Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 56
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On further collisions, it is further accelerated by the speed of the brick.
Andrei, does your solution use lasers?
(4)
80 megabrains stood in the form of a rectangle 10×8. In each longitudinal row the tallest one was found, and the lowest one was a megamogon with a dog. Then they found the lowest one in each transverse row, and the tallest among them was a megamorg wearing a hat. The question is who is taller: a megamograve with a dog or a megamograve with a hat?
In the general case it is not known.
Here is one case.
And here's a different one.
Don't tell me megabrains can't be that different. Megabrains can do anything!
And besides, you can add 1500 mm to all the numbers.
It is not clear, explain.
On impact, a weightless, perfectly elastic ball bounces off a brick with velocity V-ball-after = V-ball-before + V-brick-after
If we assume that the internal vibrations of the ball are already so messed up that they can randomly either work for a little gain or for a little loss, then we will neglect them. It is only in the squashing impact (first) that they were essential.
In the general case it is not known
...
Don't tell me megabrains can't be that different from each other. Megabrains can do anything!
Could-could. But there is an answer, and it makes perfect sense.
Andrei's original problem was this:
A perfectly elastic ball bounces on a perfectly elastic surface with an amplitude of 1 cm. How can a brick thrown from a height of 1 m lift the ball to a height of 30 m? (you can draw a picture).
I take it that the brick has to be thrown once?
(4)
80 megabrains stood in a 10×8 rectangle. In each longitudinal row they found the tallest, among them the lowest was a megamorg with a dog. Then they found the lowest one in each transverse row, and the tallest among them was a megamorg wearing a hat. The question is, who is taller: the megamoggle with the dog or the one with the hat?
Answer - with a dog.
I can't prove it, logic intuitively says so... :)
(3)
Two armies of mega-brains go to battle: pointed and blunt-tipped. Each army has 2*N people. Each megabrain has a gun that can kill no more than one enemy when fired. Megabrains follow the rules of combat: first shoot the sharp-tipped ones, then shoot the blunt-tipped ones and then shoot the sharp-tipped ones again. After these three volleys the battle ends. Question: what is the maximum number of mega-brains that could have died in this battle? Justify that this number is the maximum.
Answer: 2*N.(shoot the survivors) and(can kill no more than one enemy)
The number of volleys doesn't matter... i.e. they can continue up to 2*N victims
Answer - with a dog.
I can't prove it, the logic is so hard... :)
You have to prove it, Comrade Kolmogorov.
Answer: 2*N.(shoot survivors) and(can kill no more than one enemy).
No, wrong.
But that "shoot the survivors" - it's in the Annals, but there is none :) ... so far.
You have to prove it, comrade Kolmogorov...
:) Can't do it now, tomorrow, with a fresh head, maybe you can....
No, wrong.
But that "survivors shoot" - it's in the Annals, but there's none here :) ... so far.
Agreed.
Started to write an algorithm and saw that more could be..... :(
Andrei's original objective was this:
I take it that the brick has to be thrown once?
About the brick:
Throw the brick strictly from above. The ball gets caught between the brick and the plane and accelerates sharply. Theoretically, it can reach first cosmic velocity. At the desired speed and direction of motion from the plane, we sharply fire off the brick with a laser, and the ball flies away and hits the moon.
The main thing is not to hit the ball with the brick while it is still strictly on the surface of the plane.