Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 33
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(5 points)
Two mega-brains are playing a game. Each takes turns taking 1, 2 or 3 cakes from a pile of cakes and eating them. They can't take as many as their opponent took in the previous turn. The winner is the one who eats the last pie or after whose move the opponent can't make his move. Which of them will win if the game is played correctly, if there were 2000 pies in the pile first?
I'll see you tonight. I hope there are enough problems (7 of them accumulated, see a bit earlier) not to get bored.The first will win because the second won't be physically able to eat twice as many cakes... :)
(3 points)
With probability 1/2 a letter was placed in one of the eight drawers of the table (chosen at random). Then 7 drawers were opened one by one - all empty. What is the probability that the letter is in the last drawer?
Probability 1/2
The first will win, as the second will not physically be able to eat twice as many cakes... :)
Exactly. And not twice, but three times. All you have to do is start eating one pie each, and the second will have to eat three pies each. As long as the weight categories are about the same, the first will win. You won't even have to finish them all...
It's a cruel thing, the obligation to win. It's the trouble with it.
I'm sad.
It's not the whole error. The intersection will be, just in another place - outside the triangle.
You need to find the specific place where the error is.
P.S. I also wrote about this at first, but I was told that the error has not yet been found. And they showed me a second picture, an alternative one:
Probability 1/2
What's wrong with my decision?)
Actually point E is on the same side of point C as point A (not different, as in the picture), unlike point D, which really is on different sides of point A from point B. (admittedly, you still have to prove it, but that's a matter of technique.) With this construction, all reasoning holds, except for one - from AD=AE and BD=CE it no longer follows that AB=BC.
Alexei, you stay here with us already. We've missed you.
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There's also the blots that need to be spelled out. It seems to be solvable, but I can't prove it.
alsu:
It follows that every point in a cell that is not filled with ink corresponds to at least one point outside the cell that is filled with ink. Hence, in turn, it follows that the area of the ink cannot be smaller than the area of the cell. Come to a contradiction, the theorem is proved.
Suppose the statement of the theorem is wrong, i.e. for any grid shift at least one node is covered by the blot.
Let fix some position of the grid. Let node 1 of some cell be under the ink. Since the area of the blots is smaller than the area of the cell, there must be an area inside the cell that is not covered by the blot. Consider all possible shifts of the grid such that node 1 moves into a clean region. By our assumption, at least one of the nodes 2,3,4 of the same cell must move under the blot, and necessarily outside the cell (since node 1 has moved inside). Hence, each point of the cell, not filled with ink, corresponds to at least one point outside the cell, filled with ink. Hence, it follows that the area of the ink cannot be smaller than the area of the cell. Come to the contradiction, the theorem is proved.
Grifter. Can you elaborate on that?
According to our assumption, at least one of the nodes 2,3,4 of the same cell must move under the blot,
The blot problem, I take it, does not interest anyone. Is the solution interesting or not? Or will you try? It's really very simple (although it's 5 points).
On a plane with a rectangular grid with step n, ink is poured in the form of lots of blots of different size and shape. The total area of the ink spots is less than n². Prove that it is possible to shift the grid in such a way that no node of the grid is flooded with ink.