7.3. There are 55 boxes standing in a row, numbered in order from 1 to 55. Each box contains no more than 10 balls, and the number of balls in any two adjacent boxes differs by exactly 1. It is known that the boxes numbered $1,4,7,10, \ldots, 55$ contain a total of 181 balls. What is the minimum total number of balls that can be in all 55 boxes?
Solution. Since the number of balls in two adjacent boxes differs by 1, the parity of this number always differs. Therefore, throughout the entire row of boxes, the parity alternates - the number of balls in boxes with different parity numbers also has different parity. We will divide the boxes into pairs: 1 with 4, 7 with 10, 13 with 16, ..., 49 with 52 - a total of 9 pairs. In each pair, one of the quantities is even, the other is odd. Therefore, in two boxes of one pair, there are no more than $10+9=19$ balls. Then in the 55th box, there are no fewer than $181-19 \cdot 9=10$ balls. Therefore, the condition of the problem can only be met in one situation: in the 55th box, there are exactly 10 balls, and in the specified pairs - exactly 19 each. Moreover, the boxes with 10 and 9 balls in the pairs alternate. Then we notice that each interval between the specified boxes can only be of two types: 9 and 10 balls or 8 and 9 balls. In the second case, there are fewer balls. We calculate the minimum number of balls: $181+18 \cdot(8+9)=487$.

Answer. 487 balls.