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?

Please forgive this technical proof.  I could not find a more elementary proof.

Let the eleven real numbers be z_1,...,z_11.  The z_i span a finite-dimensional subspace of the real numbers considered as a vector space over the rationals.  Let c_1,...,c_k be a basis for this subspace.  Then there are rational numbers a_i,j, for i=1,...,11 and j=1,...,k such that

z_i=a_i,1*c_1+...+a_i,k*c_k, for i=1,...,11

We know that, if any one of the z_i is left out, the remaining 10 can be split into two sets of five with the same sum.  To reduce ugliness, let's reorder the z_i so that the number left out is z_11 and the two sets of five with the same sum consist of the first five and the second five in the reordering.  Hence

z_1+...+z_5=z_6+...+z_10.

In terms of the basis, we have

a_1,1*c_1+...+a_1,k*c_k+...+a_5,1*c_1+...+a_5,k*c_k=
=a_6,1*c_1+...+a_6,k*c_k+...+a_10,1*c_1+...+a_10,k*c_k.

Collecting terms with the same basis element, we get

(a_1,1+...+a_5,1)c_1+...+(a_1,k+...+a_5,k)c_k=
=(a_6,1+...+a_10,1)c_1+...+(a_6,k+...+a_10,k)c_k.

Therefore, since the c_i form a basis,

a_1,j+...+a_5,j=a_6,j+...+a_10,j for j=1,...,k.

This implies that the eleven rational numbers a_1,j,...,a_11,j satisfy the same conditions as the z_i.  By earlier posts, it follows that the a_i,j are all equal, for each j.  Hence the z_i are all equal.

Posted by McWorter on 2005-07-15 00:20:25