Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 177
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This is in Humour
First one first. Got the sequence wrong. ))
Mischek2014.01.10 16:54#RU
...
Give me something, anything, anywhere.
Just don't give me anything anywhere. What if it's not what you want? ))
In front of Leopold the cat has five mouse burrows, arranged in a row. There is a mouse hiding in one of them. Leopold can stick his paw into any of the holes and try to catch the mouse. The mouse is afraid of the cat and after each try it will move to the next den on the right or on the left. In what minimum number of attempts can the cat be guaranteed to catch the mouse? Proof of the minimum is not required.
The weight of the problem on the Mind Games is 3. But it's a very nice puzzle.
If you have a complete solution, send it to me in a private message.
It is desirable to post here only the answer itself with a faint hint of the solution. And explanatory questions, of course.
FAQ:
- the mouse will definitely run over, and only to some neighboring hole. It might even run over to where the cat has just put its paw.
- The "guaranteed catch" means "calculate the burrow, put the paw in and immediately get the mouse out". No "probabilistic" solutions such as "you'll probably catch it someday if you're lucky" are accepted.
- Leopold is one-legged (from the front) handicapped, so he can only use one paw to catch the mouse.
Once upon a time there was a homeless man in a cellar among other homeless men like him. He collected ten steers. He uses three of them to make a joint and smokes it. Question: how many cigarettes does the homeless man smoke?
The weight of the problem is 2.
FAQ:
- you only get a self-circulation from 3 steers, no more and no less.
- after a minibus is smoked, the equivalent of any other minibus is left.
- the first answer that comes to mind is almost certainly wrong. Keep in mind that the bum is quite savvy and aims for total stoners.
There is one litre of hot coffee (t=95°C), one litre of cold tea (t=5°C) and a set of vessels of different sizes. Is it possible, by heating one liquid with another and without using any other heat/cold sources, to make the final temperature of all the tea higher than the final temperature of all the coffee? The heat capacity of the vessels and the heat loss to the environment are neglected.
The weight of the problem is 3.
...Leopold can stick his paw into any of the burrows...
There is one litre of hot coffee (t=95°C), one litre of cold tea (t=5°C) and a set of vessels of different sizes. Is it possible, by heating one liquid with another and without using any other heat/cold sources, to make the final temperature of all the tea higher than the final temperature of all the coffee? The heat capacity of the vessels and the heat loss to the environment are neglected.
If you look at it from a heat exchange perspective, then (ideally) the tea and coffee temperatures should equalize.
And if we look at it from the point of view of radiant energy exchange, we can speculate...
After all, coffee should be darker than tea.
Can Leopold use two paws at the same time?!
No, just one. He's a forefooted invalid with no sense of smell, but plenty of intelligence. And he loves mice to death, too.
Three joints can be made from as many as nine steers.
The fourth from the remains of the first three and the tenth steer.
Few.
By the way, I wrote in the FAQ for the problem that you can only make a self-twist from three steers.
Think again. This is too simple an answer to be correct.
If you look at it from a heat exchange perspective, then (ideally) tea and coffee temperatures should equalise.
And if you look at it from the point of view of radiant energy exchange, you can speculate...
After all, coffee should be darker than tea.
Well, if you wait for the temperatures to equilibrate, they do. But that's not the answer to the problem question. It's not about the vessels for nothing.
There is no radiative heat transfer involved.
Tea and coffee can only change temperature as a result of non-radiant heat exchange and only with each other (tea-tea, tea-tea, coffee-coffee), but not with the environment.
About the darkness of coffee - not a fact. I drink tea, which is about as dark as insoluble coffee.
No, just one. He's a forefooted invalid with no sense of smell, but plenty of intelligence. Also loves mice to death.
I can't think of anything shorter than six tries:
2,2,3,3,4,4. (according to mink numbers).
In front of Leopold the cat has five mouse burrows, arranged in a row. There is a mouse hiding in one of them. Leopold can stick his paw into any of the holes and try to catch the mouse. The mouse is afraid of the cat and after each try it will move to the next den on the right or on the left. In what minimum number of attempts can the cat be guaranteed to catch the mouse? Proof of the minimum is not required.
The weight of the problem on the Mind Games is 3. But it's a very nice puzzle.
If you have a complete solution, write me in a private message.
It is desirable to post here only the answer itself with a faint hint of the solution. And explanatory questions, of course.
FAQ:
- the mouse will definitely run over, and only to some neighboring hole. It might even run over to where the cat has just put its paw.
- The "guaranteed catch" means "calculate the burrow, put the paw in and immediately get the mouse out". No "probabilistic" solutions such as "you'll probably catch it someday if you're lucky" are accepted.
- Leopold is one-legged (from the front) handicapped, so he can only use one paw to catch the mouse.
So far, no shorter option than six tries comes to mind:
2,2,3,3,4,4. (by the numbers of the burrows).