10.3. Find all natural numbers $n$ such that a square of size $n$ by $n$ cells can be cut along the grid lines into a single-cell square and four rectangles, all nine side lengths of which are pairwise distinct.
Answer: All $n \geq 11$.

Solution. We will show that the cutting scheme must look like the one shown in the left figure (note that in the solution, this proof is not necessary; it is sufficient to simply indicate the scheme itself! (
![](https://cdn.mathpix.com/cropped/2024_05_06_f9a2383f252244264691g-2.jpg?height=294&width=1128&top_left_y=374&top_left_x=294)

Indeed, if a square is adjacent to a side or vertex of the larger square, then next to it will be two or one rectangles with sides larger than 1 (see the central and right figures), and to the side of the square that does not adjoin these rectangles and the sides of the larger square, nothing with larger sides can adjoin, according to the condition.

After this, note that each vertex of the larger square is contained in its own rectangle of the partition. Otherwise, one of the sides of one of the rectangles coincides with a side of the larger square, and the second side of the rectangle is smaller. By cutting it off, we obtain a partition of the remaining rectangle into a square and three rectangles with different sides. Repeating the previous reasoning, we get that the square again must not adjoin the sides of the remaining rectangle, and thus must adjoin four different rectangles of the partition with its four sides, which cannot be, since there are only three left.

From the above, it follows that the cutting scheme must look like the one shown in the left figure:

Let's find the minimum possible area of the required square. Let the lengths of the nine sides of the parts of the partition in increasing order be $a_{1}=1, a_{2} \geq 2, \ldots a_{9} \geq 9$. To find the minimum total area of the square and 4 rectangles with such sides, we will use the widely known "transinequality," which, in particular, states that for any numbers $a_{1} \leq a_{2} \leq . . \leq a_{n}$ and $b_{1} \leq b_{2} \leq . . \leq b_{n}$, the minimum sum of the form $S(\sigma)=\sum_{i=1}^{n} a_{i} b_{\sigma(i)}$, where $\sigma(1), \sigma(2), \ldots, \sigma(n)$ is any permutation of the numbers $1,2, \ldots, n$, is achieved for the sum $S(\sigma)$ with $\sigma(i)=n+1-i, i=1, \ldots, n$, that is, when each $i$-th number in increasing order of the first set is multiplied by the $i$-th number in decreasing order of the second set. From this, it follows that the minimum area of the square and 4 rectangles is $1+a_{2} a_{9}+a_{3} a_{8}+a_{4} a_{7}+a_{5} a_{6} \geq 1+18+24+28+30=101>10^{2}$. Therefore, the minimum possible side length of the larger square is 11. We will indicate a way to cut a square with side $n=11$ in the required manner, with the side lengths of the corresponding rectangles indicated by the numbers in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_06_f9a2383f252244264691g-3.jpg?height=597&width=645&top_left_y=210&top_left_x=274)

The cutting method for any natural $n \geq 11$ is obtained from the given method by expanding the square and 3 rectangles to the right and down by $n-11$ cells. All sizes will remain different, as the lengths equal to $6,7,8$ and 9, which are already larger than the remaining unchanged lengths $2,3,4$ and 5, will increase equally.

Grading criteria. Proof of the minimality of $n=11: 3$ points. Method of cutting for $n=11: 2$ points. Method of cutting for all $n \geq 11$: another 2 points. Absence of justification that the minimum area of the square and 4 rectangles is $1+a_{2} a_{9}+a_{3} a_{8}+a_{4} a_{7}+a_{5} a_{6} \geq 1+18+24+28+30=101>10^{2}$: minus 1 point.