For analytical calculations in astronomy are extremely important sets of numbers $\left\{A_1, A_2, \ldots, A_n\right\}$ such that: $$ \sum\limits_k {A_K} = -2.5 \ln \left(\sum\limits_k { 10^{-0.4*A_k} } \right)/ \ln(10). $$ It is easy to show that the task of finding such sets is to find numbers whose product and sum are equal.
So, the question is: is there a set of integer powers of $10^{0.4}$ such that their product is equal to their sum?
There is a finite set of numbers $A_k$ satisfying your condition. First of all, let me reformulate the condition.
$$\prod_k 10^{-0,4 \ A_k} = \sum_k 10^{- 0,4 \ A_k}$$
An example is given by \begin{align} A_1=A_2 = -5; \quad A_3 = \dots = A_{9802}=0. \end{align} Then $$\prod_{k=1}^{9802} 10^{-0,4 \ A_k} = 100 \times 100 = 10000$$ $$\sum_{k=1}^{9802} 10^{- 0,4 \ A_k} = 100+100+9800= 10000$$