3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{60}$ be natural numbers greater than 1 (not necessarily distinct). In a $60 \times 60$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{8}$ is written. Find the smallest possible value of the sum of all numbers in the table.

Answer: -7200
Solution: $\log _{x_{k}} \frac{x_{i}}{8}=\log _{x_{k}} x_{i}-\log _{x_{k}} 8$. 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 $\log _{x_{k}} x_{i}-\log _{x_{k}} 4$ is not less than the number of cells, i.e., $60^{2}$. The subtrahends are maximal when $x_{k}=2$ is minimal and equal to 3. Therefore, the sum of the numbers in the entire table is not less than $60^{2}(1-3)=-60^{2} \cdot 2$ and equality is achieved when all $x_{k}=2$.