We are given a polynomial \(P(x)\) and numbers \(a_1\), \(a_2\), \(a_3\), \(b_1\), \(b_2\), \(b_3\) such that \(a_1a_2a_3 \ne 0\). It turned out that \(P (a_1x + b_1) + P (a_2x + b_2) = P (a_3x + b_3)\) for any real \(x\). Prove that \(P (x)\) has at least one real root.
A monkey, donkey and goat decided to play a game. They sat in a row, with the monkey on the right. They started to play the violin, but very poorly. They changed places and then the donkey was in the middle. However the violin trio still didn’t sound as they wanted it to. They changed places once more. After changing places 3 times, each of the three “musicians” had a chance to sit in the left, middle and right of the row. Who sat where after the third change of seats?
There is a group of 5 people: Alex, Beatrice, Victor, Gregory and Deborah. Each of them has one of the following codenames: V, W, X, Y, Z. We know that:
Alex is 1 year older than V,
Beatrice is 2 years older than W,
Victor is 3 years older than X,
Gregory is 4 years older than Y.
Who is older and by how much: Deborah or Z?
2012 pine cones lay under the fir-tree. Winnie the Pooh and the donkey Eeyore play a game: they take turns picking up these pine cones. Winnie-the-Pooh takes either one or four cones in each of his turns, and Eeyore – either one or three. Winnie the Pooh goes first. The player who cannot make a move loses. Which of the players can be guaranteed to win, no matter how their opponent plays?
In front of a gnome there lie three piles of diamonds: one with 17, one with 21 and one with 27 diamonds. In one of the piles lies one fake diamond. All the diamonds have the same appearance, and all real diamonds weigh the same, and the fake one differs in its weight. The gnome has a cup weighing scale without weights. The dwarf must find with one weighing a pile, in which all the diamonds are real. How should he do it?
Ladybirds gathered in a sunny clearing. If the ladybird has \(6\) spots, then it always speaks the truth, and if it has \(4\) spots, then it always lies. There are no other types of ladybirds in the meadow. The first ladybird said: “We each have the same number of spots on our backs.” The second one said: “Everyone has \(30\) spots on their backs in total.” “No, we all have \(26\) spots on their backs in total,” the third objected. “Of these three, exactly one told the truth,” – said each of the other ladybirds. How many ladybugs were gathered in the meadow?
Thirty girls – 13 in red dresses and 17 in blue dresses – led a dance around the Christmas tree. Subsequently, each of them was asked if her neighbour on the right was in a blue dress. It turned out that those girls which answered correctly were only those who stood between two girls in dresses of the same color. How many girls could have said yes?
Here’s a rather simple rebus:
\(EX\) is four times larger than \(OJ\).
\(AJ\) is four times larger than \(OX\).
Find the sum of all four numbers.
The numbers 25 and 36 are written on a blackboard. Consider the game with two players where: in one turn, a player is allowed to write another natural number on the board. This number must be the difference between any two of the numbers already written, such that this number does not already appear on the blackboard. The loser is the player who cannot make a move.
Consider a chessboard of size (number of rows \(\times\) number of columns): a) \(9\times 10\); b) \(10\times 12\); c) \(9\times 11\). Two people are playing a game where: in one turn a player is allowed to cross out any row or column as long as there it contains at least one square that is not crossed out. The loser is the player who cannot make a move. Which player will win?