In the diagram, there are exactly nine $1 \times 1$ squares. What is the largest number of $1 \times 1$ squares that can be shaded so that no two shaded squares share a side?

![](https://cdn.mathpix.com/cropped/2024_04_17_b6707dc71c8b7ccd30afg-1.jpg?height=315&width=523&top_left_y=1634&top_left_x=861)
In the diagram below, there are 6 squares shaded in such a way that no two shaded squares share an edge, which shows that the answer is at least 6 .

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-07.jpg?height=323&width=523&top_left_y=2175&top_left_x=861)

We will now argue that it is impossible to shade more than 6 square in such a way that no two shaded squares share an edge, which will show that the answer to the question is 6 .

If four or five squares in the bottom row are shaded, then at least two shaded squares will share an edge.

Therefore, at most 3 squares can be shaded in the bottom row.

If all three squares in the middle row are shaded, then there will be shaded squares that share an edge.

Therefore, at most 2 squares can be shaded in the middle row.

There is only one square in the top row, so at most 1 square can be shaded in the top row.

Thus, the number of squares that can be shaded so that no two shaded squares share an edge is at most $3+2+1=6$.

ANSWER: 6