That is an excellent analysis. I will try it with some more iterations on that solution and see whether I can eek out some more GB, but that is one fine analysis that gets me thinking on those lines. appreciated.
On Tue, 27 Nov 2012 21:35:34 -0000, "Mike Terry" <firstname.lastname@example.org> wrote:
>"Stone Bacchus" <email@example.com> wrote in message >news:firstname.lastname@example.org... >> My daughter and I were solving a math trivia and I could not come up >> with any answer other than zero. Would be interesting to see if >> somebody has a different opinion. The problem follows: >> >> You are at the start of a 1000 mile road with 3000 gummybears and a >> donkey. At the end of the road is a supermarket. You want to find >> the greatest number of gummy bears you can sell. Unfortunately, your >> donkey has a disease and can only carry 1000 gummybears at 1 time. >> Also, the donkey must eat 1 gummybear per mile. >> >> - You can drop off gummybears anywhere on the road >> - You can't carry gummybears while walking >> - No loopholes >> >> Again, this was a math trivia question and I could not ask anybody for >> clarification about what some the caveats meant or what the "no >> loopholes" meant, therefore I got zero. >> >> (If I were to guess about the "no loopholes", I would think they meant >> no "carrying the donkey" like I suggested to my daughter :) ) >> >> Thanks for your time. >> >> s >. >. >. >. >. >. >spoiler >. >. >. >. >. >. >. >. >. >. >. >. >spoiler >. >. >. >. >. >. >. >. >. >. >. >. >spoiler >. >. >. >. >. >. > >Here is my first thought, which is based on dividing the 1000 mile journey >into stretches that require different numbers of trips to transport the >entire 3000 gummybears. > >With the donkey only able to carry 1/3 of the total, there would be at least >5 trips required for the first stretch, and assuming we arrange this to get >down to 2000 gummmybears, the next stretch would require only 3 trips, >getting down to 1000 gummybears, and the final stretch is just one journey >the remainder of the distance to the market. So the max distance for the >first stretch would be 1000/5 = 200 miles, the second stretch would be >1000/3 = 333 1/3 miles, and the the remainder for the last stretch. > >So, putting this together into a plan: > >Starting from Base1... >1. Take 1000 gummybears 200 miles [to "Base2"] >2. Drop off 600 gummybears >3. Travel back to Base1 >4. Take 1000 gummybears 200 miles [to Base2] >5. Drop off 600 gummybears [Base2 now has 1200 gb] >6. Travel back to Base1 >7. Take 1000 gummybears 200 miles [to Base2] >8. Load up 200 more gummybears [Base2 now has 1000 gb] >9. Travel 333 1/3 miles [to "Base3"] >10. Drop off 333 1/3 gummybears [Base3 now has 333 1/3gb] >11. Travel back to Base2 >12. Load up 1000 gummybears [Base2 now empty] >13. Travel 333 1/3 miles to Base3 >14. Load up 333 1/3 gummybears [Base3 now empty] >15. Travel remaining distance to market. > >The remaining distance to market is 466 2/3 miles, so you'll arrive with 533 >1/3 gummybears to sell. (Maybe a bit less if we can't divide up >gummybears!) > >I don't know if this is best, as I haven't tried any alternatives. I have >tried to minimise total travel distance, but maybe I've not done this in the >best way - anyway, it's a starting point to compare with what others come up >with... > > >Regards, >Mike. > >