Consider a large square grid with side length 10, divided into small squares with side length 1. Two small squares are said to be neighbors if they share a common side. On each of the small squares, a positive real number is inscribed. Furthermore, 5 frogs move on the grid and can each cover a small square. Two frogs never cover the same square. Between two moments, each frog jumps from the square where it is to a neighboring square.

It is assumed that the sum of the visible numbers is 10 at instant 1, then $10^{2}$ at instant 2, then $10^{3}$ at instant 3, and so on until instant $k$ where the sum is $10^{k}$. What is the largest possible value of $k$?
We will show that the largest possible value of $k$ is 6. First, let's verify that 6 is indeed achievable. For this, we isolate a $2 \times 3$ rectangle in the top left corner of the grid. On all the cells that are not in this rectangle, we write the number 0. On the cells of the rectangle, we write the numbers $10, 100, \ldots, 10^{6}$ as follows:

| $10^{3}$ | $10^{4}$ |
| :---: | :---: |
| $10^{2}$ | $10^{5}$ |
| 10 | $10^{6}$ |

At instant 1, the frogs occupy the 5 cells bearing the numbers $10^{2}, \ldots, 10^{6}$, so only the number 10 is visible (along with all the 0s). Then at instant 2, the frogs rotate clockwise, so that $10^{2}$ is visible (along with the 0s), and so on... We then have a configuration as desired with $k=6$.

Now let's show that $k>6$ is impossible. Suppose $k \geqslant 7$. When moving from instant 1 to instant 2, at most 5 new numbers are revealed (the 5 that were covered at instant 1), and the sum of the visible numbers increases by $10^{2}-10=90$. Therefore, the sum of the numbers discovered between instant 1 and instant 2 is at least 90. Since there are at most 5 numbers discovered, at least one is greater than $90 / 5=18$. On the other hand, the sum of the visible numbers at instant 2 is 100, so the discovered numbers do not exceed 100, hence there is a cell bearing a number between 18 and 100. By reasoning similarly between instants 2 and 3, then 3 and 4, and so on until 6 and 7, we find a cell bearing a number between 180 and 1000, then between 1800 and 10000, and so on until $18 \times 10^{5}$ and $10^{7}$. We have thus found 6 cells greater than 18. Since there are only 5 frogs at instant 1, at least one of these cells was visible at instant 1, so the sum was at least 18, leading to a contradiction.