Author: A.K. Tolpygo
An irrational number \(\alpha\), where \(0 <\alpha <\frac 12\), is given. It defines a new number \(\alpha_1\) as the smaller of the two numbers \(2\alpha\) and \(1 - 2\alpha\). For this number, \(\alpha_2\) is determined similarly, and so on.
a) Prove that for some \(n\) the inequality \(\alpha_n <3/16\) holds.
b) Can it be that \(\alpha_n> 7/40\) for all positive integers \(n\)?
There was a football match of 10 versus 10 players between a team of liars (who always lie) and a team of truth-tellers (who always tell the truth). After the match, each player was asked: “How many goals did you score?” Some participants answered “one”, Callum said “two”, some answered “three”, and the rest said “five”. Is Callum lying if it is known that the truth-tellers won with a score of 20:17?
For all real \(x\) and \(y\), the equality \(f (x^2 + y) = f (x) + f (y^2)\) holds. Find \(f(-1)\).
Suppose that in each issue of our journal in the “Quantum” problem book there are five mathematics problems. We denote by \(f (x, y)\) the number of the first of the problems of the \(x\)-th issue for the \(y\)-th year. Write a general formula for \(f (x, y)\), where \(1 \geq x \geq 12\) and \(1970 \geq y \geq 1989\). Solve the equation \(f (x, y) = y\). For example, \(f (6, 1970) = 26\). Since \(1989\), the number of tasks has become less predictable. For example, in recent years, half the issues have 5 tasks, and in other issues there are 10. Even the number of magazine issues has changed, no longer being 12 but now 6.
Prove that for any odd natural number, \(a\), there exists a natural number, \(b\), such that \(2^b - 1\) is divisible by \(a\).
On a lottery ticket, it is necessary for Mary to mark 8 cells from 64. What is the probability that after the draw, in which 8 cells from 64 will also be selected (all such possibilities are equally probable), it turns out that Mary guessed
a) exactly 4 cells? b) exactly 5 cells? c) all 8 cells?
Author: L.N. Vaserstein
For any natural numbers \(a_1, a_2, \dots , a_m\), no two of which are equal to each other and none of which is divisible by the square of a natural number greater than one, and also for any integers and non-zero integers \(b_1, b_2, \dots , b_m\) the sum is not zero. Prove this.
Prove that for any positive integer \(n\), it is always possible to find a number, consisting of the digits \(1\) and \(2,\) that is divisible by \(2^n\). (For example, \(2\) is divisible by \(2\), \(12\) is divisible by \(4,\) \(112\) is divisible by \(8,\) \(2112\) is divisible by \(16\) and so on...).
A sequence of natural numbers \(a_1 < a_2 < a_3 < \dots < a_n < \dots\) is such that each natural number is either a term in the sequence, can be expressed as the sum of two terms in the sequence, or perhaps the same term twice. Prove that \(a_n \leq n^2\) for any \(n=1, 2, 3,\dots\)
Out of the given numbers 1, 2, 3, ..., 1000, find the largest number \(m\) that has this property: no matter which \(m\) of these numbers you delete, among the remaining \(1000 - m\) numbers there are two, of which one is divisible by the other.