Problem 4. A hundred boxes are numbered from 1 to 100. Each box contains at most 10 balls. The numbers of balls in any two boxes numbered with consecutive numbers differ by 1. The boxes numbered $1,4,7,10, \ldots, 100$ contain, in total, 301 balls. What is the maximum number of balls in the 100 boxes?
## Solution.

The numbers of balls in two adjacent boxes differ by 1, so two adjacent boxes can contain, in total, at most $10+9=19$ balls.

The pairs of boxes $(2,3),(5,6), \ldots,(98,99)$ can contain, in total, at most $33 \cdot 19=627$ balls, so the total number of balls in the 100 boxes cannot exceed $627+301=928$. .............. 3 p

There is a way to arrange 928 balls such that the conditions of the problem are satisfied. For example, the following distribution of balls is suitable:

$$
\underbrace{[91098910] \ldots[91098910]}_{10 \text { groups }} \underbrace{[910910910] \ldots[910910910]}_{6 \text { groups }}[91098]
$$

$\qquad$
Work time 2 hours. An additional 30 minutes is given for questions. Each problem is scored out of 7 points.