Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 193
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Man, the fact that these groups shouldn't have 1,000 balls each I somehow missed. :(
But, there's something wrong with the result. Let's say we have piles of 335 balls each. Where's the guarantee that, for example, each of them doesn't consist of 2 heavy and 333 light balls?
Aha. I seem to have a problem with the constraints (the generalized formula is wrong). I'll think about it some more.
OK, in point 5 the weight is different.
It is guaranteed to be different there, we could have not weighed it, and since (as it is clear to me now) we need to get 2 groups with the same amount, but different weight, after point 4 we can already get the different groups.
I.e. 4 weighing is enough.
I was proceeding from the way I understood the condition: the decision is made on the basis of weighing. That is, clause 5 is needed.
If it is known for certain that the weight is different, why this extra weighing?
Can the previous answer (about the chessboard) be posted now? Somehow everyone forgot about the chess problem :(
Aha. I seem to have a flaw with the constraints (the generalized formula is wrong). I'll think about it.
I can see the solution for 2 weighings, I can't do it in one.
I see the solution in two weighings, I can't do it in one.
Yes. It looks like there's no way around it without two. One solution is certain, the others are still unclear, I'll keep digging.
--
Found this solution:
1. Separate the two balls. Weigh them. If the weight is different, problem solved. If it's the same:
2. We divide the remaining group into three equal heaps X, Y, Z (1998/3 = 666). Weigh the two heaps (X and Y). If different - problem solved, if identical - problem also solved [X and Z] and [Y and Z] are guaranteed different.
Comment: The logic here is simple, if the weights of the balls in the first weighing are the same, then the remaining group contains 1000 balls of one weight and 998 of another. These numbers are not divisible by 3, so you cannot make three groups of the same weight from them.
As a practitioner, what is the fastest way to get results?
ZS: I'm talking about the balloon problem
Yeah. Looks like it's a two-way street. There's definitely one solution, I'm still not sure about the others, so I'll keep digging.
--
Found this solution:
1. Separate the two balls. Weigh them. If the weight is different, problem solved. If it's the same:
2. Divide the remaining group into three equal piles X, Y, Z (1998/3 = 666). Weigh the two piles (X and Y). If different - problem solved, if identical - problem also solved [X and Z] and [Y and Z] are guaranteed different.
Comment: The logic here is simple, if the weights of the balls in the first weighing are the same, then the remaining group contains 1000 balls of one weight and 998 of another. These numbers are not divisible by 3, so groups of the same weight cannot be formed from them.
There is definitely more than one solution.
In general: divide into groups A, B, X, Y, Z.
By number:
A+B+X+Y+Z=2000;
A=B;
A+B<1000;
X=Y=Z.
Further the same reasoning as in the special case: A=B=1 and X=Y=Z=666.