Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 3

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The last time I had such a brain rupture was when reading Nietzsche's Zarathustra. But it was a long time ago...
 

The weight of the problem is 5:

One day 23 mega-brains decided to play a game of football. While they were selecting teams, they noticed an interesting feature: whichever one was chosen to referee the match, the other 22 players could split into two teams of 11 with the same total weight of all the players. It is known that the weight of each megabrain was expressed as a whole number of kilograms. Is it possible that not all megabrains had the same weight?

Note: solved only recently, but I am sure that the solution is correct. The solution cannot but be beautiful.
 
Mathemat:

The weight of the problem is 5:

One day 23 mega-brains decided to play a game of football. While they were selecting teams, they noticed an interesting feature: whichever one was chosen to referee the match, the other 22 players could split into two teams of 11 with the same total weight of all the players. It is known that the weight of each megabrain was expressed as a whole number of kilograms. Is it possible that not all megabrains had the same weight?

Note: solved only recently, but I am sure that the solution is correct. The solution can't help but be beautiful.
If you plot the distribution of players-players by weight (without referee), its average value coincides with the median - based on the condition that the players can be divided into teams equal in weight and number of people. So the distribution is symmetric. So the weight of the referee must coincide with the average value of the weight of the other 22 players (otherwise, when the referee is replaced with one of the players, the distribution becomes asymmetrical). And since any of the 23 can be a referee, the weight of any of them must coincide with the average weight of the other players. This is only possible if the weights of all players are equal.
 
Avals:
......So the weight of the referee must be the same as the average weight of the other 22 players (otherwise when the referee is replaced by one of the players, the distribution becomes asymmetrical)............
Bewilderment... When a referee is replaced - teams can (and should) be reshuffled...
 
MetaDriver:
An oversight... When changing referees - teams can (and should) be re-shuffled...
If the distribution is asymmetrical, you can't divide into 2 equal teams with equal weights (the median is not the same as the mo)
 
Avals: If we plot the distribution of players by weight (without a referee), then its mean is the same as the median - assuming that the players can be divided into teams of equal weight and number of people. So the distribution is symmetrical.

What an awesome conclusion. So all distributions that have a median equal to the mean are symmetric?

P.S. My proof is based on infinite descent. Probably too convoluted again...

 
Mathemat:

What a nice conclusion. So, all distributions where the median equals the mean are symmetric?

I think so. Although they also look at the mode for unimodal. As in right-hand asymmetry Xsr>Me>Mo, in left-hand asymmetry Xsr<Me<Mo. But it could be a bimodal or multimodal distribution. And asymmetry coefficient = 3*(Mean - Median) / RMS.

At least, the counter-example when distribution is asymmetrical and median and mean coincide does not come to mind.

 
 


Help the hunter find the fox

 
Vitriba:


Help the hunter find the fox

You were already told everything in the last thread, why duplicate it?
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