ForumSections

Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 191

New comment
barabashkakvn:

And here is the 5x5 field matrix:

And a Word file with pictures of the king and queen in the trailer...

not solved myself, wrote a script with random combinations - quickly finds
option one, + its mirror variations
barabashkakvn:
8 weightings. I can prove.

Less, and a lot more.

The way I see it, the minimum possible number of attempts to find 2 different balls is one.

maximum 1000


It's about the minimum number of weighings for which the two groups are guaranteed to be formed. If the answer is N, then that means no more than N attempts are possible.

Mathemat:

Less, and by far more.

For two ;) Only without proof, maybe not.
Mathemat:

Less, and by far more.

It is about the minimum number of weighings for which the two groups are guaranteed to form. If the answer is N, it means that at any rate it is possible to manage in no more than N attempts.

And such a question - weighed - and then what? Are the weighed pieces back into the pile or can they be put aside and weighed again accordingly?
barabashkakvn:
What about this question - are they weighed - and then what? Do the weighed parts go back into the pile or can they be put aside and weighed again accordingly?

You can do whatever you like, which is usually done in such tasks. Return to the pile, weigh again, mix with another pile, set aside, break into smaller piles, etc.

But this may not be done during the weighing itself. The weighing itself is putting both piles on the bowls, looking at the scale and taking the piles off the bowls.

Integer:
For two ;) Only without proof, maybe not.
It is possible without weighing, but not guaranteed ;)
Mathemat:

Less, and by far more.

It is about the minimum number of weighings for which the two groups are guaranteed to form. If the answer is N, it means that at any rate it is possible to manage in no more than N attempts.

in 6 weighing attempts.
Contender:
for six weigh-ins.
A thought process in the studio, please. I wonder how such problems are solved in principle.
barabashkakvn:
A way of thinking, please. I wonder how such problems are solved in principle.

First, you have to divide the balls into 2 groups of 1,000 and weigh them. If the weight is different, that's it :)

If, the weight is the same, then ... (Still, let those who wish to think more, after lunch I will write an answer)

Contender:

First, you have to divide the balls into 2 groups of 1,000 and weigh them. If the weight is different, that's it :)

If, the weight is the same, then ... (Still, let those who want to think more, I will write an answer after lunch)

Got it! Well, then the fifth weighing will have 125 balls on both parts of the scales and the scales are guaranteed to be unbalanced.
1...184185186187188189190191192193194195196197198...229
New comment