A merchant has bough some 3000 banana from market and want to bring them to his home which is 1000 km far from Market. He have a elephant which can carry maximum of 1000/- banana at time. That could have been easy but this elephant is very shrewd and he charges 1 banana for every one kilometer he travel.

Now we have to maximise number of banana which should reach to home.

Solution:

At present we are at 0th km with 3000 banana and elephant.

Lets see if elephant have to shift all the 3000 banana and elephant by 1 km.

For 1 km shift:

Take first 1000 banana and drop at 1st km. 2 banana consumed. One banana each for to and fro.

Second 1000 banana and drop at 1st km. 2 banana consumed.

Third 1000 banana and reach at 1st km. 1 banana consumed.

So all in all total 5 banana will be consumed if we shift a kilometer.

But that's not all, our total banana number are also reducing so we may have to reduce the number of turns too. And this will happen once we have reduced the banana by 1000 number. That means to reduce the count of banana by 1000 for first move we have to shift base to 200 km. See 5 banana for 1 km ratio i.e. 1000 banana / 5 = 200 km.

So we are now at 200th KMs with 2000 banana and elephant.

Post 200 km, for every 1 km shift in base;

Take first 1000 banana and drop at 1st km. 2 banana consumed. One banana each for to and fro.

Third 1000 banana and reach at 1st km. 1 banana consumed.

So all in all total 3 banana will be consumed if we shift a kilometer post 200 kms.

Now the second phase will be to reduce the 1000 banana. So we can move to 1000/3 = 333.3 km maximum in this phase.

So we will be moved to (200 + 333.3) 533.3 km base with 1000 banana.

Now onward we need only 1 banana for every km shift as elephant can carry all the 1000 banana and need not to come back.

So out of 1000 banana, elephant will consume (1000 - 533.3) more banana to reach home.

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**And total saved banana will be (1000 -(1000 -533.3)) = 533.3 banana. **