The problem
A jeep can carry one tank of fuel, enough for 1 unit of distance. It starts at a depot with n tanks available and may cache fuel anywhere in the desert. How far can it get?
The structure
Work in segments, from the depot outwards.
With n tanks at the start, moving all of them forward requires n trips out and n - 1 trips back (you do not return on the last one). So each unit of distance in the first segment costs 2n - 1 tanks' worth of fuel per unit.
To move all n tanks forward by distance d costs (2n - 1)d. You want to advance until you are down to n - 1 tanks, which happens at
d = 1/(2n - 1)
Then repeat with n - 1 tanks, which costs 2(n-1) - 1 = 2n - 3 per unit, and so on. The final tank travels its full 1 unit.
The answer
Total distance:
1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n - 1)
A sum of reciprocals of odd numbers - roughly (1/2) ln n plus a constant.
What it means
Distance grows logarithmically in fuel. Doubling your fuel does not double your range; it adds a fixed increment. Getting far into the desert becomes brutally expensive.
With 4 tanks you reach about 1 + 0.333 + 0.2 + 0.143 = 1.68 units. With 8 tanks, about 2.02. Four times the fuel for 20% more distance.
Where the same shape appears
Multi-stage rockets face exactly this: fuel must carry fuel, so payload fraction collapses with required delta-v. It is also the reason electric-vehicle relay schemes and long-range drone caching hit hard limits.
An interviewer who asks this is usually testing whether you can set up a cost-per-segment recursion, not whether you know the harmonic series.
Variants
Round trip required. Costs roughly double the fuel per segment; the distance for a there-and-back journey is about half.
Fuel available at the far end. Changes the problem entirely into a two-sided version.
More in brainteasers.