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?
Upon the installation of a keypad lock, each of the 26 letters located on the lock’s keypad is assigned an arbitrary natural number known only to the owner of the lock. Different letters do not necessarily have different numbers assigned to them. After a combination of different letters, where each letter is typed once at most, is entered into the lock a summation is carried out of the corresponding numbers to the letters typed in. The lock opens only if the result of the summation is divisible by 26. Prove that for any set of numbers assigned to the 26 letters, there exists a combination that will open the lock.
Reception pupil Peter knows only the number 1. Prove that he can write a number divisible by 2001.
Around a table sit boys and girls. Prove that the number of pairs of neighbours of different sexes is even.
Could the difference of two integers multiplied by their product be equal to the number 1999?
a) There are 21 coins on a table with the tails side facing upwards. In one operation, you are allowed to turn over any 20 coins. Is it possible to achieve the arrangement were all coins are facing with the heads side upwards in a few operations?
b) The same question, if there are 20 coins, but you are allowed to turn over 19.
From the set of numbers 1 to \(2n\), \(n + 1\) numbers are chosen. Prove that among the chosen numbers there are two, one of which is divisible by another.
The number of permutations of a set of \(n\) elements is denoted by \(P_n\).
Prove the equality \(P_n = n!\).
Write in terms of prime factors the numbers 111, 1111, 11111, 111111, 1111111.
Let \(a\), \(b\), \(c\) be integers; where \(a\) and \(b\) are not equal to zero.
Prove that the equation \(ax + by = c\) has integer solutions if and only if \(c\) is divisible by \(d = \mathrm{GCD} (a, b)\).