According to problem 4283 of Lapunk B., if a $23 \times 23$ square is divided into $1 \times 1$, $2 \times 2$, and $3 \times 3$ squares, then we need at least one $1 \times 1$ square for this. Where can the $1 \times 1$ square be placed if we know that exactly one appears in the division?
Let's number the rows of the square from top to bottom and the columns from left to right, from 1 to 23.

First, color the square such that the odd-numbered rows are black and the even-numbered rows are white. Then, there are $12 \times 23$ black squares and $11 \times 23$ white squares. Every $3 \times 3$ square covers either 3 or 6 black squares, and every $2 \times 2$ square covers exactly 2 black squares. Therefore, the difference in the number of black and white squares covered by $3 \times 3$ squares is divisible by 3, and the difference in the number of black and white squares covered by $2 \times 2$ squares is also divisible by 3. We know that the total difference in the number of black and white squares is 23. If the $1 \times 1$ square is on a black square, then the difference in the number of remaining black and white squares is 22, which is not divisible by 3, so the remaining area cannot be covered by $2 \times 2$ and $3 \times 3$ squares. Therefore, the $1 \times 1$ square must be on a white square, which is in an even-numbered row.

Now, color the $23 \times 23$ square such that the odd-numbered columns are black and the even-numbered columns are white. By a similar reasoning, we find that the $1 \times 1$ square must be on a white square, which is in an even-numbered column.

Now, color the $23 \times 23$ square such that the rows whose row numbers are divisible by 3 are red, and the other rows are blue. Then, every $3 \times 3$ square covers 6 blue squares, and every $2 \times 2$ square covers either 2 or 4 blue squares. Therefore, every $3 \times 3$ and $2 \times 2$ square covers an even number of blue squares. We know that there are $16 \times 23$ blue squares in total, which is an even number. If the $1 \times 1$ square is on a blue square, then an odd number of blue squares remain, so the remaining area cannot be covered by $2 \times 2$ and $3 \times 3$ squares. Therefore, the $1 \times 1$ square must be on a red square, which is in a row number divisible by 3.

Now, color the $23 \times 23$ square such that the columns whose column numbers are divisible by 3 are red, and the other columns are blue. By a similar reasoning, we find that the $1 \times 1$ square must be on a red square, which is in a column number divisible by 3.

Therefore, the $1 \times 1$ square can only be in a row and column number divisible by 6, which means it can be in 9 positions on the board.

Now, we will show that these 9 possibilities can indeed be realized.

First, we show that if $k \geq 1$, then a $(2k+1) \times 6n$ rectangle can be covered by $2 \times 2$ and $3 \times 3$ squares. A $3 \times 6n$ rectangle can be covered by only $3 \times 3$ squares. A $5 \times 6$ rectangle can be covered as shown in Figure 1. A $5 \times 6k$ rectangle can also be covered by cutting it into $k$ $5 \times 6$ rectangles, which can already be covered. A $(2k+1) \times 6n$ rectangle (for $2k+1 > 5$) can be covered by cutting off a $5 \times 6n$ rectangle from the edge. This can be covered, and the remaining $(2k+1-5) \times 6n = (2k-4) \times 6n$ rectangle can be covered by only $2 \times 2$ squares.

![](https://cdn.mathpix.com/cropped/2024_05_02_d3ae732e6af2ce039777g-1.jpg?height=217&width=266&top_left_y=1407&top_left_x=927)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_d3ae732e6af2ce039777g-1.jpg?height=219&width=225&top_left_y=1674&top_left_x=945)

Figure 2

As shown in Figure 2, a $11 \times 11$ square can be covered by one $1 \times 1$ square and 4 $5 \times 6$ rectangles, with the $1 \times 1$ square at the center. Let the square in the $i$-th column and $j$-th row be the $(i, j)$ square. Let the $1 \times 1$ square be at the $(6l, 6m)$ square, where $l \geq 1$ and $m \geq 1$. Then, cover this $11 \times 11$ square with 4 $5 \times 6$ rectangles and the $1 \times 1$ square, with the $1 \times 1$ square at the center. The coordinates of the corners of the $11 \times 11$ square are: $(6l-5, 6m-5)$, $(6l+5, 6m-5)$, $(6l-5, 6m+5)$, $(6l+5, 6m+5)$. Since $l > 0$ and $m > 0$, the $11 \times 11$ square is within the $23 \times 23$ square.

Cut the $23 \times 23$ square along the sides of the $11 \times 11$ square (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_d3ae732e6af2ce039777g-1.jpg?height=414&width=585&top_left_y=2292&top_left_x=759)

Then, we get 8 rectangles: one $(6l-6) \times (6m-6)$ rectangle, one $11 \times (6m-6)$ rectangle, one $(18-6l) \times (6m-6)$ rectangle, one $(18-6l) \times 11$ rectangle, one $(18-6l) \times (18-6m)$ rectangle, one $11 \times (18-6m)$ rectangle, one $(6l-6) \times (18-6m)$ rectangle, and one $(6l-6) \times 11$ rectangle. Since $6l < 23$ and $6m < 23$, we have $l \leq 3$ and $m \leq 3$. Therefore, all eight rectangles have sides that are either zero or positive numbers. Among these, four rectangles have all sides even, so they can be covered by $2 \times 2$ squares, and four are of the form $(2k+1) \times 6n$, which can be covered. Therefore, we have covered the $23 \times 23$ square. We have shown that the $1 \times 1$ square can be at the $(6l, 6m)$ coordinates.