moonbeam wrote:Soooooooooooooooooo close ....... you're prob on the right track - I think I know where you went wrong. Explain your answer and I'll show you where you went wrong (plus others may wanna see the solution too

).
I was in a bit of a hurry, so didn't want to post the working if I'd managed to confuse myself thoroughly. I had a feeling I might have gone slightly wrong at one stage.
My first attempt ended exactly like Nightfall's answer. As I knew that was wrong, I left it (places to go...).
I came back and theorised that, the further the donkey travels before its first "drop-off point", the more carrots it eats on the journey out and the journey back. So, I took it to the opposite extreme, and started working out what would happen if the donkey went only one unit of distance before dropping off a batch of carrots.
It came to this: To go one unit of distance, with the whole batch of carrots, takes 5 carrots. Pick up 1000, eat one, drop off 998, keeping one more to eat on the way back. Pick up 1000, drop off 998, go back for the last 1000 carrots. Travel one unit, leaving you one unit away from the start, with 2995 carrot left.
You keep repeating this process until you have gone 200 units of distance. At this point, 1000 carrots have been consumed (200 x 5). For the next stage, each unit of distance only costs 3 carrots, because you now only need to make two trips instead of three (you have 2000 carrots left).
Ahh. I did indeed go wrong at the point I thought I did! Something didn't feel quite right about the uneven division at this stage. The catch is that the donkey eats a carrot
before moving, isn't it?
So, what
should happen is to divide 999 by 3 (the number needed to carry two maximum loads of carrots), giving 333. At 533 units of distance along, you have 1001 carrots. You feed the donkey one carrot, then move the 1000 one unit. You then use up 467 more carrots, getting to the end of your journey with 53
3 carrots left.
Believe it or not - Stoner has the correct answer

.
The deuce, you say!