Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2006-11-18 00:35:37

Beanny
Member
Registered: 2006-11-18
Posts: 3

Please Help. Much Appreciated.

Hi I got this question in an interview, and I'm just dying to know how to do it, because I only had 10 mins to complete it!! My best answer is 533, but the most is 567. Please show me how.

Problem:
A man has a camel with a carrying capacity of up to 1000 bananas. He has to cross 1000km of desert to get to the market to sell his bananas. He starts with 3000 bananas. For each km the camel travels, it requires 1 banana. What is the most he can sell?

My working:
He carries 1000 bananas for 200km and deposits 600. He goes back and does the same. He goes back again and does the same again, but now he has no bananas left at the start, and 2000 at 200km. Next he carries 1000 for 333km and deposits 333 at 533 km. Does this again until 0 at 200km and 1000 at 533km. Then he carries 1000 bananas for 467km to market, which leaves 533 bananas to sell.

Please try to improve on this and get 567 if you can.
Thanks in advance.

Offline

#2 2006-11-18 04:17:28

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Please Help. Much Appreciated.

I get exactly the same answer as you. It's certainly if the best answer you can get if you make the assumptions that:

• The delivery must be made in stages, where each stage involves going from back and forth from point A to point B, until all the carrots have been moved from A to B.

• After each stage, the total number of carrots left must be a multiple of 1000, for efficient delivery.

If we believe these things, then at the first stage he can either go the the 200km point with 2000 carrots left, or the 400km point with 1000 carrots left.

If he goes to the 400km point, then all he can do afterwards is go to the finish and be left with 400 carrots at the end.

If he goes to the 200km point, then he can either go straight to the finish (which is silly because then he'd only have 200 carrots and leave 1000 of them behind) or he can do another intermediate stage where he's left with 1000 carrots at the end of it.

It turns out that he can either get to the 533km point with 1001 carrots, or the 534km point with 998 carrots.

From the 534km point, he can take all 998 carrots with him, he would use 466 of them on the journey, leaving him with 532 to sell.
From the 533km point, he needs to leave one of the carrots in the desert, but he can still carry 1000 to the finish, using up 467 of them, and so leaving him with 533 to sell.

This seems to be the best answer, and it's the same one that you have.

So if we're going to beat that, we need to somehow find a way to move the carrots that doesn't follow the assumptions above. I can't see any way to do that though.


Why did the vector cross the road?
It wanted to be normal.

Offline

#3 2006-11-18 06:42:19

Beanny
Member
Registered: 2006-11-18
Posts: 3

Re: Please Help. Much Appreciated.

Well I would definitely agree with assumption one, because the furthest he could go in one journey is to the market. However, he would have left back 2000 bananas, and he would arrive with nothing.

It would also seem logical to assume the second, though I am not completely sure. Has anyone had any success with this?

Offline

#4 2006-11-18 07:59:10

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Please Help. Much Appreciated.

The first assumption assumes more than that the delivery must be made in stages though. If that wasn't true then the problem would be trivial.

It also assumes that all bananas must be picked up at A and dropped at B until every uneaten banana is at B, with no dropping anywhere else (except the camel's tongue).

There might be a weird and clever way of improving on the solution by dropping some of your surplus (as in, not needed for food) bananas somewhere, then dropping the rest somewhere else.


Why did the vector cross the road?
It wanted to be normal.

Offline

#5 2006-11-18 10:24:37

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Please Help. Much Appreciated.

This is a very interesting problem.  Got me thinking...


igloo myrtilles fourmis

Offline

#6 2006-11-18 18:50:52

pi man
Member
Registered: 2006-07-06
Posts: 251

Re: Please Help. Much Appreciated.

Like you Beanny, I came up with the answer of 533 but also suspected there must be a more efficient way.   But now after putting a lot of thought and effort into it, I'm 90% convinced that 533 is the best you can do.  This is a great problem - certainly one you can spend more than 10 minutes on!  (Which also leads me to believe 533 is the correct answer.  If 567 is the correct answer, I suspect it's a pretty complicated solution which would require more than 10 minutes of effort.)   Were you interviewing for a job?   If so, what type of job?   

Some of my failed efforts to come up with 567: 

Figuring the best way to maximize the use of the camel was to make sure he was carrying as close to 1000 bananas as possible, I tried making a whole lot of 1 km trips.   Below is the start of that journey.   The camel eats a banana at the end of each 1 km trip.   The "*" marks where that segment of the journey ended and the "<" and ">" indicate the direction of travel for that segment.

   3000
> 2000    999* 
< 1999*   999   
>   999   1998*
>   999     998     999*
<   999     997*   999 
<   998*   997     999   
>            1994*   999 
>      0      994      1998*
>             994       998     999*

Working this all the way through still yields 533 banananas at the end.   

I also tried making the first trip 200km but in addition to eating 1 banana every km, I dropped off 4 bananas at each 1 km marker for "snacks" on future trips.  So after returning from the 200 km trip (each way), I still had 2000 bananas at the start and 3 bananas littering each 1km marker along the way up to 200km.   That enabled the camel to carry 1000 bananas all the way to the 200 km marker on the next 2 trips.  So I ended up having 2000 bananas at the 200 km marker - the same thing you have in your original answer.

Offline

#7 2006-11-18 20:22:18

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Please Help. Much Appreciated.

Perhaps approaching the problem backwards would help? Start with 567 at the destination and work backwards to get the lowest quantity at the source.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Offline

#8 2006-11-18 23:11:44

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: Please Help. Much Appreciated.

533+1/3 at most

I proved 533+1/3 is the most in my home forum. But I lack time to translate it into English currently.

Stangely enough, someone did say 567 could be an answer at that time.


X'(y-Xβ)=0

Offline

#9 2006-11-18 23:40:12

Beanny
Member
Registered: 2006-11-18
Posts: 3

Re: Please Help. Much Appreciated.

Thanks for all the input guys and gals!

I didn't realise that you could leave 'extra' bananas along the way for a future trip, but there's nothing in the wording of the question that forbids it. However, I did forget to mention that the bananas are integers, e.g. you cannot travel 333 1/3 km and deposit 333 1/3 bananas. Someone on another forum tried this, but still couldn't better 533.

As for the interview, I was trying to get a actuarial internship with Deloitte.

By the way, George Y, if you have proved that 533 is the maximum, could you please try to translate it or give me the URL to that page.

I have tried MathsIsFun's suggestion of working backwards, but there are still too many possibilities. Thanks again for all your contributions, but if there is anyone from the Deloitte recruitment dept out there, please share the answer!

Offline

Board footer

Powered by FluxBB