Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 110

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Mathemat:
OK, let the pole consist of a thousand bricks. Can you solve it for that height?
that's the same - half a brick // that's why you brought up mucik ))
 
Mathemat:
OK, let the column consist of one thousand bricks. Can you solve it for this height?

Let's try this. Let's say that the pole is solid and stands upright. Its centre of gravity will be in the middle.

Now shift all the bricks so that the uppermost brick is shifted by a full width relative to the lowest one.

The centre of gravity is now half a brick to the left of the lowest brick.

Conclusion: such a post will stand as long as its centre of gravity is no further than one of the boundaries of the base (the lowest brick).

Yes, it turns out as in the first solution, the answer is correct. Or not. I'm confused.

 
Mischek:
so the same thing - half a brick // that's why you brought up muzik ))
Nah. He remembered Muzik because it's not half a brick, it's a brick.
 
TheXpert:
Nah. Mucik he remembered because it's not half a brick, but a brick.
Then for a post made of 2 bricks, the offset per brick solution does not pass.
 
Mischek: // that's why you brought up mucik ))

Because the solution is very similar, the same law comes out there.

fyords: Then for a post of 2 bricks, the displacement per brick solution does not pass.
Try a post of five bricks. How much more can you shift the topmost one relative to the bottom one?
 
TheXpert:
Nah. He remembered Muzik because it's not half a brick, it's a brick.
Shit, maybe it's a brick. With the number of bricks tending towards infinity and the shift towards zero.
 
Where am I going to find so many bricks at half past two in the morning, eh?)
 
Mischek:
Shit, maybe it's a brick. When the number of bricks tends to infinity
The problem does not seem to have an explicit solution because the condition is not explicit.
 
Mischek:
Where am I gonna find so many bricks at 2:30 in the morning, huh?)
I've got a couple, I'm just keeping them in reserve.)
 
fyords:
The problem does not seem to have an explicit solution, because the condition is not explicit.
The explicit one is not needed, it is enough to show that if the number of bricks tends to infinity and the shift to zero, the answer is one brick
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