Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 32
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If the fuel-to-length ratio is less than 1.0, then it is a fuel deficient section.
If the ratio of fuel quantity in a section to section length is greater than 1.0, then it is a fuel surplus section.
This is not proof, just plausible reasoning.
Even starting from a fuel surplus site, you can run into a shortage later on - if you choose the wrong starting point.
Give a rigorous normal proof - if you think it's possible. (I have it as an algorithm for choosing the only possible one out of several possible starting points).
(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 they play correctly, if there were 2000 pies in the pile first?
I'll see you tonight. I hope there are enough problems (7 have accumulated, see a bit earlier) to keep you entertained.(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 there is a letter in the last drawer?
Eh)) Strict solution for 1st year technical universities:
Event A is "letter in desk", a priori P(A) = 1/2
event B - "the first 7 drawers of the table are empty", total probability P(B) = P(B/A)*P(A) + P(B/~A)*P(~A) = 1/8*1/2 + 1*1/2 = 9/16
(Explanation 1: P(Q/A) is the probability that the first 7 boxes are empty, if the letter is exactly in the box. Since there are exactly 8 ways to choose the box where the letter is placed, this probability is 1/8)
(Explanation 2: P(B/~A) is the probability that the first 7 drawers are empty if there is no letter in the drawer. It is obviously a credible event)
By Bayes' theorem P(A/B) = P(B/A)*P(A)/P(B) = 1/8*1/2:9/16 = 1/9 - that is the answer.
There is another, more illustrative way:
We have a possible series:
00000000 - 1/2
10000000 - 1/16
01000000 - 1/16
00100000 - 1/16
00010000 - 1/16
00001000 - 1/16
00000100 - 1/16
00000010 - 1/16
00000001 - 1/16
Those series, which remain after 7 boxes are opened, are shown in bold. As we see, their a priori probability ratio is 1:8; since there is no reason to change this ratio, the probability of the last outcome is 1/(1+8) = 1/9.
5 points is a bit much for such a task))
Strategy for the second player: If the first player takes 1 pie, take 3, if 3, take 1. Thus, the second player makes sure that after his turn the number of pies is divisible by 4. If the 1st player took 2 pies, then the 2nd player should take 1 pie, on the next move the 1st player has to take 2 or 3, then the 2nd player with his move (3 or 2 pies respectively) achieves the multiple of 4. On the last move (when only 4 pies are left) the same rules: 3>1 (eaten), 1>3 (eaten), 2>1 (player 1 has no moves).
Everything fits. Well done.
It all adds up. Well done.
The game and the principle of winning are similar, so the solution came to mind almost immediately.
It's more complicated than that. A multiple of four is achieved in either one cycle or two. It's beautiful.
Strictly speaking, the last step starts with either four or eight. But still the second one wins in the same way.