*Originally posted by davegage*

**The distance between Cairo and Damascus is 1000 miles. You have 10,000 kilograms of grass you want to move from Cairo to Damascus using your camel. However, two difficulties arise:
**

1. Your camel will not budge unless you continuously allow him to eat grass, and he consumes 1 kilogram of grass per mile.

2. The maximum carrying capacity of the camel is ...[text shortened]... t any of the grass to Damascus? If so, what is the maximum amount you will be able to transfer?

If I take the camel half the distance in one go, I lose all my grass to the camel (loss of 100% per iteration, 1 iteration).

If I take it one quarter of the distance, and then another quarter once it's all been taken to 250 miles, I lose half my grass, and then half of that, so I am left with one quarter of my supply at 500 miles, and one eighth at 750 miles, or 1250 at 750 miles, and 625 at 1000 miles (loss of 50% per iteration, 4 iterations).

If I take it one tenth of the distance at a time, I lose 20% per iteration, and there are ten iterations; 8000, 6400, 5120, 4096, 3276, 2620, 2096, 1670, 1336 at 900 miles; and then 1236 at 1000 miles.

It looks as though the smaller the distance at one go, the more you can get there. There must be some kind of limiting problem; the limit as distance per iteration approaches zero of the amount that reaches the end after all the iterations.

*Busts out the calculator program*

Let's say the distance per iteration were one fiftieth of the total distance, or 20 miles per. The loss would be 40/1000 or 4% per iteration for fifty iterations (multiply by 96% 50 times).

Ten iterations 6648; 20 iterations; 4420; 30, 2939...it looks like we're losing a third per 10 iterations approximately. 40, 1953; 50, 1298. The smaller each iteration is, the more gets there in the end, looks like. I wonder what the limit is, and how would I find it? It would be some sort of calculus problem.