Prove that for \(n> 0\) the polynomial \(x^{2n + 1} - (2n + 1)x^{n + 1} + (2n + 1)x^n - 1\) is divisible by \((x - 1)^3\).
A class contains 33 pupils, who have a combined age of 430 years. Prove that if we picked the 20 oldest pupils they would have a combined age of no less than 260 years. The age of any given pupil is a whole number.
In a one-on-one tournament 10 chess players participate. What is the least number of rounds after which the single winner could have already been determined? (In each round, the participants are broken up into pairs. Win – 1 point, draw – 0.5 points, defeat – 0).
Specify any solution of the puzzle: \(2014 + YES =BEAR\).
The height of the room is 3 meters. When it was being renovated, it turned out that more paint was needed on each wall than on the floor. Can the area of the floor of this room be more than 10 square meters?
A castle is surrounded by a circular wall with nine towers, at which there are knights on duty. At the end of each hour, they all move to the neighbouring towers, each knight moving either clockwise or counter-clockwise. During the night, each knight stands for some time at each tower. It is known that there was an hour when at least two knights were on duty at each tower, and there was an hour when there was precisely one knight on duty on each of exactly five towers. Prove that there was an hour when there were no knights on duty on one of the towers.
In the entry \({*} + {*} + {*} + {*} + {*} + {*} + {*} + {*} = {*}{*}\) replace the asterisks with different digits so that the equality is correct.
Hannah placed 101 counters in a row which had values of 1, 2 and 3 points. It turned out that there was at least one counter between every two one point counters, at least two counters lie between every two two point counters, and at least three counters lie between every two three point counters. How many three point counters could Hannah have?
A chequered strip of \(1 \times N\) is given. Two players play the game. The first player puts a cross into one of the free cells on his turn, and subsequently the second player puts a nought in another one of the cells. It is not allowed for there to be two crosses or two noughts in two neighbouring cells. The player who is unable to make a move loses.
Which of the players can always win (no matter how their opponent played)?
We are given 111 different natural numbers that do not exceed 500. Could it be that for each of these numbers, its last digit coincides with the last digit of the sum of all of the remaining numbers?