Prove that in a game of noughts and crosses on a \(3\times 3\) grid, if the first player uses the right strategy then the second player cannot win.
a) There are 10 coins. It is known that one of them is fake (by weight, it is heavier than the real ones). How can you determine the counterfeit coin with three weighings on scales without weights?
b) How can you determine the counterfeit coin with three weighings, if there are 27 coins?
Vincent makes small weights. He made 4 weights which should have masses (in grams) of 1, 3, 4 and 7, respectively. However, he made a mistake and one of these weights has the wrong mass. By weighing them twice using balance scales (without the use of weights other than those mentioned) can he find which weight has the wrong mass?
There are some coins on a table. One of these coins is fake (has a different weight than a real coin). By weighing them twice using balance scales, determine whether the fake coin is lighter or heavier than a real coin (you don’t need to find the fake coin) if the number of coins is: a) 100; b) 99; c) 98?
A family went to the bridge at night. The dad can cross it in 1 minute, the mum in 2 minutes, the child in 5 minutes, and the grandmother in 10 minutes. They have one flashlight. The bridge only withstands two people. How can they cross the bridge in 17 minutes? (If two people cross, then they pass with the lower of the two speeds. They cannot pass along the bridge without a flashlight. They cannot shine the light from afar. They cannot carry anyone in their arms. They cannot throw the flashlight.)
Several stones weigh 10 tons together, each weighing not more than 1 ton.
a) Prove that this load can be taken away in one go on five three-ton trucks.
b) Give an example of a set of stones satisfying the condition for which four three-ton trucks may not be enough to take the load away in one go.
Prove that multiplying the polynomial \((x + 1)^{n-1}\) by any polynomial different from zero, we obtain a polynomial having at least \(n\) nonzero coefficients.
A daisy has a) 12 petals; b) 11 petals. Consider the game with two players where: in one turn a player is allowed to remove either exactly one petal or two petals which are next to each other. The loser is the one who cannot make a turn. How should the second player act, in cases a) and b), in order to win the game regardless of the moves of the first player?
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?
Of 11 balls, 2 are radioactive. For any set of balls in one check, you can find out if there is at least one radioactive ball in it (but you cannot tell how many of them are radioactive). Is it possible to find both radioactive balls in 7 checks?