This is probably one of the tougher puzzles, and the responses indicate the same. I got only one answer for the puzzle, and once again, from Tishyaa – Thank you Tishyaa!
1. For the simpler puzzle, One needs to to set up a depot at 200 miles, leave the truck with a full load, empty 1/3rd at the 200 mile post and come back. In the next trip, leave again with a full load, reach 200 miles. By this time, 1/3rd of the fuel tank is empty as the total capacity is for 600 miles. At the 200 miles, therefore, it refills the 1/3rd of the full load available and again becomes full. With this load, it can cross the remaining 600 miles.
2. The solution to the tougher version is derived from an analysis presented in an issue of Eureka, a publication of mathematics students at The University of Cambridge. To make terminology simple, lets call 500 miles a “unit” and gasoline for 500 miles as a “load” and a “trip” is a journey of the truck in either direction from one stopping point to another.
Two loads will carry the truck a maximum distance of 1 and 1/3 units (as we have seen in the children’s version).
Three loads will carry the truck 1 and 1/3 plus 1/5 units in a total of 9 trips. The first deposit point (call it a cache) is 1/5 units from the start. Three trips put 6/5 in the cache (duming 3/5 load in each stop). The truck returns, picks the remaining full load and arrives at the first cache with 4/5 load in the tank. This, together with the fuel in the cache, makes two full loads, sufficient to take the truck another 1 and 1/3 as we have seen.
We are asked for the minimum amount of fuel required to take the truck 800 miles. Three loads will take it 766 and 2/3 miles (1 + 1/3 + 1/5), so we need a third cache at a distance of 33 and 1/3 miles (which is 1/15 units). In five trips, the truck can build up this cache so that when the truck reaches this cache at the end of 7th trip, the combined fuel of the truck and the cache will be three loads. As we have seen, this will then be sufficient to take the truck the remaining 766 and 2/3 miles. Seven trips are made between starting point and first cache, using 7/15 load of gasoline. The three loads of fuel that remain are just sufficient to cross the rest of the way, and hence the total gas used will be 3+ 7/15 loads, which is ~3.46 loads.
As you can also see, the truck can successively cross 1, 1/3, 1/5, 1/7…and so on. This is a diverging series and hence the truck can cross any distance.
The problem has some more variations, will encourage the more mathematically oriented reader to read Martin Gardner’s My Best Mathematical and Logic Puzzles, puzzle #25. The original puzzle, btw, appeared in 1946/ 1947 in a few publications.
Apologies for a text only solutions, my iPad is not with me this weekend 😦