2. In a square table $11 \times 11$, we have written the natural numbers $1,2, \ldots, 121$ sequentially from left to right and from top to bottom. Using a square tile $4 \times 4$, we covered exactly 16 cells in all possible ways. How many times was the sum of the 16 covered numbers a perfect square?
SOLUTION. Let the number covered by the top left corner of the tile be denoted as $z$. The entire tile must lie within the given table, so the values of $z$ can only be the numbers written in the first eight rows and the first eight columns of the table (for example, if $z=10$, the tile would protrude, and thus could not cover 16 numbers in the table).

The first 8 rows of the table contain numbers from 1 to 88, from which we must exclude the numbers in the last three columns. Notice that the numbers in each column give the same remainder when divided by eleven. The last three columns from the left (the first three from the right) thus contain numbers that give remainders of 9, 10, and 0 when divided by eleven; these are the numbers 9, 10, 11 (first row), 20, 21, 22 (second row), and so on up to 86, 87, 88 (eighth row).

With this preparation, we can calculate the sum of the numbers that the tile will cover. The covered numbers are $z, z+1, z+2, z+3$ (first row of the tile), $z+11, z+12, z+13, z+14$ (second row of the tile), $z+22, z+23, z+24, z+25$ (third row of the tile), and $z+33, z+34, z+35, z+36$ (fourth row of the tile), and their sum is

$$
16 z+288=16(z+18)=4^{2}(z+18)
$$

If this sum is a perfect square of some integer, then $z+18$ must be a perfect square of some integer $n$. We already know that $1 \leqq z \leqq 88$, and thus $19 \leqq z+18=$ $=n^{2} \leqq 18+88=106$. This ensures that the top left corner of the tile is placed on a cell in the first eight rows. For the natural number $n$, where $19 \leqq n^{2} \leqq 106$, the possible values are $n \in\{5,6,7,8,9,10\}$. By calculating the values $z=n^{2}-18$, we get the corresponding $z \in\{7,18,31,46,63,82\}$.

We must also check whether any of these $z$ values lie in the last three columns of the table. Therefore, we calculate the remainders of these numbers when divided by eleven and find that we must additionally exclude the value $z=31$ with a remainder of 9.

The tile can be placed in five different positions as required, characterized by the number covered by the top left corner of the tile, which is $z \in\{7,18,46,63,82\}$. In these cases, the sum of the numbers covered by the tile will be $16(z+18) \in\{16 \cdot 25,16 \cdot 36,16 \cdot 64, 16 \cdot 81,16 \cdot 100\}$.