|
The answer is 533 1/3 cells.
To achieve this, load the car with 1000 cells, drive out 200 miles, drop 600 cells, and return back with the 200 cells you have left (200 + 200 + 600 = 1000).
Repeat this once more, and you will have 1200 cells at the 200 mile mark.
Now, load the remaining 1000 cells and drive out to 200th mile. By the time you get there, you will have 800 left, making 2000 with the 1200 that are there already.
Load 1000 onto the car, and drive to mile 533 (a 333 mile drive from mile 200). Unload 334 cells there, and drive back to 200. Pick up the remaining 1000 cells and go to mile 533 again. (You will have 667 cells in your car, and 334 on the ground.)
Pick up 333 of the cells, leaving one on the ground. Your car is now full.
You could now drive straight to the finish line (467 miles) and get there with 533 cells remaining, but here's a trick that will let you transport a whole extra 1/3 of a cell:
a: With your 1000 cells, drive 1/3 of a mile to 533 1/3, and unload one battery there.
b: Drive back 1/3 of a mile to 533, and pick up the cell you have left before.
c: Drive to 533 1/3 again. In steps a, b and c, you have driven a total of one mile, so you have used up one cell. You now have the room on board for the cell you dropped at mile 533 1/3 in step (a). Grab it.
You are now at mile 533 1/3, with 1000 full power cells. This means that if you drive straight to mile 1000, you will use up 466 2/3 cells, leaving 533 1/3 intact for delivery.
Thanks to Ender for submitting this solution |
