Four numbers (from 1 to 9) have been used to create two numbers with four-digits each. These two numbers are the maximum and minimum numbers, respectively, possible. The sum of these two numbers is equal to 11990. What could the two numbers be?
Prove that there are no natural numbers \(a\) and \(b\) such that \(a^2 - 3b^2 = 8\).
Prove there are no integer solutions for the equation \(x^2=y^2+1990\).
Given a board (divided into squares) of the size: a) \(10\times 12\), b) \(9\times 10\), c) \(9\times 11\), consider the game with two players where: in one turn a player is allowed to cross out any row or any column if there is at least one square not crossed out. The loser is the one who cannot make a move. Is there a winning strategy for one of the players?
Prove that for a real positive \(\alpha\) and a positive integer \(d\), \(\lfloor \alpha / d\rfloor = \lfloor \lfloor \alpha\rfloor / d\rfloor\) is always satisfied.
Prove that if \(p\) is a prime number and \(1 \leq k \leq p - 1\), then \(\binom{p}{k}\) is divisible by \(p\).
Prove that if \(p\) is a prime number, then \((a + b)^p - a^p - b^p\) is divisible by \(p\) for any integers \(a\) and \(b\).
Prove that if \((p, q) = 1\) and \(p/q\) is a rational root of the polynomial \(P (x) = a_nx^n + \dots + a_1x + a_0\) with integer coefficients, then
a) \(a_0\) is divisible by \(p\);
b) \(a_n\) is divisible by \(q\).
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 \(n\)?
What has a greater value: \(300!\) or \(100^{300}\)?