $2015$ prisinoers.The king gives everyone a hat coloured in one of $5$ colors.Everyone sees all hats expect his own.Now,the King orders them in a line(a prisioner can see all guys behind and in front of him).The king asks the prisinoers one by one does he know the color of his hat.If he answers [b]NO[/b],then he is killed.If he answers [b]YES[/b],then answers which color is his hat,if his answers is true,he goes to freedom,if not,he is killed.All the prisinors can hear did he answer [b]YES[/b] or [b]NO[/b],but if he answered [b]YES[/b],they don't know what did he answered(he is killed in public).They can think of a strategy before the King comes,but after that they can't comunicate.What is the largest number of prisinors we can guarentee that can survive?

Fill in the grid with the numbers 1 to 6 so that each number appears exactly once in each row and column. A horizontal gray line marks any cell when it is the middle cell of the three consecutive cells with the largest sum in that row. Similarly, a vertical gray line marks any cell when it is the middle of the three consecutive cells with the largest sum in that column. If there is a tie, multiple lines are drawn in the row or column. A cell can have both lines drawn, with the appearance of a plus sign. [asy] // Change this to see the solution bool DRAW_SOLUTION = true; int n = 6; real LINE_WIDTH = 0.3; void drawHLine(int x, int y) { fill((x,y+0.5-LINE_WIDTH/2)--(x,y+0.5+LINE_WIDTH/2)--(x+1,y+0.5+LINE_WIDTH/2)--(x+1,y+0.5-LINE_WIDTH/2)--cycle, gray(0.8)); } void drawVLine(int x, int y) { fill((x+0.5-LINE_WIDTH/2,y)--(x+0.5+LINE_WIDTH/2,y)--(x+0.5+LINE_WIDTH/2,y+1)--(x+0.5-LINE_WIDTH/2,y+1)--cycle, gray(0.8)); } void drawNum(int x, int y, int num) { label(scale(1.5)*string(num), (x+0.5,y+0.5)); } void drawSolNum(int x, int y, int num) { if (DRAW_SOLUTION) { drawNum(x, y, num); } } drawHLine(2,0); drawHLine(4,1); drawHLine(1,2); drawHLine(3,2); drawHLine(4,3); drawHLine(2,4); drawHLine(3,5); drawVLine(0,4); drawVLine(1,3); drawVLine(2,1); drawVLine(2,3); drawVLine(3,4); drawVLine(4,1); drawVLine(5,2); drawNum(0, 0, 5); drawNum(4, 0, 3); drawNum(1, 2, 2); drawNum(3, 3, 4); for(int i = 0; i <= 6; i += 1) { draw((i,0)--(i,6)); draw((0,i)--(6,i)); } [/asy]

On a mountain, three shepherds cyclically alternate shearing the same herd of sheep. The shepherds agreed to obey the following rules: (i) Every day a sheep can be shorn* on one side only; (ii) Every day at least one sheep must be shorn; (iii) No two days the same group of sheep can be shorn. The shepherd who first breaks the agreement will have to accompany the herd in the valley next fall. Can anyone of the shepherds shear the sheep in such a way to make sure that he will avoid this punishment? *shorn is the past tense of shear

In a room, we have 2019 aligned switches, connected to 2019 light bulbs, all initially switched on. Then, 2019 people enter the room one by one, performing the operation: The first, uses all the switches; the second, every second switch; the third, every third switch, and so on. How many lightbulbs remain switched on, after all the people entered ?

Elfs and trolls are seated at a round table, 60 creatures in total. Trolls always lie, and all elfs always speak the truth, except when they make a little mistake. Everybody claims to sit between an elf and a troll, but exactly two elfs made a mistake! How many trolls are there at this table?

Let $ABCD$ be a tetrahedron such that $AB = CD = \sqrt{41}$, $AC = BD = \sqrt{80}$, and $BC = AD = \sqrt{89}$. There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt{n}}{p}$, when $m$, $n$, and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$.

Fernando has six coins, one of which is fake. We do not know what the weight of a fake coin is or the weight of a real coin, we only know that real coins all have the same weight and that the weight of the fake coin is different. Using a two-pan scale, show that it is possible to discover the fake coin using just $3$ weighings.

The NIMO problem writers have invented a new chess piece called the [i]Oriented Knight[/i]. This new chess piece has a limited number of moves: it can either move two squares to the right and one square upward or two squares upward and one square to the right. How many ways can the knight move from the bottom-left square to the top-right square of a $16\times 16$ chess board? [i] Proposed by Tony Kim and David Altizio [/i]

Given a $ 1 \times 25$ rectangle divided into $ 25$ "boxes" ($ 1 \times 1$), is it possible to write integers $ 1$ to $ 25$ so that the sum of any two adjacent "boxes" is a perfect square?