On all cells of an $8 \times 8$ chessboard, natural numbers are placed. It is allowed to select any square of size $3 \times 3$ or $4 \times 4$ and increase all numbers in it by 1. We want to achieve, after several such operations, that the numbers in all cells are divisible by 10. Is it always possible to do this?
We will record in the table only the last digits (remainders from division by 10). Note that the order of performing the operations is not important, but the set of operations is. Moreover, since performing an operation ten times in one square does not change anything, it makes sense to perform an operation in each square no more than 10 times. There are a total of $(8-3+1)^{2}=36$ squares of $3 \times 3$ and 25 squares of $4 \times 4$ on the board. Therefore, no more than $10^{25+36}=10^{61}$ different tables can result from each table. This means that a table of all zeros can be obtained from no more than $10^{61}$ tables (if a table $A$ can be transformed into a zero table, then a zero table can also be transformed into table $A$), which is less than the total number of tables $10^{64}$.

## Answer

Not always.

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