шестое чувство

[barzel]

В эксперименте участвовало 30 слепых. Каждого слепого привели в закрытую комнату и поставили его напротив 5 цветных ящиков(цвета разные). Каждому слепому были выданы 5 записок в соответствии с цветом ящика, и попросили каждого из них подобрать правильный ящик к записке.
Перед вами результаты эксперимента:
Каждый из них сумел подобрать по крайней мере одну записку.
10 смогли ровно одну
6 смогли ровно две
4 смогли ровно 3

Вопрос: сколько людей смогло подобрать ровно 4? и сколько ровно 5?

Comments

[(Anonymous)]

I don't think it's possible to determine the answer based on the facts presented.

[barzel.livejournal.com]

be sure, this problem has solution. it has sufficient facts to solve it.

[(Anonymous)]

Ok, several questions, if I may:

1. Is the solution based on the theory of probability?

2. Were these experiments done in some sequence, e.g. the first blind person was in the room, then the second, then the third or each blind person was in his/her own room? Do the results of the previous blind person affect the next in a row? Do these blind people communicate between themselves?

3. Do all these blind people receive their paper slips in the same sequence, e.g., for instance:

1- red
2 - blue
2 - black
4 - white
5 - pink

4. Do the experimentators use the same paper slips for each experiment?

Thanks in advance.

[barzel.livejournal.com]

why do you not authorithed?

answers:

1. not exactly, the promlem is based on pure logic, nothing probability

2. whole experiment occurs simultenusly

3. if it'll help you can propose it

4. it's doesn't matter

[(Anonymous)]

Ok, my former work colleague explained it to me. How stupid I now feel

I didn't authorize because I don't have an account. I'm reading my husband's journal His nick is d******t.

[barzel.livejournal.com]

So do you know the solution now?
say, it was very simple, therefore i choiced it to post here. usually people tend to think in hard way, often solution lay on surface

[(Anonymous)]

Exactly. I often can not solve the obvious puzzles, but can crack the more complicated ones. But I can not solve really complex. Do you know more puzzles? I like them. I also know quite a lot.

[markure.livejournal.com]

don't over think it, the answer is very simple, it's pure Logic.

[lifesoul.livejournal.com]

ti uveren 4to usloviy dostato4no?

[barzel.livejournal.com]

абсолютно уверен, я знаю решение

Re: a ya ponyal! :)

[barzel.livejournal.com]

молодец, ты не на сайте асата прочитал ответ?

Re: a ya ponyal! :)

[markure.livejournal.com]

net

[barzel.livejournal.com]

абсолютно правильно!!!!!!!!!!!!

[lifesoul.livejournal.com]

eto bilo poleg4e 4em zada4a so 100 kolpakami :))

[(Anonymous)]

I know about 100 hats and this one I solved myself. But for not solving this simple one I can blame only that years of programming made me even more stupid than I was...

[barzel.livejournal.com]

хотите еще задачу?

В классе 100 папок, входят люди и меняют их положение,если папка открыта ее закрывают, если закрыта, то открывают. первый меняет все папки, второй меняет только четный, третий- только кратные трем и т.д. сколько папок будет открыто после 100 человек?

[(Anonymous)]

This one seems more complex. I'll think at home, right now I need to concentrate on work

[barzel.livejournal.com]

good luck in your work

[(Anonymous)]

Ок. 1-ая, 4-ая, 9, 16, 25, 36, 49, 64, 81 папки поменяют свой статус. Остальные папки останутся в исходном состоянии.

[(Anonymous)]

страшно, но вы опять правы., у квадратов не четное число делителей

[(Anonymous)]

Анонимус на анонимуса :) А как теперь разберешься, кто есть кто? Первый - это я, с работы писала на английском, а из дома - на русском.

[barzel.livejournal.com]

второй это я, просто не было сил регистрироваться

[(Anonymous)]

Ouch, forgot to choose Anonymous...

[(Anonymous)]

This problem has several right answers: N+/-9, N+/-7, etc.

[barzel.livejournal.com]

it isn't right, think more, here isn't any probability

[(Anonymous)]

Ok, we know that only 9 folders will change their original state, e.g. 1, 4, 9, 16, etc.

Now, let's assume that all these 9 were open originally. Obviously, the number of open books would be now N-9.

Now, say 1st was closed originally and other 8 were open. After we change the state, the 1st would be open and 8 closed. So, the number of open folders will decrement by 7.

Repeating the same train of thoughts will have +-5, +-3, etc.

Do you see a logic error in these explanations?

[barzel.livejournal.com]

the answer is that folders no. 1, 4, 9, 16,25 etc will change their status, because square of number has odd number of multiplies, that is why it change, rest folder will return to initial status. you solved id right.

[(Anonymous)]

I actually gave the wrong answer, because I forgot 100th person. E.g. the difference would be 10.

[markure.livejournal.com]

hochu zametit chto vi tak i ne otvetili na vopros "сколько папок будет открыто после 100 человек?"

i vobse podazrevau chto barzel tut pereputal i na eto nelzya otvetit neznaya nachalnogo polazheniya papok.

[barzel.livejournal.com]

сначало они все открыты, ответ был правильный

[(Anonymous)]

You didn't say that they were all open before. In other words, I assumed the random distribution. In this case my solution is correct. Otherwise, of course, there is only one right answer.

[(Anonymous)]

Here is a new problem:

You have three quarters of a square Split them in four parts, all equal in shape and size.

[barzel.livejournal.com]

you split along two diagonals of former square

[(Anonymous)]

This is wrong. Do you want a link to correct answer?

[barzel.livejournal.com]

in such way?
http://foto.mail.ru/mail/murysh/124/156.html

[(Anonymous)]

No. Here it is
http://www.profox.ro/threequarterssolved.gif

[(Anonymous)]

BTW, do you play in WWW team?

[ak-74u.livejournal.com]

Насколько я понимаю, оставшиеся 10 человек верно подобрали ровно 5?
Ведь если верно подберёшь 4, пятая поневоле отправится верно.

[barzel.livejournal.com]

молодец, это и есть ответ