B2. All 2012 camels in the Netherlands must be distributed over 40 pastures. No two pastures may have the same number of camels. The pasture in the center of Amsterdam must receive the largest number of camels.

How many camels must be placed there at a minimum?
B2. 70 We can place 23 camels in one meadow and in the other 39 meadows sequentially 32, 33, 34, ..., 70 camels, with the last meadow being in the center of Amsterdam. The total number of camels is then indeed equal to \(23 + (32 + 33 + \cdots + 70) = 23 + 39 \cdot \frac{32 + 70}{2} = 2012\). Thus, there is a distribution with 70 camels in the meadow in the center of Amsterdam.

With fewer than 70, it is not possible. Suppose that at most 69 camels are placed in the meadow in the center of Amsterdam. Since the number of camels is different everywhere, the second busiest meadow gets at most 68 camels, the third busiest gets at most 67 camels, and so on. In total, there are then at most \(69 + 68 + \ldots + 30 = 40 \cdot \frac{30 + 69}{2} = 1980\) camels distributed over the meadows, leaving at least 32 camels remaining. Therefore, there is no solution with fewer than 70 camels in the meadow in the center of Amsterdam. The minimum number is thus 70.