Problems

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Found: 16

In the number \(1234096\dots\) each digit, starting with the 5th digit is equal to the final digit of the sum of the previous 4 digits. Will the digits 8123 ever occur in that order in a row in this number?

The Babylonian algorithm for deducing \(\sqrt{2}\). The sequence of numbers \(\{x_n\}\) is given by the following conditions: \(x_1 = 1\), \(x_{n + 1} = \frac 12 (x_n + 2/x_n)\) (\(n \geq 1\)).

Prove that \(\lim\limits_{n\to\infty} x_n = \sqrt{2}\).

What will the sequence from the previous problem 61297 be converging towards if we choose \(x_1\) as equal to \(-1\) as the initial condition?

The sequence of numbers \(a_n\) is given by the conditions \(a_1 = 1\), \(a_{n + 1} = a_n + 1/a^2_n\) (\(n \geq 1\)).

Is it true that this sequence is limited?

We call the geometric-harmonic mean of numbers \(a\) and \(b\) the general limit of the sequences \(\{a_n\}\) and \(\{b_n\}\) constructed according to the rule \(a_0 = a\), \(b_0 = b\), \(a_{n + 1} = \frac{2a_nb_n}{a_n + b_n}\), \(b_{n + 1} = \sqrt{a_nb_n}\) (\(n \geq 0\)).

We denote it by \(\nu (a, b)\). Prove that \(\nu (a, b)\) is related to \(\mu (a, b)\) (see problem number 61322) by \(\nu (a, b) \times \mu (1/a, 1/b) = 1\).

Problem number 61322 says that both of these sequences have the same limit.

This limit is called the arithmetic-geometric mean of the numbers \(a, b\) and is denoted by \(\mu (a, b)\).

On a calculator keypad, there are the numbers from 0 to 9 and signs of two actions (see the figure). First, the display shows the number 0. You can press any keys. The calculator performs the actions in the sequence of clicks. If the action sign is pressed several times, the calculator will only remember the last click.

a) The button with the multiplier sign breaks and does not work. The Scattered Scientist pressed several buttons in a random sequence. Which result of the resulting sequence of actions is more likely: an even number or an odd number?

b) Solve the previous problem if the multiplication symbol button is repaired.

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.

Prove that every pair of consecutive Fibonacci numbers are coprime. That is, they share no common factors other than 1.