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Special Similar Set (Posted on 2005-06-27) Difficulty: 3 of 5
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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Some Thoughts re: The real number case -- a better proof? | Comment 10 of 11 |
(In reply to The real number case by McWorter)

Another proof (still a bit technical).  Given any special similar set of 11 real numbers, there is an 11x11 rational matrix A with zeros on the main diagonal, 5 1's in each row, and 5 -1's in each row such that Ax=o, where x is the column vector whose entries are the numbers in the special similar set; because the similar set conditions determine 11 linear equations in 11 unknowns.  Thus there exists a special similar set with not all 11 numbers equal if and only if some such matrix A has rank less than 10 (because the column vector of all 1's is also a solution of Ay=o linearly independent of x).

By earlier posts, there cannot exist 11 rational numbers x_1,...,x_11, not all equal, such that Ax=O, x being the column vector with x_i its i-th entry, for i=1,...,11.  Hence, relative to the rational numbers, the rational matrix A has rank 10.  But then A also has rank 10 relative to the real numbers; whence there are no 11 real numbers with not all 11 numbers equal which form a special similar set.


  Posted by McWorter on 2005-07-15 22:05:30
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