Let $A$ be one of the lightest weights, and $B$ be one of the weights that follow $A$ in weight. Clearly, the pair of weights $\{A, B\}$ can only be balanced by an identical pair.

Therefore, there are at least two weights $A$ and two weights $B$. The pair $\{A, A\}$ can also only be balanced by an identical pair. Therefore, there are at least 4 weights $A$. For similar reasons, there are at least 4 of the heaviest weights $E$ and at least two weights of the next heaviest weight $D$.

In addition, by the condition, there is at least one weight $C$ that is heavier than $A$ and $B$ and lighter than $D$ and $E$.

Thus, the total number of weights in our set is no less than $4+4+2+2+1=13$.

On the other hand, it is easy to verify that a set of 13 weights: $\{1,1,1,1,2,2,3,4,4,5,5,5,5\}$ satisfies the condition of the problem.

## Answer

13.00
Rubanovo I.S.

There is a set of weights with the following properties:

a. It contains 5 weights, all of different weights.

b. For any two weights, there are two other weights with the same total weight.

What is the smallest number of weights that can be in this set?