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Hello everyone,

I amnew to this forum. I saw this puzzle on internet. I liked it.

The distance between the towns A and B is 1000 miles. There is 3000 apples in A, and the apples have to be delivered to B. The available car can take 1000 apples at most. The car driver has developed an addiction to apples: when he has apples aboard he eats 1 apple with each mile made. Figure out the strategy that yields the largest amount of apples to be delivered to B.

Generalize the strategy for an arbitrary amount of apples.

The site did not give a solution. I have one but not sure if it is the best one. I will post my solution soon.

Imran

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Will he eat cooking apples too ?

I would say 1000 apples are eating every trip, so unless there is a short cut or he can use a trailer - answer is always ZERO

If they are all identical in which they case they are pairs/pears then = 3.000

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Will he eat cooking apples too ?

I would say 1000 apples are eating every trip, so unless there is a short cut or he can use a trailer - answer is always ZERO

If they are all identical in which they case they are pairs/pears then = 3.000

No there are no pairs/pears. All apples are single. There is a solution to this so dont give up just yet.

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Hello everyone,

I amnew to this forum. I saw this puzzle on internet. I liked it.

The distance between the towns A and B is 1000 miles. There is 3000 apples in A, and the apples have to be delivered to B. The available car can take 1000 apples at most. The car driver has developed an addiction to apples: when he has apples aboard he eats 1 apple with each mile made. Figure out the strategy that yields the largest amount of apples to be delivered to B.

Generalize the strategy for an arbitrary amount of apples.

The site did not give a solution. I have one but not sure if it is the best one. I will post my solution soon.

Imran

he would manage to transport 800 apples from A to B.

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Well I can get him there with 1000 apples and I think that's the best he can do.

Assuming that he can leave them at some point in between the towns, then he should load up with 1000 apples and go half way. He then dumps his 500 apples left and comes back to town A. He loads up with 1000 more apples and goes half way, again. By now he's down to 500 apples on board, so he picks up the 500 he left from before and gets to town B with 500 apples left. Now 1500 apples are gone from the original 3000. He can do this whole procedure 1 more time to bring the total at town B to 1000 apples.

I think this is the best he can do since no matter what he's going to have to travel the 1000 miles from A to B at least twice. This means that 2000 apples will be consumed, and 1000 is the most that can be saved. :D

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I need to keep thinking about this one, but so far I am centering on the fact he only eats an apple per mile if the apple is on board

Thus,

it'd take a while, but he could keep throwing the apple ahead of himself, pick it up and throw it again, then, the apples would never actually be on board... - I know, I need to keep thinking, but I'll come up with something reasonable soon ...

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Well I can get him there with 1000 apples and I think that's the best he can do.

Assuming that he can leave them at some point in between the towns, then he should load up with 1000 apples and go half way. He then dumps his 500 apples left and comes back to town A. He loads up with 1000 more apples and goes half way, again. By now he's down to 500 apples on board, so he picks up the 500 he left from before and gets to town B with 500 apples left. Now 1500 apples are gone from the original 3000. He can do this whole procedure 1 more time to bring the total at town B to 1000 apples.

I think this is the best he can do since no matter what he's going to have to travel the 1000 miles from A to B at least twice. This means that 2000 apples will be consumed, and 1000 is the most that can be saved. :D

I think your second part has some flaw. Could you please explain it.

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Assuming that he can leave them at some point in between the towns, then he should load up with 1000 apples and go half way. He then dumps his 500 apples left and comes back to town A. He loads up with 1000 more apples and goes half way, again. By now he's down to 500 apples on board, so he picks up the 500 he left from before and gets to town B with 500 apples left. Now 1500 apples are gone from the original 3000. He can do this whole procedure 1 more time to bring the total at town B to 1000 apples.

I think this is the best he can do since no matter what he's going to have to travel the 1000 miles from A to B at least twice. This means that 2000 apples will be consumed, and 1000 is the most that can be saved. :D

Well, it seems to me there is some flaw there ... It is 1000 apples left from the original 3000 (NOT 1500) -- because the driver already made two rounds (each round 1000).

