Humphrey Thou Sand
A camel has to transport as many as possible of an available 3000 bananas. For every comlete mile it walks, it must consume 1 banana (regardless of its load). It can only carry up to 1000 bananas at a time. What is the maximum number of bananas it can transport to a place a 1000 miles away?
Labels: mathschallenge, thinktank
posted by Chris at
11:06 AM
17 Comments:
I see a deficit, but it doesn't seem right. There's a trick somewhere.
All bananas!
he only walks in .5 mile steps so its never a complete mile and he doesn't has to eat =)
I think its more to zero bananas because he can only walk a maximum of 500 miles when he has to return, and eat the other 500 bananas on its way back.
1st trip: Take 1000 bananas, leave 500 of them 250 miles down the track, consume the other 500 getting there and back.
2nd trip: Take 1000 bananas, top up at the 250 mark, leave 500 at the 500 mile mark, consume the rest getting back.
3rd trip: Take 1000 bananas, top up at 500 miles, get to the destination with 500 left over.
true...
you probably would end up with more by more iterations
Hi all. I seem have woken a few people up:) I think that the "official" version of this problem says that the camel can only drop the bananas on 1 mile boundaries. I won't insist on that. Looks like at least 500 bananas is possible.
I'll also allow that the camel doesn't need to be carrying a banana for the last mile, as long as he arrives somewhere where there is a banana to eat and then it must eat 1. Also add, the camel isn't allowed to eat more than one banana at a time, because that would give the trivial result of 1000 bananas.
PS I don't know the answer, te-he.
First take 3000 bananas 200 miles. That'll be 5 legs of 200 mile. So get 2000 bananas with 800 miles to go. Now take 2000 bananas 333.333... miles (3 legs). Then end up with 1000 bananas and 466.666... miles to go. Take the 1000 to the end, end up with 533.333... bananas;) I expect only a small correction if stick to whole miles (and bananas). Can anyone beat that?
I've got a horrible idea, that that's the best that can be done. If so, that was a lucky first try.
Hi Chris,
Your first legs would be 4 lots of 400 miles (200 each way) then one of 200, i.e. 1800 miles. So you'd only have 1200 bananas left for the remaining 800 miles.
WoO. You nearly fooled me then. My first stage was 5 lots of 200 miles. I only needed 3 outward and 2 returns. 1000 miles and 1000 bananas used.
All of them would still be in his belly! It never said the bananas had to get there whole!
I thought I'd beaten 533.333...., but I'd made the mistake of saying 5*333.333... = 1333.333... rather than 1666.666... I'm almost certain that 533.333... is the best that can be done.
Oops! You're right about the first legs, Chris. 600 bananas deposited from each of the first 2 round trips to the 200 mile mark, then 800 for the third, makes 2000. Sorry!
the maximum is 1000.. cuz it can only carry up to 1000 bananas.. ^^
Hi WoO. I had actually posted a reply to the effect that I was "very embarrassed" at having goofed up. But I deleted it when I realised that I hadn't.
I am surprised that I got the correct answer on my first go. I thought it was going to be diffcult. I got the puzzle from:
http://mathforum.org/dr.math/faq/faq.camel.html
I thought e=2.71828... might enter it.
Ragknot pointed out that the solution was easy to Google for in his "Even upside down?" blog. Greetz.
Hi WoO. I forgot to say that I thought your answer was very nice and I thought it was the best. I only tried my way on a hunch plus knowing that to move the first 3000 bananas a distance M would result in 3000-5M remaining bananas with a distance 1000-M to go. I then realised that if I made 5M=1000 that I would then have exactly 2000 bananas and would only need 3 legs to move them to the next stage. So I tried repeating the strategy with 3M=1000. blah, blah. I was delighted when I found that I'd got more than 500 bananas across. Greetz.
bananas.
If the camel walks one mile with 1000 banabas, he eats one and drops the other 998, saving one for the way back. He then comes back and picks up another 1000, again eating two including the return trip. He continues this pattern, one mile at a time, eating five per mile, until he only has two thousand left (200 miles) then, he eats only three, until he has 1000 left (533 miles) after which he eats one. He then eats 477 bananas, leaving 533.
That's as good as it gets. Yes, I know you already posted this, Chris, but I had a different thought that ended up with the same answer.
Hi Anonymous. I'm pleased that you noticed that there is another equivalent solution and that you considered the problem worth your time.
In fact I was already aware of your result. It's partly why I was sure that I'd accidently got the right strategy.
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