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The following words/sounds are given: look, yar, yell, lean, lease. Determine what will happen if the sounds that make up these words are pronounced in reverse order.

Prove that the polynomial \(P (x)\) is divisible by its derivative if and only if \(P (x)\) has the form \(P(x) = a_n(x - x_0)^n\).

Prove that for \(n > 0\) the polynomial \[P (x) = n^2x^{n + 2} - (2n^2 + 2n - 1) x^{n + 1} + (n + 1)^2x^n - x - 1\] is divisible by \((x - 1)^3\).

Prove that for \(n> 0\) the polynomial \(x^{2n + 1} - (2n + 1)x^{n + 1} + (2n + 1)x^n - 1\) is divisible by \((x - 1)^3\).

Prove that the polynomial \(P (x) = a_0 + a_1x + \dots + a_nx^n\) has a number \(-1\) which is a root of multiplicity \(m + 1\) if and only if the following conditions are satisfied: \[\begin{aligned} a_0 - a_1 + a_2 - a_3 + \dots + (-1)^{n}a_n &= 0,\\ - a_1 + 2a_2 - 3a_3 + \dots + (-1)^{n}na_n &= 0,\\ \dots \\ - a_1 + 2^{m}a_2 - 3^{m}a_3 + \dots + (-1)^{n}n^{m}a_n &= 0. \end{aligned}\]

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).

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.