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?

(In reply to Solution for part 1 by e.g.)

It is true that all the weights must be all odd or all even.  For, if w is any weight and T is the total of all weights, then T-w=2x, where x is the weight of the of the two five weight sets that balance each other.  Hence each weight has the same parity as the total of all weights.

On the other hand, I don't see how shrinking the size of the weights can produce a contradiction.  However, scaling and/or translating the problem may contradict the condition that not all weights are equal.  So let me start an argument focusing on the difference between two weights.

Suppose there is a solution with not all the weights equal.  Let d>0 be the smallest difference between two weights among all solutions.  Not all weights of this minimal solution are even; for, by dividing by 2 we get a solution with a smaller minimum difference between weights.  Hence at least one weight is odd; and so, by the opening remarks, all weights are odd.  Now subtract (or add) 1 to each weight to get a solution with all weights even and minimum difference between weights equal to d.  But then we can divide all weights by 2 and get a solution with minimum difference d/2, (well, I'll be dinged!) contradiction.  Hence it is not possible that the man offered Oskar 11 weights not all equal satisfying the special similar set condition.

Posted by McWorter on 2005-06-27 22:33:59