Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 195
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So the problem was solved manually. A crossword puzzle with large squares was used as a matrix. And then did it quickly - I have MS Office 2013, as it were.
Well, didn't you write that it was a brute force solution?
No, not you, sorry )
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Pure Math, Physics, Logic (braingames.ru): Tasks for brains, not related to trading
maxfade, 2014.06.23 22:14
not solved myself, wrote a script with random combinations - quickly findsone option, + its mirrored variations.
Well, didn't you write that the problem was solved by brute force?
aren't the moderators just being silly with their posts? (only "-to", "-either", "-anything", "whether" is written without a hyphen)
If something does not suit you, correct it with an answer, I am not a fool, I will understand if something is wrong.
There is exactly more than one solution.
In general terms: 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.
Also incomplete. Counterexample: 4+4+664+664+664. If groups of 4 weigh the same, it doesn't mean that groups of 664 are different.)
For example, it can turn out that we have separated exactly four balls from each of thousands of luminescent and duraluminous ones, and the remaining 996 balls in them will split up exactly into 332 in the X-Y-Z groups.
My general formula is: A+B = 2 + n*6. Respectively, X+Y+Z = 2000 - ( 2 + n*6 ). Where n 0...332 // Limitation A+B < 1000 is unnecessary (think about it).
Also incomplete. counterexample: 4+4+664+664+664. If groups of 4 weigh the same, then it's not a fact that groups of 664 are different. :)
For example, it might turn out that we separated exactly four balls from each of thousands of luminescent and duralumin balls, then the remaining 996 balls in them would break down exactly into 332 X Y Z piles.
Yes, it seems indeed that the short solution is the only one:
1+1+666+666+666 and 2 weighings.
Yes, the short solution does indeed seem to be the only one:
1+1+666+666+666 and 2 weights.
Not really, see above, I added there.
I'll copy it, though:
I got a general formula like this: group A+B = 2 + n*6. Accordingly group X+Y+Z = 2000 - ( 2 + n*6). where n 0...332 // Restriction A+B < 1000 you have extra (think about it).
Not really. See above, I added there.
I'll copy it, though:
Six as a multiplier ensures that the set of light balls and the set of heavy balls in the second group do not divide by 3 at the same time.Take, for example, n=332 (you can do that based on your constraints).
We get: A=B=997. Where's the guarantee that A and B don't take the same type of balls entirely? I.e. A and B may contain 500 balls of one type and 497 of another, and the remaining 6 identical (!) balls are distributed over X,Y,Z.
Take, for example, n=332 (you can do this based on your constraints)
We get: A=B=997. Where is the guarantee that A and B do not take the same type of balls? I.e. A and B may contain 500 balls of one type and 497 of another, and the remaining 6 identical (!) balls are distributed over X,Y,Z.
I think I've got it, so n must be in the range 0...166
Total: group A+B = 2 + n*6. Correspondingly, group X+Y+Z = 2000 - ( 2 + n*6). where n is in the range 0...166
It means that we have exactly 167 solutions.
I think I've got it. So n must be in the range 0...166
So: group A+B = 2 + n*6. Correspondingly, group X+Y+Z = 2000 - ( 2 + n*6). Where n is in the range 0...166
So we have exactly 167 solutions.
I also found a loophole. Six (2*3) as a manifold is weak. 18 (=2*3*3) is needed. // Counterexample for the upper formula: n = 2;
There seems to be no holes left now: group A+B = 2 + n*18. Correspondingly, group X+Y+Z = 2000 - ( 2 + n*18). Where n is in the range 0...55
This leaves a total of 56 solutions.