Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 43

 
Avals:
In general, the series itself is clear - each next term is equal to the sum of the previous three, not two as in fiba. But we can come up with a lot of such series depending on the first terms of the series, and we need to make it generally infinite when tending to zero. To do this, we need to find an analogue of pfi number for this series - it will be the ratio of lengths of two neighboring numbers in the series. In general, these are the roots of the characteristic equation X^3-X^2-x-1=0. I.e. 1.839... Therefore, taking the first term of the series as 1 and extending right and left this series multiplying/week by this number, we obtain a series by taking any 3 consecutive terms and we have the sticks with the desired property

Yes, that figure has cleared up.

--

The uniqueness of the solution is not yet obvious.

How about another 'solution': x^3-2*x^2-2*x-1=0

Remove the inverted commas?

 
MetaDriver:

How about another 'solution': x^3-2*x^2-2*x-1=0

Justify.
 
TheXpert:
Justify.

Well, this is the variant where "If the length of the stick has not turned to zero and the triangle cannot be folded again, then the megabrain repeats the operation,"

More precisely - if it has to repeat the variation twice (with the same stick) If it cannot fold a triangle, it shortens the longest of the sticks by the sum of the lengths of the other two .

--

The question here is whether there are valid solutions

 
MetaDriver:

The question here is whether there are valid solutions.

There are.

2.83117721

In short, there are infinitely many solutions :) let's leave it at that.

 
TheXpert:
Bingo.

2.83117721

In short, there are endless solutions :)

Cool. (chuckles)


That's a deal breaker.

I'm in.
 
alsu:
Two ships launch simultaneously from the north pole. At the moment they cross the equator, one of the ships takes the passengers and the rest of the fuel (exactly half, just enough to fit) from the other ship. At the moment they reach the South Pole, the third ship sails south and meets the travelers at the equator, after which everyone amicably returns home))
It doesn't work like that. The ship they took the fuel from will be lost. Obviously, all three ships have to start. When a quarter of the way through, one of the ships keeps a quarter of the tank, and one quarter to give the other 2 and go back. Two ships have full tanks. When they reach the equator, they have 3/4 left. One ship gives a quarter to the other and goes back. The second one has a full tank and can run down to the south and return to the equator. The one that came back second refuels and he and the one that came back first go towards the equator. But before they reach it, it's refuelled again at the point where one of them has exactly enough fuel to get back - 1/3 of the way to the south pole. And in general, at the equator 2 ships meet - one already empty, and the other with 2/3 of the tank. They split up fraternally, coming back and waiting for a refueller for 1/3 of the way back
 
Avals:
And anyway, at the equator, two ships meet - one already empty and the other with a full tank - and split the fuel in half to get back

And how will he get to the equator with a full tank?

--

It will only be with half a tank.

So half a tank is spilled in half, after which they drive half the remaining distance to the homeland, where they are met by a third one with 3/4 of a tank, which is already enough for everyone.

 
MetaDriver:
And how will he get to the equator with a full tank I wonder?

Yes, corrected - will hit with 2/3 and split in half. A third ship will come towards it.

P.S. not 2/3, but 5/6 :)

 
Avals:

Yes, corrected - will hit with 2/3 and split in half. A third ship will come towards it.

Z.Y not 2/3 but 5/6 :)

;)

MetaDriver:

Taki he will only be with half a tank.

So half a tank is spilled in half, then they drive half of the remaining distance to the homeland, where they are met by the third one, with 3/4 of a tank, which is already enough for everyone.

 
MetaDriver:

that's the end of it.

I agree.

And I shouldn't have agreed. There are only four [groups] that are valid solutions.

For multiplier >=5, only complex roots. For example, for x^3-5x^2-5x-1=0


Still cool. Even cooler.