As long as more than 2000 oranges are left, the camel needs to return twice to pick each one up. For every mile 5 oranges are lost. This continues until only 2000 oranges are left. The first 1000 oranges will have been lost after 1000/5=200 miles. Therefore,
1. The camel takes 1000 oranges to the 200-mile mark, leaves 600 there and returns.
2. The camel repeats this with another 1000 oranges.
3. The camel takes the remaining 1000 oranges to the 200-mile mark. At this point, 600+600+800=2000 oranges are at the 200-mile mark.
As long as more than 1000 but fewer than 2001 oranges are
left, the camel needs to return once to pick each one up. For every mile 3
oranges are lost. This continues until only 1000 oranges are left. The second
1000 oranges will have been lost after an additional 1000/3=333 miles. Therefore,
4. The camel takes 1000 oranges from the 200 to the 533-mile mark, leaves 333 oranges there and returns to the 200-mile mark.
5. The camel takes the remaining 1000 oranges to the 533-mile mark. At this point, 333+667=1000 oranges are at the 533-mile mark.
6. These 1000 oranges the camel can now transport one way the remaining 467 miles to the market. When the camel arrives at the market, he will bring 533 oranges.
533 is probably not the maximum number, but at least it sets a lower bound on this maximum.
Suppose the actual maximum is 533+x oranges. How big can x be? It seems reasonable to assume that the camel carries the maximum 1000 oranges on each forward trip (towards the market). Then his last forward trip must have started x miles after the 533- mark. This implies x≤467, so the actual maximum is ≤1000 oranges, which sets an upper bound.