Competition problem — Combinatorics

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Combinatorics Difficulty 5.6 AIME, harder Prove it Belarus

Given a $20 \times 20$ table with one of two signs "+" or "-" in any of its cells. Per move one can replace the signs in all cells of some row (or of some column) by the opposite signs. At the beginning there are $8$ minuses in the table (all other signs are pluses). After some moves the table with exactly $50$ minuses is obtained.
Prove that exactly one initial minus is in the same cell.

Solution

than $10$ (since $0 \le x \le 10$ , $0 \le y \le 10$ ). It follows that exactly one number, namely $72$ , may be presented as the product of two positive integers no greater than $10$ ( $72 = 8 \cdot 9$ , e.g., $x = 1, y = 2$ ). It corresponds to the case when $7$ initial minuses are changed. This means that exactly one initial minus keeps its position.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.