
Re: Interesting trivia  anybody has a different answer than I keep getting  zero
Posted:
Nov 27, 2012 4:35 PM


"Stone Bacchus" <x@x.com> wrote in message news:03q9b8d8de65oj4cninvsj3q42v3ira9hf@4ax.com... > 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.

