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You can get an ice cream from me if you get this right...

A farmer has 3000 melon to deliver to the market which is 1000m away from where he is now. His donkey can only carry 1000 melons at a time and will eat 1 melon per meter. You are not allowed to drop the melon on the road and pick up later. What is the maximum number of melons the farmer can deliver to the market? The min is 500.

This is supposed to be a pre-calculus question. I am appalled when I couldn't solve it at all after so many years of formal Math education. Just walking back and forth to carry the other 2000s, the farmer will never reach the marketplace.. since he is not allowed to drop some melons on the roads for later pick-up.... Because of this, I can't even write a deterministic program to calculate the maximal subarray or as a series... (some sequence of numbers)

Anyone has any idea how to get this started? Is this even computable..?

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closed as off-topic by Jonas Meyer, Lord_Farin, Did, Fundamental, graydad Jan 18 at 19:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Jonas Meyer, Lord_Farin, Did, graydad
If this question can be reworded to fit the rules in the help center, please edit the question.

The min is actually zero :) –  process91 Jun 12 '12 at 2:32
Are these numbers correct? –  user17794 Jun 12 '12 at 2:34
@TimDuff They can't be, right? I'm seeing that the donkey can take 1000 melons, eat all of them by the time it gets to the market, and then presumably die of hunger if it tries to get back. Maybe the maximum is 3000 - the farmer just takes as many as he can carry over and over, and the donkey is a red herring. –  process91 Jun 12 '12 at 2:35
@JohnWong The answer to a similar riddle is here: edurite.com/kbase/answer-to-4th-grade-math-problem. This one does not stipulate that you cannot drop the melons, however, and clearly the same strategy will not apply to your question. There is a loose upper bound of 3000 - miles walked, so 2000 melons, however my contention is that the upper bound equals the lower bound, which is zero. –  process91 Jun 12 '12 at 2:50
Without the "not allowed to drop the melon" provision, this is usually posed as "cross the desert" or "the jeep problem", and many discussions can be found by searching those terms. With that provision, it's hard to see how you can get anywhere. –  Gerry Myerson Jun 12 '12 at 3:17

1 Answer 1

I say the answer is that it doesn't matter what the farmer and donkey do, they will not be able to get any melons to market. Assuming there's no "lateral thinking" trickery here, which there probably is (the farmer had a car the whole time, we just didn't tell you!) then we can assume that at any point in time the farmer and donkey can chose to (a) not move (b) move 1m toward the market or (c) move back toward the place with the melons. The stipulation on (b) and (c) is that the donkey must have 1 melon before making the movement. They can also (d) give melons to the donkey subject to two stipulations: the number of melons on the donkey cannot exceed 1000 and they may only add melons to the donkey when they are at the place with the melons. There's a final possible action (e) sell melons at the market, but we will never get to that point.

So at the start our options are (a) or (d). In general, (a) will not change anything, so let's just omit that option from consideration. Then our only choice is to perform (d). If we add only one melon then our next options are (b) and (d). If we choose (d), then the donkey will move 1m toward the market and then be stuck because it doesn't have any more melons, so let's choose (b) again. Now the donkey has two melons, so we could choose (d) (d), and then get stuck, or choose (d) (c) but this only has the total effect of reducing our supply of melons by 2. Our last option is to choose (b) again.

This argument repeats itself until we get to 1000 melons, but then we choose (d) until we get to the market where we are, once again, stuck.

I think there is a high chance that the premise of the question is incorrect. Either wrong numbers or wrong conditions. In particular, if you were able to drop melons along the way, the problem would be much more interesting. For instance, here's one way to get one melon to market that way: take three melons, move 1m toward market, drop one melon, move back, take 1000 melons, move 1m, pick up the melon you dropped, go to market. The question of the maximum number of melons is much more interesting in that case. As Gerry Myerson pointed out, in that case the problem is known as "crossing the desert" or "the jeep problem".

I ask again - where did this riddle come from? A book? If so, what is the name/author? In any event, the claim at the end of the problem that the min is 500 makes no sense - even if you omit the restriction of not being able to drop melons on the road you can still easily make it so all the melons are eaten by the donkey by taking 1000 melons and then walking halfway to the market and then halfway back, so wherever you got that information from it is incorrect.

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