Solving the problem: “What is the solution of the expression \(x^{2000} + x^{1999} + x^{1998} + 1000x^{1000} + 1000x^{999} + 1000x^{998} + 2000x^3 + 2000x^2 + 2000x + 3000\) (\(x\) is a real number) if \(x^2 + x + 1 = 0\)?”, Vasya got the answer of 3000. Is Vasya right?
Prove that amongst the numbers of the form \[19991999\dots 19990\dots 0\] – that is 1999 a number of times, followed by a number of 0s – there will be at least one divisible by 2001.
A game with 25 coins. In a row there are 25 coins. For a turn it is allowed to take one or two neighbouring coins. The player who has nothing to take loses.
There are three piles of rocks: in the first pile there are 10 rocks, 15 in the second pile and 20 in the third pile. In this game (with two players), in one turn a player is allowed to divide one of the piles into two smaller piles. The loser is the one who cannot make a move. Which player would be the winner?
In the first pile there are 100 sweets and in the second there are 200. Consider the game with two players where: in one turn a player can take any amount of sweets from one of the piles. The winner is the one who takes the last sweet. Which player would win by using the correct strategy?
Let \(M\) be the point of intersection of the medians of the triangle \(ABC\), and \(O\) an arbitrary point on a plane. Prove that \[OM^2 = 1/3 (OA^2 + OB^2 + OC^2) - 1/9 (AB^2 + BC^2 + AC^2).\]
Three non-coplanar vectors are given. Is it possible to find a fourth vector perpendicular to the three vectors given?
Find the volume of an inclined triangular prism whose base is an equilateral triangle with sides equal to a if the side edge of the prism is equal to the side of the base and is inclined to the plane of the base at an angle of \(60^{\circ}\).
The grandad is twice as strong as the grandma, the grandma is three times stronger than the granddaughter, the granddaughter is four times stronger than the dog, the dog is five times stronger than the cat and the cat is six times stronger than the mouse. The grandad, the grandma, the granddaughter, the dog and the cat together with the mouse can pull out the pumpkin from the ground, which they cannot do without the mouse. How many mice should be summoned so that they can pull out the pumpkin themselves?
In the dense dark forest ten sources of dead water are erupting from the ground: named from #1 to #10. Of the first nine sources, dead water can be taken by everyone, but the source #10 is in the cave of the dark wizard, from which no one, except for the dark wizard himself, can collect water. The taste and color of dead water is no different from ordinary water, however, if a person drinks from one of the sources, then he will die. Only one thing can save him: if he then drinks poison from a source whose number is greater. For example, if he drinks from the seventh source, then he must necessarily drink poison from the #8, #9 or #10 sources. If he doesn’t drink poison from the seventh source, but does from the ninth, only the poison from the source #10 will save him. And if he originally drinks the tenth poison, then nothing will help him now. Robin Hood summoned the dark wizard to a duel. The terms of the duel were as follows: each brings with him a mug of liquid and gives it to his opponent. The dark wizard was delighted: “Hurray, I will give him poison #10, and Robin Hood can not be saved! And I’ll drink the poison, which Robin Hood brings to me, then ill drink the #10 poison and that will save me!” On the appointed day, both opponents met at the agreed place. They honestly exchanged mugs and drank what was in them. However, afterwards erupted the joy and surprise of the inhabitants of the dark forest, when it turned out that the dark wizard had died, and Robin Hood remained alive! Only the Wise Owl was able to guess how Robin Hood had managed to defeat dark wizard. Try and guess as well.