Problems

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

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.

In a row there are 2023 numbers. The first number is 1. It is known that each number, except the first and the last, is equal to the sum of two neighboring ones. Find the last number.

In good conditions, bacteria in a Petri cup spread quite fast, doubling every second. If there was initially one bacterium, then in \(32\) seconds the bacteria will cover the whole surface of the cup.

Now suppose that there are initially \(4\) bacteria. At what time will the bacteria cover the surface of the cup?

Among the first \(20\) Fibonacci numbers: \(F_0 = 0,F_1 = 1,F_2 = 1, F_3 = 2, F_4 = 3,..., F_{20} = 6765\) find all numbers whose digit-sum is equal to their index. For example, \(F_1=1\) fits the description, but \(F_{20} = 6765\) does not, since \(6+7+6+5 \neq 20\).

Among the first \(20\) Fibonacci numbers: \(F_0 = 0,F_1 = 1,F_2 = 1, F_3 = 2, F_4 = 3,..., F_{20} = 6765\) check whether the numbers with prime index are prime. The index is another name for a number’s place in the sequence.

Consider Pascal’s triangle: it starts with \(1\), then each entry in the triangle is the sum of the two numbers above it. Prove that the diagonals of Pascal’s triangle add up to Fibonacci numbers.

Prove for any integers \(m,n\ge0\) that \(F_{m+n} = F_{m-1}F_n + F_mF_{n+1}\).
Corollary: if \(k\mid n\), then \(F_k\mid F_n\). This can be proven by induction if we write \(n=sk\) for a natural \(s\), then \[F_{k+(s-1)k} = F_{k-1}F_{(s-1)k} + F_kF_{(s-1)k+1}.\]

Denote by \(\gcd(m,n)\) the greatest common divisor of numbers \(m,n\), namely the largest possible \(d\) which divides both \(n\) and \(m\). Prove for any \(m,n\) that \[\gcd(F_n,F_m) = F_{\gcd(m,n)}.\]

Simplify \(F_0-F_1+F_2-F_3+...-F_{2n-1}+F_{2n}\), where \(n\) is a positive integer.