## Task 25/62

An apprentice is supposed to count 1000 balls with a diameter of $1 \mathrm{~cm}$ in a ball bearing factory. To speed up this task, he takes a container with inner dimensions of $10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 10$ $\mathrm{cm}$; he lays the first layer neatly and then continues to fill it.

In the end, he finds that, contrary to his expectations, the inner space of the container is not completely filled. He therefore counts the balls. Surprisingly, there are more than a thousand.

How many were there, and how many centimeters were missing from the top layer of balls to the edge of the container?
The 2nd, 4th, 6th, ... layer consists of only 81 balls each, as each ball in these layers fits into the depression between four balls of the previous layer. The 1st, 3rd, 5th, ... layers, on the other hand, consist of 100 balls each, corresponding to the inner dimensions of the container.

The distance between two adjacent planes through the centers of the balls is obtained by a consideration based on the diagram, which yields \(\frac{1}{2} \sqrt{2} \, \text{cm}\). Thus, \(n\) layers have a total thickness of \(\left(1 + \frac{n-1}{2} \sqrt{2}\right) \, \text{cm}\).

We need to find the largest (natural) number \(n\) for which

\[
\left(1 + \frac{n-1}{2} \sqrt{2}\right) \leq 10
\]

or, equivalently, \(n \leq 9 \sqrt{2} + 1 \approx 13.73\).

This means there are 13 layers in the container, 7 layers with 100 balls each and 6 layers with 81 balls each, totaling 1186 balls.

The total height of these 13 layers is \(\left(1 + \frac{12}{2} \sqrt{2}\right) = (1 + 6 \sqrt{2}) \approx 9.48 \, \text{cm}\). From the top layer of balls to the rim of the container, there is still approximately \(0.52 \, \text{cm}\) missing.