There are an infinite number of $1\times 1, 2\times 2,$ and $3\times 3$ squares available. If we want to completely cover a $23\times 23$ square with these elements, what is the fewest $1\times 1$ squares that can possibly be used?

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Details and Assumptions:
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- The squares are not allowed to overlap or stick out of the $23\times 23$ board.
- We get to choose the placement of the squares.

This problem is from the Hungarian KöMal (High School Mathematical Pages).

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It is possible to cover the $23\times 23$ square with only one $1\times 1$ square. But is this the minimum? We will prove that 1 is the minimum!

Suppose that we covered the $23\times 23$ square with only $2\times 2$ and $3\times 3$ squares. If we colour the $1.,\space 4.,\space 7., \space 10.,\space 13.,\space 16.,\space 19.,\space 22.$ rows black, then there are $8*23$ black, and $15*23$ white cells. Every $2\times 2$ square covers $4$ or $2$ white cells, and every $3\times 3$ square covers exactly 6 white cells. So every square covers an even number of white cells, but $15*23$ is odd.

So the minimum of $1\times 1$ squares is $\boxed{1}$ .