Is this question difficult, or easy?

I'd say "difficult" -- but not impossible.

Although I would not completely rule out "impossible."

And I wouldn't rule out easy either -- because the answer may be "0" or "3" -- and someone much brighter than I may be able inutituvely to see that.

I'm working on it.

Hi Frank, I did not know you visited the brainy peoples section. :wink:
I know the answer. It is the degree of difficulty I seek. I thank you for your comment. I think.

It appears the answer is zero. If not, the question is difficult. Neil

That's just it, the answers to riddles are usually not what they first appear to be so I'd say this one is fairly difficult, but easy for riddle/math buffs.

I think the camel can get several hundred bananas to market, but I haven't done the math yet. (Think small trips.)

It's easy (because of it's great similarities to other multiple-step-trip riddles).

I won't post the answer (I haven't actually done the calculation) but a hint is that it'll be similar to the South Pole expedition.

Is the grower riding the camel?

If not, how many bananas can the grower carry?

Would someone please answer!

These questions are weighing heavily on my mind.

I've PM'd Try with my answer for how many bananas can get to market by PM -- and it is wrong by a bit. I'm looking over my calculations again.

My instincts told me the answer would be about 1/3 of the bananas can get there -- and if the final answer comes in at around that point -- I will rate the puzzle as fairly easy even if I don't get the exact answer myself.
(Hope my reasoning on that makes sense.)

1/3 is pretty far off, Frank. Think about the clues Craven & I gave.

A simple zero seems very obvious. I'll check in to see if I'm wrong.

The answer is clearly not zero or three. I can get 250 of the bananas to market using this process.


1. I load up 1000 bananas and go 250 miles.
bananas at start = 2000
bananas eaten = 250
bananas at 250 mile mark = 750

2. I load 250 bananas (for eating and go back to the start point
bananas at start 2000
bananas eaten = 500
bananas at 250 mile mar = 500

3. I load 1000 more bananas and go 250.
bananas at start = 1000
bananas eaten = 750
bananas at 250 mile mark = 1250

4. I now load 1000 bananas And go the 750 miles to market
During this time my camel eats 750 bananas leaving me with
250 to sell. (Unfortunately this leaves me with no bananas to return with meaning I have to sell my camel. But this was not part of the problem.)

I believe I can do better. When I have time I will do the algebra.


Nice problem.

I can do 332. I'm trying for 333, 334, 335.

I think more than 335 is impossible -- but I am not yet sure of 333 - 335.

Much tougher than I thought. How about 400?

2 round trips to the 400 marker leaves 400 bananas there. On the third trip you fill up there and walk the remaining 600 miles. = 400 Correct?

ebrown may be on the right traack I think, but the steps should be one mile each(resulting in rotton bananas less than half way unless this is miilion mile per hour camel) 2 bananas get 998 bananas to mile marker one, plus the return trip empty, so 5 bananas gets 3000 bananas to the 1 mile marker (no need to return empty following the third trip) repeat 999 times using 4995 bananas, which is almost 2000 more bananas than you have. We can use some what larger steps and abandon some of the bannanas at some of the stops, but I suspect he still gets zero to market, even if he carries some of the bananas himself (assuming he will eat some of the bananas if he carries any)
Why not expend all this effort analzing rotating tethers that have million to one probability of getting us cheap acsess to space? Aagh... I'm still analyzing. he can only use one camel, but he can use a different camel each trip avoiding return trips. It might get a few rotten bananas to market, but I can't handlle leaving camels in the desert to starve. Neil

If you go to mile 400 and drop off 200 bananas you will still be left with 400 bananas to get you back to start.

You pick up 1000 -- go back to the 400 mile post and drop off 200 bananas -- and use the 400 remaining to get you back to start.

You pick up the last 1000 -- and go to the 400 mile post. You will have 600 left plus you pick up the 400 waiting.

You then travel the remaining 600 miles -- and end up with 400 bananas to market.

Pity camels don't have thumbs;
'hitching a ride' would save it a lot of walking, and hundreds of bananas!

This is now an optimization problem. I won't be happy untill we have an algebraic answer to prove that we have the "best" answer.

The expression for the number of Bananas we can move from the start to a checkpoint D miles away is...

BM = 1000T - D(2T-1) where BM is the number of bananas moved and T is the number of trips.

I thinkg this problem is going to turn into an optimization problem with at least 2 variables.

anyone care to continue...

P.S. Do we allow for the camel to eat a fractional banana to go a fraction of a mile?

Ok I started working this the other way round. The last leg of the journey I think I want to have exactly 1000 bananas left (so I don;t need to leave any bananas behind.

That being said I solved for the distance I can go to have a given number of bananas left.

D = (1000T - BM ) / (2T-1)

Assuming one stopping point the best you can do is set the number of trips to 3 (since you have 3000 bananas to start). This leaves D (the distance of the best stopping point) as 400 as Bill said above.

But you can do better by using 2 stops. used the equation below to find the distance I can go to have 2000 bananas left. This distance is 200 miles (still with 3 trips).

This now I do another step to see how far I can go to have 1000 bananas left. I used the eqation above to find I could go 333 more miles in 2 trips and have 1000 bananas left.

This leaves me with 1000 bananas and only 467 miles to go.

I will now end up at market with 533 bananas to sell.

I think thats the best there is.