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 Special Similar Set (Posted on 2005-06-27)
A man offered me a set of eleven weights, not all them equal, each an integer number of pounds, which he said had the following property: if you removed any of the eleven weights, the other ten could form two five weights sets that balanced each other. Is this possible?

And if the weights didn't weigh an integer number of pounds each?

 No Solution Yet Submitted by Old Original Oskar! Rating: 3.3333 (3 votes)

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 More Part 2 thoughts | Comment 6 of 11 |
Part 1 seems to be impossible, but Part 2 is not necessarily so.  It seems unlikely that it would work.  At least some weights would have to be irrational.

To make sure of that, I'm trying to first understand the concept of this special similar set.  What if there were five weights instead of eleven?

a, b, c, d, e

d and e must each equal one of the following:
a+b-c
b+c-a
c+a-b

What if d and e were the same: a+b-c without loss of generality?

c would be equal to one of these:
a+2b-2c=(a+2b)/3
a
and one of these:
b+2a-2c=(b+2a)/3
b

This is only possible if c=a=b, which is only possible if a=b=c=d=e, but this case was precluded by the puzzle.

Therefore each weight must be different.

So without loss of generality, let's say d=a+b-c and e=b+c-a

b must equal one of the following:
3a-2c
2c-a
and one of the following:
3c-2a
2a-c

It seems no matter what combination of choices I make for b, a=c all the time (though I did not list the choices where b=a or b=c)

Therefore, the 5 weight case is proven impossible.

I wish I could say the same of the 11 weight case.

 Posted by Tristan on 2005-06-27 23:32:55
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