I listen to a podcast called Math Factor. During the episode, they usually give you math puzzle to ponder. I like to figure it out before I go onto the next episode. However, this time I think I have it right, but I'm not sure. I'd like to see if anyone else can get a better answer. If so, I'll think about it some more. This problem was apparently part of a 10th grade honors math test to add insult to injury:

A man has 3000 bananas he'd like to sell at market. The market is 1000 miles away. He has a camel (probably a yellow one ) that can carry 1000 bananas. However, for every mile he travels, his camel eats 1 banana. He can set up depots of bananas along the route. What's the maximum amount of bananas he'll be able to sell at market if he plans accordingly?

My answer? If you'd like the details, just ask and I'll post another spoiler thread.

Spoiler (click to reveal)

533

If you get an answer higher than this, post it, but without the details! Or the details in a separate spoiler tag. I'd like to solve it myself, but I'd like to know if it's possible to get higher. Have fun!

Specious arguments are not proof of trollish intent.

There's nothing whatsoever to indicate that you're not supposed to optimize fully. Which requires traveling fractions of miles, and eating fractions of bananas....

"that's a smith and wesson, and you've had your six"

MisterCranky wrote:

There's nothing whatsoever to indicate that you're not supposed to optimize fully. Which requires traveling fractions of miles, and eating fractions of bananas....

There is nothing to indicate that you can split a banana up either. A lack of a rule doesn't always means that it is assumed acceptable. Come on, you're a gamer right?

I listen to a podcast called Math Factor. During the episode, they usually give you math puzzle to ponder. I like to figure it out before I go onto the next episode. However, this time I think I have it right, but I'm not sure. I'd like to see if anyone else can get a better answer. If so, I'll think about it some more. This problem was apparently part of a 10th grade honors math test to add insult to injury:

A man has 3000 bananas he'd like to sell at market. The market is 1000 miles away. He has a camel (probably a yellow one ) that can carry 1000 bananas. However, for every mile he travels, his camel eats 1 banana. He can set up depots of bananas along the route. What's the maximum amount of bananas he'll be able to sell at market if he plans accordingly?

My answer? If you'd like the details, just ask and I'll post another spoiler thread.

Spoiler (click to reveal)

533

If you get an answer higher than this, post it, but without the details! Or the details in a separate spoiler tag. I'd like to solve it myself, but I'd like to know if it's possible to get higher. Have fun!

I've seen this puzzle in the 'IMP' series of high school textbooks. Yes, your answer is best (unless you allow the possibility of fractions, and even then, you only gain a third; not like you're selling a third of a banana, anyway).

Specious arguments are not proof of trollish intent.

Yeah, I'm a gamer, and in the absence of any rule that says you can't have fractions, I'm taking them. Any true banana merchant is also a gamer, and is going to go for the extra third of a banana profit EVERY TIME!

Thanks everyone for your input. Now I can listen to the next episode guilt free! I'm interested in what he has to say about this problem because in the description of the problem, he said it had something to do with Euler's number (e). Hmmm... I certainly didn't use e to solve the problem but we'll see. Thanks again! I need to get back to figuring out how many bananas a unicorn eats per mile.

I don't understand how this can work. For each round trip, the camel will eat 2000 bananas, no matter how you fractionally divide the movements. And 3 trips to market are necessary....

Well, since you didn't state any restrictions on where the banana depots could be, and assuming you have to feed the camel every mile, here's what I'd do:

Spoiler (click to reveal)

Load 1000 bananas on the camel and go 999 miles (1 uneaten banana left). Here I stop at my one and only banana depot and load 999 bananas back on the camel (back to 1000) and travel the last mile to market, arriving with 999 bananas.

Am I missing something? Nevermind. I was assuming that the caches could be setup ahead of time. That defeats the whole point of the puzzle, obviously.

Last edited Mon Jan 29, 2007 3:57 pm (Total Number of Edits: 1)

I don't understand how this can work. For each round trip, the camel will eat 2000 bananas, no matter how you fractionally divide the movements. And 3 trips to market are necessary....

Someone enlighten me.

Sure: Little Hint:

Spoiler (click to reveal)

The key is with the depots. You travel for a while, leave some bananas, then return to get some more. The difficult part is reasoning where those depots should be.

Detailed description of solution:

Spoiler (click to reveal)

Load up your camel with 1000 bananas and head out 200 miles. You'll have 800 bananas left. Drop off 600 and save 200 for the return trip. Repeat. For the third trip you don't need to return so you'll have 800 bananas by the time you get to the 200 mile mark. So that's 600 + 600 + 800 = 2000 bananas at the 200 mile mark.

Next, load up 1000 bananas again. Head out 333 miles to the 533 mile mark. Drop off the bananas (334 of 'em) and return. Pick up the remaining 1000 bananas and head off. When you get to the 533 mile mark you'll have 334 bananas (those left on the first trip) + 667 (those you still have from the second trip) = 1001. See why some people say that I could get fractionally better? Anyways, it's time for the final trip. Load up 1000 bananas, have one for a snack and travel the final 467 miles leaving you with 533 bananas.