Edited by brhan
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Well I can get him there with 1000 apples and I think that's the best he can do.

Assuming that he can leave them at some point in between the towns, then he should load up with 1000 apples and go half way. He then dumps his 500 apples left and comes back to town A. He loads up with 1000 more apples and goes half way, again. By now he's down to 500 apples on board, so he picks up the 500 he left from before and gets to town B with 500 apples left. Now 1500 apples are gone from the original 3000. He can do this whole procedure 1 more time to bring the total at town B to 1000 apples.

I think this is the best he can do since no matter what he's going to have to travel the 1000 miles from A to B at least twice. This means that 2000 apples will be consumed, and 1000 is the most that can be saved. :D

According to your plan, he picks up 1,000 and goes halfway, then goes back and picks up another 1,000 and goes halfway, which means he's taken 2,000 from the original 3,000, not the 1,500 that you suggest. If he went back again, as there are only 1,000 left he would eat them all on the way there, so this solution only yields 500 apples.

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I think your second part has some flaw. Could you please explain it.

After he drops 500 apples in the middle of nowhere, he comes back to town A. He picks up 1000 more apples. When he gets to the halfway point again, he has eaten 500 apples on board but still has 500 left. This means he can pick up the 500 apples that he had dumped from before to bring his total to 1000 in the car. There are only 500 more miles to go to town B, so he gets there with 500 apples.

After typing it all out that time I do realize my flaw. There are 2000 apples gone from the original 3000 instead of 1500 like I thought before.

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Care to explain how it's 833?

ya, here is what I did.


Distance (from A) Original Apples Apples Eaten Remaining
---------------------------------------------------------------------------------
0 3000 0 3000
334 1200 1998
833 998 1000
1000 167 833

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Hello everyone,

I amnew to this forum. I saw this puzzle on internet. I liked it.

The distance between the towns A and B is 1000 miles. There is 3000 apples in A, and the apples have to be delivered to B. The available car can take 1000 apples at most. The car driver has developed an addiction to apples: when he has apples aboard he eats 1 apple with each mile made. Figure out the strategy that yields the largest amount of apples to be delivered to B.

Generalize the strategy for an arbitrary amount of apples.

The site did not give a solution. I have one but not sure if it is the best one. I will post my solution soon.

Imran

Theoretically, isn't it possible to carry all 3000 apples to town B. It's not practical, but it's possible.

What if he goes .9 miles with 1000 apples, drops them, then comes back to town A. Then he goes back and picks them up. He then drives another .9 miles and drops them. He keeps doing this until all 3000 are there.

Edited by Enlightened
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Theoretically, isn't it possible to carry all 3000 apples to town B. It's not practical, but it's possible.

What if he goes .9 miles with 1000 apples, drops them, then comes back to town A. Then he goes back and picks them up. He then drives another .9 miles and drops them. He keeps doing this until all 3000 are there.

I'm not sure if it's a legit answer, but either way, after his first three trips, as soon as he drives the next tenth, he'll eat an apple

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I'm not sure if it's a legit answer, but either way, after his first three trips, as soon as he drives the next tenth, he'll eat an apple

Yeah I thought about that, too. I wasn't sure how that would work.

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Theoretically, isn't it possible to carry all 3000 apples to town B. It's not practical, but it's possible.

What if he goes .9 miles with 1000 apples, drops them, then comes back to town A. Then he goes back and picks them up. He then drives another .9 miles and drops them. He keeps doing this until all 3000 are there.

probably the driver would eat 90% of an apple when he travels 0.9 miles ...

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This is the how the math works out ...

It's all about how many apples you "lose" per mile. This is a function of total apples to move divided by max carrying capacity. In this case, at the beginning, we are trying to move 3000 apples, but can only carry 1000, for a total number of "trips" of 3000/1000 = 3. This remains true until there are only 2000 apples left. Thus, the generalization works like the following:

Loss rate of 3 per mile (1 apple per mile * number of trips) till 2000 left = (3000-2000)/3 = 333 1/3 miles

2 per mile till 1000 left = (2000-1000)/2 = 500 miles … now at mile #833 1/3

1 per mile till done = (1000 - 833 1/3) = 166 2/3 miles

This means you lose 166 2/3 of the last lot of 1000 apples, leaving you with 833 1/3 apples at your destination. (

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Theoretically, isn't it possible to carry all 3000 apples to town B. It's not practical, but it's possible.

