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#12 |
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Mar 2004
3×127 Posts |
Ath the beginning there are 10000 pounds. That means that the donkey has to start 10 times and of yourse return 9 times. Hence he needs 19 pounds per mile. After 1000/19 the amount reached 9000 pounds. From here he just needs to start 9 times and return 8 times.
The plan is also take 1000 pounds, walk 1000/19 miles (53.63) drop 1000-(2000/19) pounds and return. The last time he does not need to return. So he has 9000 pounds at milepost 1000/19. Now he can continue with 1000/17....and so on. At the end, the donkey has 2000 pound, goes home and unloads everything minus that amount he used on the last part. Returns and takes the last part. Therefore, he optimal solution is 3 * ( 1000/19 + 1000/17 + 1000/15 + 1000/13 + 1000/11 + 1000/9 + 1000/7 + 1000/5 ) - 1000 8 depots on the way are necessary. Last fiddled with by biwema on 2006-12-10 at 13:07 Reason: add information about depots |
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#13 |
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Jan 2005
Transdniestr
7678 Posts |
Yep, that's my approach
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#14 |
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Aug 2002
Portland, OR USA
2×137 Posts |
I also get 1399.76659 lbs. of sugar.
There is one way to improve on this solution, and that is to do the work yourself:
Code:
The donkey eats 1,000 lbs. The donkey carries 1,000 lbs. You carry 8,000 lbs. You deliver 9,000 lbs of sugar. * This distance can be anything; from leading the donkey home with 1,000 lbs and then doing the rest yourself; to loading the donkey on one side and unloading it on the other. Bruce
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#15 | |
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Bronze Medalist
Jan 2004
Mumbai,India
22·33·19 Posts |
Quote:
Well to settle the problem with a round figure he can be allowed to eat 0.76659 lbs to the last grain. Not more or he will end up with Type 2 diabetes in the long run. Mally
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