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The original condition is:
In all cells of a chessboard the natural numbers are placed. You can select a square 3 by 3 or 4 by 4 and add 1 to all numbers in the squares. Is it possible to make a chain of such operations that will make all numbers in all cells of the board divisible by 10? (The problem is taken from the book https://mirtitles.org/2013/01/01/combinatorial-mathematics-vilenkin/ by author https://en.wikipedia.org/wiki/Naum_Ya._Vilenkin)

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closed as off-topic by Eevee Trainer, mrtaurho, Abcd, Cesareo, Riccardo.Alestra Jan 10 at 9:55

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  • $\begingroup$ Can any naturals be placed in starting setup? Do they need to be distinct for different squares? Or even more, are they the numbers from $1$ to $64$? $\endgroup$ – coffeemath Jan 10 at 5:27
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    $\begingroup$ There is nothing about numbers being distinct or from 1 to 64. Just any natural numbers. $\endgroup$ – Melina Jan 10 at 5:37
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Unfortunately I don't have enough reputation points to comment, so I guess I'll have to make it as an answer.

Anyway, observe that it is sufficient to use the mod-10 integers to denote a board position and to count how many times you added a 1 to a 3x3 or 4x4 sub-board, and observe that the order of the adding moves has no overall effect.

If we identify each solution of a board position with a $5^2+6^2=61$ tuplet of mod-10 integers corresponding to how many times we added a 1 to each of the 61 3x3 or 4x4 sub-boards, we find that in fact there are $10^{61}$ possible solutions in total.

But observe that one solution can only correspond to at most one board position, thus there are at most $10^{61}$ solvable board positions in total. However as there are $10^{64}$ possible board positions, we may conclude that not all of them are solvable.

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Are the 3x3 and 4x4 squares allowed to overlap the edges of the board?

If no: I'll let another user answer this one.

If yes: Then it is possible to make all numbers divisible by 10. Start at the bottom left cell of the board. Add ones until it is divisible by 10 using a 3x3 square whose bottom left corner coincides with this cell. Next do the same thing one cell to the right. Again, this cell coincides to the bottom left corner of the 3x3 square. Complete the entire bottom row like this. Then do the next row up in the exact same way. Do all the rows like this.

This strategy works because once we have done each cell, we don't ever touch it again. But it requires some of our squares to extend off the edge of the chess board.

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  • $\begingroup$ Yes,the squares are allowed to overlap the edges of the chessboard. Thank you:)) $\endgroup$ – Melina Jan 10 at 11:48

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