What if he goes .9 miles with 1000 apples, drops them, then comes back to town A. Then he goes back and picks them up. He then drives another .9 miles and drops them. He keeps doing this until all 3000 are there.

As the question states that the driver eats 1 apple for every mile made, I think it's reasonable to assume it takes him the full mile to eat it, rather than eating it instantaneously after the full mile is completed. So this would result in lots of mostly eaten apples nearly 1 mile down the road, which would be promptly finished off as soon as he picks them back up again.

The general strategy here is to break the journey down into parts and then reduce the number of times the driver has to travel over each part. This is because each time the driver travels over a certain bit of road with apples in the car, he will eat them.

As long as there are more than 2,000 apples there needs to be at least 3 trips. So we need to figure out how far the driver needs to travel in 3 trips in order to lose 1,000 apples - 1,000/3 = 333 miles. That means he can get 2,001 apples as far as 333 miles (and is welcome to eat 1 apple for free as a bonus!). Carrying on with the same logic, he now only needs to make 2 trips, so to lose 1,000 apples he will go 1,000/2 = 500 miles. We now have 1,000 apples at the 833 mile mark. One final journey of 167 miles will leave him at the end with 1000-167 = 833 apples.

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This is the how the math works out ...

It's all about how many apples you "lose" per mile. This is a function of total apples to move divided by max carrying capacity. In this case, at the beginning, we are trying to move 3000 apples, but can only carry 1000, for a total number of "trips" of 3000/1000 = 3. This remains true until there are only 2000 apples left. Thus, the generalization works like the following:

Loss rate of 3 per mile (1 apple per mile * number of trips) till 2000 left = (3000-2000)/3 = 333 1/3 miles

2 per mile till 1000 left = (2000-1000)/2 = 500 miles … now at mile #833 1/3

1 per mile till done = (1000 - 833 1/3) = 166 2/3 miles

This means you lose 166 2/3 of the last lot of 1000 apples, leaving you with 833 1/3 apples at your destination. (

Oh, I forgot to add that while you do end up with only a few apples left (from what you started), you now probably weight about 600 pounds or so - literally! (I assume a starting weight of about 100 pounds, additional weight of 4 apples per pound, and then the slow loss of weight from sitting and driving, since you aren't walking these apples the 1000 miles!

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Well you people got it right.

Let the number of apples be X. it can be writen as X = 1000Y+Z where Y and Z are integers and 1<=Z<=1000.

let f=floor(Z/(Y+1))

So make Y+1 trips(1000 apples each) till f miles.

Now you get a new number of apples. Repeat the process unless you reach the destination.

Here is my general solution.

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This is the how the math works out ...

It's all about how many apples you "lose" per mile. This is a function of total apples to move divided by max carrying capacity. In this case, at the beginning, we are trying to move 3000 apples, but can only carry 1000, for a total number of "trips" of 3000/1000 = 3. This remains true until there are only 2000 apples left. Thus, the generalization works like the following:

Loss rate of 3 per mile (1 apple per mile * number of trips) till 2000 left = (3000-2000)/3 = 333 1/3 miles

2 per mile till 1000 left = (2000-1000)/2 = 500 miles … now at mile #833 1/3

1 per mile till done = (1000 - 833 1/3) = 166 2/3 miles

This means you lose 166 2/3 of the last lot of 1000 apples, leaving you with 833 1/3 apples at your destination. (

I'm having trouble applying your mathematical explanation to the problem :(

Could you please state what the driver did, in terms of apples/trips (if it's not too complicated) so i could understand the solution?

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I'm having trouble applying your mathematical explanation to the problem :(

Could you please state what the driver did, in terms of apples/trips (if it's not too complicated) so i could understand the solution?

probably you might find my table format on Post # 16 easier to understand.

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