Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 198

[Sceptic Philozoff]

VNIK: There must be an extra trick here, as the initial condition is too vague:

No tricks. "Some of them burn" should be understood to mean that not all of them burn (or even all are extinguished).

[Vladimir Karputov]

So far, I see it coming down to this sequence: all the other candles in a row are lit, except:

• one not lit.
• two in a row not lit
• three in a row no light (well, it's easy to change the state of the middle candle)
• and four in a row no light

There is a solution, all that is left to do is to formulate it clearly.

[Sceptic Philozoff]

A brute-force analysis of all the combinations is the first stage of reflection. A seemingly exhaustive analysis is possible, but it is difficult to call such a solution beautiful.

There is a very short and even, one might say, elegant solution without any trying at all. Try looking at the properties of the magic operation itself.

[Sceptic Philozoff]

Below is my moderator's solution to the weighing problem (not on the first try). I will delete it in a few hours.

/deleted by me/.

Another problem:

There is an ordinary chessboard with 4 knights on the main diagonal (fields h1, g2, f3, e4). It is required to divide the board into 4 equal, equally shaped pieces so that each has one knight. Each piece must be connected (consist of one piece).

Weight is 4. The problem is here.

Try to solve it without computer attack, just with your brain.

[Vladimir Karputov]

Mathemat:

There are 13 candles in the magical candlestick, arranged in a circle. Some of them are lit. The magic is that if you light or extinguish one candle, two neighbouring candles will also change their state: the unlit ones will light up and the burning ones will go out. Is it always possible to get all the candles to burn at the same time?

Weight - 3. The task is here.

The problem comes down to the variant: all candles but one burn. Then it's easy.

[Sergey Gridnev]

Mathemat:

Prove that in a single weighing it is impossible. Zadachas of this type on braingames.ru must be justified - unless it is specifically stated that it is not necessary to prove minimality.

Or show how a single weighing can be done. You certainly can't do without weighing :)

I apologize for not answering earlier(someone strongly concerned about politics thought that the links to the video with Venediktov and Bykov are too political).

Yes, it is indeed guaranteed to distinguish 2 groups of balls equal in number but different in weight with the help of 1 weighing (it is probably not necessary to give a proof).

[Sergey Gridnev]

Mathemat:

Below is my moderator's solution to the weighing problem (not on the first try). I'll delete it in a few hours.

/deleted by me/.

Exactly, it's no longer worth it. :)

[Sergey Gridnev]

barabashkakvn:
The problem comes down to getting the option: all but one candle burns. Then it is simple.

The problem is solved elementary. At the first stage it is necessary to get 1 burning and 12 extinguished candles, then it is solved in 4 moves.

[Sceptic Philozoff]

Contender:
The problem is solved in an elementary way. The first step is to get 1 burning and 12 extinguished candles, then it is solved in 4 moves.

Show me how to get one burning candle out of three candles in a row :)

It is possible to do without the preliminary stages.

You don't have to reduce the problem to algorithmic decomposition of the subtleties of candle arrangement which leads to one burning candle. It would not be pretty and is unlikely to be convincing.

It is enough to find a single "complex" operation that solves all problems in one fell swoop. This is a big clue.

[Sergey Gridnev]

Mathemat:

Show me how to get one burning from three in a row :)

You have to go around in a circle.