I've read this and cannot think of a better answer, but what is the method of calculation -- other than trial and error? Is there a general equation that gives a solution for x bananas, carried y miles, consumed at rate of z per mile?

I've read this and cannot think of a better answer, but what is the method of calculation -- other than trial and error? Is there a general equation that gives a solution for x bananas, carried y miles, consumed at rate of z per mile?

C = carrying capacity of camel B = total banana quantity D = distance of journey R = consumption rate in bananas/mile

Well, you can split the problem down into a set of distances. Since we haven't used enough letters yet, let's label the waypoints as W0 (end point) to WN (starting point). Distance between W0 and W1 will be travelled once, W1...W2 will be travelled three times, W2..W3 will be travelled 5 times, etc. For maximum efficiency, at the waypoint W1, there should be exactly C bananas before the final stretch; before that, at W2 there should be exactly 2C bananas, and at WN-1 there should be exactly (N-1)C bananas.

So, B/C (rounded up) gives us the number of waypoints N we should use. Waypoint WN-1 will be at the distance where we have exactly (N-1)C bananas left, i.e. distance x where B-(2N-1)Rx = (N-1)C; x = (N-1)C/(B-(2N-1)R)

I've read this and cannot think of a better answer, but what is the method of calculation -- other than trial and error? Is there a general equation that gives a solution for x bananas, carried y miles, consumed at rate of z per mile?

Thanks. I enjoyed this problem.

That was actually the question in the following podcast. Though, I think it assumed that the number of bananas would be the same as the number of miles. And that the camel always ate 1 banana/mile. Okay, so it was a similar question.

Basically, it gets to be a lot of return trips. I'd like to try out the MP3 insert here so I'll let the math professor explain it:

I think it relates to 'e' because it's effectively the Compound Interest calculation, in reverse.

I don't see any relation with this puzzle to 'e'.

As for solutions, here's one I like:

Spoiler (click to reveal)

Initially, the bananas can be transfered in no fewer than 3 piles. Now, if you make a 'depot' each mile, you find that each mile costs you 5 bananas: two back-and-forth trips for the first 2 piles, and a one-way trip for the last pile. (Example: take 1000 bananas, travel one mile (using up one), drop off 998, return (using up one), repeat with another 1000 bananas (using two more), and take the last set to the first mile marker (using a fifth.)

So long as you have a minimum of 3 piles (i.e., over 2000 bananas), you will lose 5 bananas per mile. Thus, after 200 miles you will have used 1000 bananas exactly.

At this point, you need only 2 piles; thus, each mile costs only 3 bananas: 2 for the back-and-forth of Pile 1, and 1 for the one-way for Pile 2. This 3-per-mile cost continues until you are down to 1000 or fewer bananas. Here, you travel 333 miles at a cost of 999 bananas (leaving 467 miles and 1001 bananas). *** Note that, if fractions are allowed, you can stop at 333 1/3 miles and 1000 bananas even; this leaves 466 2/3 miles to go. ***

The rest of the trip, with only one pile (after you treat yourself to the 'extra' banana), costs 1 banana per mile; you'll have 1000-467 = 533 bananas left.

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mrraow wrote:

Well, you can split the problem down into a set of distances. Since we haven't used enough letters yet, let's label the waypoints as W0 (end point) to WN (starting point). Distance between W0 and W1 will be travelled once, W1...W2 will be travelled three times, W2..W3 will be travelled 5 times, etc. For maximum efficiency, at the waypoint W1, there should be exactly C bananas before the final stretch; before that, at W2 there should be exactly 2C bananas, and at WN-1 there should be exactly (N-1)C bananas.

This breaks down when the number of bananas gets too big, as it can't deliver more than a full load of bananas (as you only travel the last stage once). If you started with, say, a million bananas (a thousand loads - overkill is good in examples) you can do better than that if you are allowed to turn round at the finish and go back for more bananas.

I did (before seeing Stephen's post) work out where the single trip strategy breaks down as the perfect strategy (it's basically how many reciprocal odd numbers you need to add to get to one). I didn't attempt working out the new strategy.

This breaks down when the number of bananas gets too big, as it can't deliver more than a full load of bananas (as you only travel the last stage once).

True. I did work out a rough revised strategy for this - you basically turn the problem around, and calculate the distances backwards from the finish line... but they get very big, fast. That's how I calculated the (roughly) 16,000 bananas required to get 3,000 to the finish line. Once you go above 1,000 bananas at the finish, you can set aside 1,000 bananas to get your camel home again, which was troubling me in the original problem.

Aside: the sum if 1/(2x-1) diverges, so you should be able to get any number of bananas to the finish line... at a cost.