The merchant's weights

1. Being able to put weights on both plates completely changes the problem.
2. Each weight can help in three ways: adding, compensating or being left out.
3. Look for a choice of weights that will allow you to build any small amount without leaving gaps.

Answer: The weights must be 1, 3, 9 and 27 kilos. Explanation: Since the weights can be placed on both plates, each of them can contribute in three ways: - do not use it;

• put it on the plate opposite the object, which is equivalent to adding its weight;
• put it on the same plate as the object, which is equivalent to subtracting it. That means that each weight functions as a “digit” with values ​​\(-1\), \(0\) or \(1\). The natural basis for this type of representation is, therefore, that of the powers of 3: \[

1,\ 3,\ 9,\ 27.
\] With these four weights you can write any integer between 1 and 40 as a combination of the form \[
a_0\cdot 1+a_1\cdot 3+a_2\cdot 9+a_3\cdot 27,
\qquad a_i\in\{-1,0,1\}.
\] That is exactly the balanced ternary. Examples: - \(2 = 3-1\),

• \(8 = 9-1\),
• \(20 = 27-9+3-1\),
• \(40 = 27+9+3+1\). Therefore, the only weights that allow you to cover all the entire quantities from 1 to 40 kilos with only four pieces are 1, 3, 9 and 27.