Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 190
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my last answer is final with proof.
off-topic question: why aren't posts getting a rating these days?
haven't seen the proof )
that doesn't mean there isn't one )
aah, you're private )
I think the minimum possible number of attempts to find 2 different balls is one.
the maximum is 1,000.
Why are you forgetting about the previous problem?
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Pure Math, Physics, Logic (braingames.ru): problems for brains, not related to trading
Mathemat, 2014.06.23 08:38
Another problem:
Is it possible to place three white kings and five black queens on a 5x5 board in such a way that white is not under check?
Weight is 4.
...
Why didn't you mention the previous one?
And here is the 5x5 field matrix:
And a Word file with pictures of the king and queen in the trailer...
Another one:
There are 2,000 balls that look the same, half of which are aluminium and half are dural. Balls of the same material weigh the same, balls of different materials weigh differently. What minimum number of weighings on a cup scale will be needed to ensure the formation of two groups of different weight from the same number of balls?
The weight is 4.
FAQ:
- The scales are cup scales, infinitely accurate, there are no weights. Weighing is putting something on both bowls, looking at the balance, remembering the result and removing the contents from the bowls,
- Wiki says that the density of dural is about equal to that of aluminium. For this problem, it's enough to assume that it's just different from the density of aluminium,
- In formed groups of different weights of the same number of balls there can be any number of balls, even one at a time,
- proving the minimum number of weights is necessary - unless, of course, you've managed for the minimum possible number of weights.