UA-UMO 2023 Problem 81

Oleksiy placed positive integers in the cells of the $8\times 8$ chessboard. For each pair of adjacent-by-side cells, Fedir wrote down the product of the numbers in them and added all the products. Oleksiy wrote down the sum of the numbers in each pair of adjacent-by-side cells and multiplied all the sums. It turned out that the last digits of both numbers are equal to $1$ . Prove that at least one of the boys made a mistake in the calculation.

For example, for a square $3\times 3$ and the arrangement of numbers shown below, Fedir would write the following numbers: $2, 6, 8, 24, 15, 35, 2, 6, 8, 20, 18, 42$ , and their sum ends with a digit $6$ ; Oleksiy would write the following numbers: $3, 5, 6, 10, 8, 12, 3, 5, 6, 9, 9, 13$ , and their product ends with a digit $0$ .

\begin{tabular}{| c| c | c |} \hline 1 & 2 & 3 \hline 2 & 4 & 6 \hline 3 & 5 & 7 \hline \end{tabular}

Proposed by Oleksiy Masalitin and Fedir Yudin