Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 192
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Got it! Well, then at the fifth weighing, there will be 125 balls on both sides of the scale and the scale is guaranteed to be unbalanced.
Any objections?
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 weights are the same, then ... (Still, let those who want to think more, I will write the answer after lunch)
The point, naturally, is to find subgroups equal in number, but different in weight, and transfer them to the opposite 1000.
Since the groups consisting of 1000 balls are equal to each other in weight, hence, they have the same number of heavy balls (500 each), and the same number of light ones (500 each).
We divide each group of 1000 into 2 subgroups of 500. Weigh them in pairs: 500 from the first 1000 with 500 from the 1st 1000 (weighing #2); 500 from the 2nd 1000 with 500 from the 2nd 1000 (weighing #3). If any (or both) of the weighings record a weight difference, then simply exchange the balls of the light subgroup of the first 1000 with the balls of the heavy subgroup of the second 1000 (the experiment is over).
If weighing number 2 and number 3 recorded equality in weight, then all subgroups of 250 heavy balls (and light, by the way, too).
Let's divide any of the 2 subgroups (500 each) of the 1st 1000 and any of the 2 subgroups of the 2nd 1000 into subgroups of 250 balls. Let's do a pairwise weighing: 250 of the 1st 1000 with 250 of the 1st 1000 (weighing #4); 250 of the 2nd 1000 with 250 of the 2nd 1000 (weighing #5). If any (or both) of the weighings record a weight difference, exchange the balls of the light subgroup of the first 1000 with the balls of the heavy subgroup of the second 1000 (experiment terminated).
If at weighing № 4 and № 5 is fixed equality of weight, then in all subgroups on 125 heavy balls (and light, by the way, too). Now, when dividing into subgroups, we will not get equality in the number of heavy balls (and light ones too)!
Divide any of the 2 subgroups (250 each) of the 1st 1000 and any of the 2 subgroups (250 each) of the 2nd 1000 into subgroups of 125 balls. Weigh (this is the 6th) any subgroup of 125 balls of the 1st 1000 with any subgroup of 125 balls of the 2nd 1000. If the weights differ, we exchange the balls of the weighted subgroups, otherwise we exchange the balls of one weighted subgroup with the balls of the unweighted subgroup of the other 1000. The experiment is over.
Will there be objections?
There will be.
The subgroups with different weights must belong to different thousands.
And this is my thinking:
I made do with one weighing :)
The logic is as follows:
1) we separate an odd number of balls from 2000 so that the residual group is divisible by 3 without a remainder. i.e. [2 + 3*n] balls, and n must be odd (to make sure the group is odd) and less than 333, so that the residual group contains over 1000 balls to make sure it contains balls of different weight.if we correct the formula by taking into account these limits, we get [5 + 6*n] where n = 0...166, so the maximal number in the second group would be 1995 (the minimum number is 1005).
2) divide the remaining (second) pile into 3 equal parts.
3. Now for the first weighing: Weigh two piles from the second group. If they have different weights, the problem is solved. If they are the same, we take any of the weighed piles and an unweighed pile (of the same second group), their weight is guaranteed to be different, so they may not be weighed.
In this case (minimum heap size = 1005/3 = 335, maximum = 1995/3 = 665).
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 in any case it is possible to manage in no more than N attempts.
What the fuck, you said it all, but I don't get it)
you need a guaranteed to sort it into 2 piles, without any probability of that happening.
The most guaranteed option is to put one ball on the scales and compare the others to it, the minimum in this weighing is 1, the maximum is 999.
Damn mathematicians give at least some deadline after which you will give an answer, because I am still solving queens)
3. Now for the first weighing: we weigh two piles from the second group. If they have different weights, the problem is solved. If they are the same, we take any of the weighed piles and an unweighed pile (of the same second group), their weight is guaranteed to be different, so they may not be weighed.
In this case (minimum heap size = 1005/3 = 335, maximum = 1995/3 = 665).
Shit, the fact that these groups shouldn't have 1000 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 marbles?
I made do with one weighing :)
The logic is as follows:
1) we separate an odd number of balls from 2000 so that the residual group is divisible by 3 without a remainder. i.e. [2 + 3*n] balls, and n must be odd (to make sure the group is odd) and less than 333, so that the residual group contains over 1000 balls to make sure it contains balls of different weight.if we correct the formula by taking into account these limits, we get [5 + 6*n] where n = 0...166, so the maximal number in the second group would be 1995 (the minimum number is 1005).
2) divide the remaining (second) pile into 3 equal parts.
3. Now for the first weighing: Weigh two piles from the second group. If they have different weights, the problem is solved. If they are the same, we take any of the weighed piles and an unweighed pile (of the same second group), their weight is guaranteed to be different, so they may not be weighed.
In this case (minimum heap size = 1005/3 = 335, maximum = 1995/3 = 665).
And this is my thinking:
OK, in point 5, the weight is different.
It is guaranteed different there, we could have not weighed, and since (as it is now clear to me) need to get 2 groups with the same amount, but different weight, then after point 4 you can already get a balanced group.
I.e. 4 weighing is enough.