Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 199

[Vladimir Gomonov]

Contender:
We need to go around in circles.

There you go. You hacked it. Well done, though. :)

It only remains to generalize the result: the problem is solvable in any initial state for any number of candles, provided that the number of candles is greater than three and is not divisible by 3.

--

I drew it up for playing with Excel+VBA. (I wanted to see how it works with chaotic initial positions etc.)

// Especially for those writing in VBA I did not delete the second page, there is a useful trick, which is generating a large number of functions of the same type.

// This trick is useful when programming in any language, in cases when it is impossible for some reason to pass an array by reference (offset),

// Hang one handler over an array of controls, or other similar cases. Whoever likes it, can borrow it.

// I regularly use this trick in similar cases (often using Exel - I already have generator blanks there for a long time).

[Vladimir Karputov]

So I made a toy too. In MQL5.

[Vladimir Gomonov]

barabashkakvn:
So I made a toy too. In MQL5.

Yeah, I've already seen it.

[Vladimir Karputov]

MetaDriver:
Yeah, I've already seen it.

But still, I haven't found a solution yet. It's both sad and funny.

P.S. Sorry, I've erased the links, no overt advertising.

[Sceptic Philozoff]

barabashkakvn: But still, I haven't found a solution yet.

So should I publish it?

Of course, I'll delete it later.

[TheXpert]

Bring all the candles together and then go around in circles) until the result is achieved

[Vladimir Karputov]

Mathemat:

Shall I publish it?

Sure, I'll delete it later.

Yes, publish it, or I won't get any rest on holiday.

[Sceptic Philozoff]

This is the solution to the candlestick problem. The first line is the candle numbers, the second line is the initial states of the candles, and the next line is the algorithm for changing the state of one candle.

/Removed - Mathemat/.

So no preliminary information is needed together. It is enough to apply this algorithm to each unlit candle.

[Sceptic Philozoff]

Next problem.

A trapezium (arbitrary) is given. How using a single ruler (without divisions) divide the bottom base of the trapezoid into 3 equal parts?

The weight is 5.

There are no marks on the ruler and cannot be. The other side of the ruler cannot be used to draw parallel lines.

[Vladimir Karputov]

Mathemat:

This is the solution to the candlestick problem. The first line is the candle numbers, the second line is the initial states of the candles, and next is the algorithm for changing the state of one candle.

So no preliminary information is needed together. It is enough to apply this algorithm to each unlit candle.

Thanks for the reply.