8. (5 points) A $10 \times 10$ table is filled with zeros. In one operation, the smallest number in the table (if there are several, any one is chosen) is found, and one is added to it, as well as to all numbers in the cells adjacent to it by side or corner. What is the largest number that can appear in one of the cells of the table after 80 operations?
Answer: 20.

Solution: Let's call the $n$-th phase a sequence of consecutive operations applied to numbers equal to $n$. If such operations did not occur, we will say that this phase consists of zero operations. Everything starts with the zero phase.

During one phase, no number can increase by more than 4. Indeed, if we increased the number itself, then neither to it nor to its neighbors can our operation be applied during this phase, neither before nor after, since a number increased by one in some phase cannot be the minimum during the same phase. Among the neighbors of a number, no more than 4 non-adjacent ones can be chosen, so indeed, we can increase a number by no more than 4 in one phase.

Consider the first, fourth, seventh, and tenth cells in the first, fourth, seventh, and tenth rows of the table, a total of 16 cells. Note that no operation affects two of them. Therefore, to change $n$ phases, at least $16n$ operations are needed. (Important! We cannot claim that one phase lasts at least 16 operations, as some of our numbers could have been increased in previous phases)

Thus, no more than 5 phases have passed, and in each, no number could increase by more than 4. Therefore, in 80 operations, no number could become greater than 20.

The example is constructed as follows: consider the second, fourth, seventh, and tenth cells in the second, fourth, seventh, and tenth rows of the table, a total of 16 cells. We will apply operations only to them. Operations applied to one of these cells do not affect the other 15, and after applying operations to all 16 cells, all numbers in the table increase by at least 1. Meanwhile, the number in the third cell of the third row will increase by 4 each phase.

Of course, the example is not unique.