This is an interesting puzzle. Certainly the kind of puzzle which tickles your mind and get your maths come out of you. 😉

So, Suppose there is a multistory building. Which has 100 floors.

-If an egg drops from the Nth floor or above it will break.

-If it’s dropped from any floor below, it will not break.

And, You’re given 2 eggs. You need to find out the minimum number of trails you need to perform in order to find out the ‘Nth’ floor.

How many drops you need to make?

*What strategy should you adopt to minimize the number egg drops it takes to find the solution?*

Try to solve it before reading the solution, which is of course given below. 🙂

The puzzle follows the below logic.

Say, the egg breaks at floor n we try to find out by going (N-1) till the first floor by doing linear search.

For example, I throw the egg from 10th floor, and it breaks, I will go to floor 1 to 9 to find out the floor..

Then I would try the same logic for every 10 floors thereby setting a worst case scenario of 19 chances.. i.e. 10,20,30,40,50,60,70,80,90,100,91,92,93,94,95,96,97,98,99

To find optimum solution, let’s try this:

If for every n, egg doesn’t break, instead of going to next n, go to N-1, this would save us one drop as we are doing a linear search with second egg when egg 1 breaks…

So the series would look something like this..

N + (N-1) + (N-2) + (N-3) +…+ 1

Now this is a series which is equal to N(N+1)/2

Now since it is given that the egg may or may not break from 100th floor..

We can write it as..

N(N+1)/2>=100

And n=14(approx)

So we should start from 14 then move up N-1 to 13 floor I.e. 27,39…

So the floors from where the drop needs to be done are: 14,27,39,50,60,69,77,84,90,95,99,100

*So the answer is 14*