3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{100}$ be natural numbers greater than 1 (not necessarily distinct). In an $80 \times 80$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{16}$ is written. Find the smallest possible value of the sum of all numbers in the table.
Answer: -19200

Solution: $\log _{x_{k}} \frac{x_{i}}{16}=\log _{x_{k}} x_{i}-\log _{x_{k}}$ 16. Add the minuend with the similar minuend in the symmetric cell: $\log _{x_{k}} x_{i}+\log _{x_{i}} x_{k}=\log _{x_{k}} x_{i}+\frac{1}{\log _{x_{k}} x_{i}} \geqslant 2$, since this logarithm is positive. If $i=k$, then this logarithm
equals one.

Thus, the sum of the minuends in all expressions $\log _{x_{k}} x_{i}-\log _{x_{k}} 16$ is no less than the number of cells, i.e., $80^{2}$. The subtrahends are maximal when $x_{k}=2$ and equal 4. Therefore, the sum of the numbers in the entire table is no less than $80^{2}(1-4)=-80^{2} \cdot 3=-19200$ and equality is achieved when all $x_{k}=2$.