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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 for every natural number \(n > 1\) the equality: \[\lfloor n^{1 / 2}\rfloor + \lfloor n^{1/ 3}\rfloor + \dots + \lfloor n^{1 / n}\rfloor = \lfloor \log_{2}n\rfloor + \lfloor \log_{3}n\rfloor + \dots + \lfloor \log_{n}n\rfloor\] is satisfied.

\(a_1, a_2, a_3, \dots\) is an increasing sequence of natural numbers. It is known that \(a_{a_k} = 3k\) for any \(k\). Find a) \(a_{100}\); b) \(a_{2022}\).

For which natural \(n\) does the number \(\frac{n^2}{1.001^n}\) reach its maximum value?

The function \(F\) is given on the whole real axis, and for each \(x\) the equality holds: \(F (x + 1) F (x) + F (x + 1) + 1 = 0\).

Prove that the function \(F\) can not be continuous.

We consider a sequence of words consisting of the letters “A” and “B”. The first word in the sequence is “A”, the \(k\)-th word is obtained from the \((k-1)\)-th by the following operation: each “A” is replaced by “AAB” and each “B” by “A”. It is easy to see that each word is the beginning of the next, thus obtaining an infinite sequence of letters: AABAABAAABAABAAAB...

a) Where in this sequence will the 1000th letter “A” be?

b) Prove that this sequence is non-periodic.