We are given a \(100\times 100\) square grid and \(N\) counters. All of the possible arrangements of the counters on the grid which follow the following rule are considered: no two counters lie in adjacent squares.
What is the largest value of \(N\) for which, in every single possible arrangement of counters following this rule, it is possible to find at least one counter such that moving it to an adjacent square does not break the rule. Squares are considered adjacent if they share a side.
With a non-zero number, the following operations are allowed: \(x \rightarrow \frac{1+x}{x}\), \(x \rightarrow \frac{1-x}{x}\). Is it true that from every non-zero rational number one can obtain each rational number with the help of a finite number of such operations?
Find all functions \(f (x)\) defined for all positive \(x\), taking positive values and satisfying the equality \(f (x^y) = f (x)^f (y)\) for any positive \(x\) and \(y\).
On a particular day it turned out that every person living in a particular city made no more than one phone call. Prove that it is possible to divide the population of this city into no more than three groups, so that within each group no person spoke to any other by telephone.
The function \(f (x)\) is defined and satisfies the relationship \((x-1) f((x=1)/(x-1)) - f (x) = x\) for all \(x \neq 1\). Find all such functions.
Prove that for any natural number \(a_1> 1\) there exists an increasing sequence of natural numbers \(a_1, a_2, a_3, \dots\), for which \(a_1^2+ a_2^2 +\dots+ a_k^2\) is divisible by \(a_1+ a_2+\dots+ a_k\) for all \(k \geq 1\).
Is there a sequence of natural numbers in which every natural number occurs exactly once, and for any \(k = 1, 2, 3, \dots\) the sum of the first \(k\) terms of the sequence is divisible by \(k\)?
At all rational points of the real line, integers are arranged. Prove that there is a segment such that the sum of the numbers at its ends does not exceed twice the number on its middle.
Find all the functions \(f\colon \mathbb {R} \rightarrow \mathbb {R}\) which satisfy the inequality \(f (x + y) + f (y + z) + f (z + x) \geq 3f (x + 2y + 3z)\) for all \(x, y, z\).
A number set \(M\) contains \(2003\) distinct positive numbers, such that for any three distinct elements \(a, b, c\) in \(M\), the number \(a^2 + bc\) is rational. Prove that we can choose a natural number \(n\) such that for any \(a\) in \(M\) the number \(a\sqrt{n}\) is rational.