Drawing this one out and process of elimination make it a bit clearer. Instead of cutting them out, I imagined being given pieces to fit on the board.

It is clear that using only 1x1 squares is greater than 6.

Using only 2x2 squares, we can fit a maximum of 4. The remaining board can only be cut with 1x1's. After using the 4 2x2's, we have 9 1x1's tiles left on the board, so this scenario is also greater than 6.

We can only fit 1 3x3 square on the board. If we position it anywhere but the corners, we'll make scenarios that have many 1x1 squares left on the board, and all those scenarios are greater than 6.

On to the solution following this line of thinking:


For loss of generality, we'll assume we put the 3x3 square in the bottom left corner of the 5x5 board.

After putting the 3x3 in the corner, we have a 2x5 space to the right of the 3x3 and a 3x2 space above the 3x3.

In the 2x5 space, we can fit 2 2x2's on top of each other, leaving 2 1x1 spaces left.

In the 3x2 space, we can fit 1 2x2, leaving 2 1x1 spaces left.

We cannot arrange the 2x2's in such a way that the 1x1's leftover can accommodate another 2x2. Therefore, this must be the minimum.

Counting the pieces we fit (or cut):

1 3x3, 3 2x2's, and 4 1x1's.

\(1+3+4 = 8\)

Therefore, the answer is A

Would love a more formal answer. Wonder if there's a way to make the minimum 7 as well!