How Strong is an Egg? You have two identical eggs. Standing in front of a 100 floor building, you wonder what is the maximum number of floors from which the egg can be dropped without breaking it. What is the minimum number of tries needed to find out the solution?
 A: I found the solution. Here it is:
The easiest way to do this would be to start from the first floor and drop the egg. If it doesn’t break, move on to the next floor. If it does break, then we know the maximum floor the egg will survive is 0. If we continue this process, we will easily find out the maximum floors the egg will survive with just one egg. So the maximum number of tries is 100 that is when the egg survives even at the 100th floor.
Can we do better? Of course we can. Let’s start at the second floor. If the egg breaks, then we can use the second egg to go back to the first floor and try again. If it does not break, then we can go ahead and try on the 4th floor (in multiples of 2). If it ever breaks, say at floor x, then we know it survived floor x-2. That leaves us with just floor x-1 to try with the second egg. So what is the maximum number of tries possible? It occurs when the egg survives 98 or 99 floors. It will take 50 tries to reach floor 100 and one more egg to try on the 99th floor so the total is 51 tries. Wow, that is almost half of what we had last time.
Can we do even better? Yes we can. What if we try at intervals of 3? Applying the same logic as the previous case, we need a max of 35 tries to find out the information (33 tries to reach 99th floor and 2 more on 97th and 98th floor).
Interval – Maximum tries

    1  – 100
    2  – 51
    3  – 35
    4  – 29
    5  – 25
    6  – 21
    7  – 20
    8  – 19
    9  – 19
    10 – 19
    11 – 19
    12 – 19
    13 – 19
    14 – 20
    15 – 20
    16 – 21

So picking any one of the intervals with 19 maximum tries would be fine.
Instead of taking equal intervals, we can increase the number of floors by one less than the previous increment. For example, let’s first try at floor 14. If it breaks, then we need 13 more tries to find the solution. If it doesn’t break, then we should try floor 27 (14 + 13). If it breaks, we need 12 more tries to find the solution. So the initial 2 tries plus the additional 12 tries would still be 14 tries in total. If it doesn’t break, we can try 39 (27 + 12) and so on. Using 14 as the initial floor, we can reach up to floor 105 (14 + 13 + 12 + … + 1) before we need more than 14 tries. Since we only need to cover 100 floors, 14 tries is sufficient to find the solution.
Therefore, 14 is the least number of tries to find out the solution.
A: The critical strip of eggs is the interval $C:=[k,k+1]$ such that, when an egg is dropped from a height $\leq k$ it survives, and when it is dropped from a height $\geq k+1$ it brakes.
The numbers
$$h_r(n)\qquad(r\geq0,\ n\geq0)$$
("allowed length for $r$ eggs and $n$ trials") are defined as follows: If we know that $C$ lies in a certain interval of length $\ell\leq h_r(n)$ we can locate $C$ with $r$ eggs in $n$ trials, but if  we know only that $C$ lies in a certain interval of length $\ell>h_r(n)$ then there is no deterministic algorithm that allows to locate $C$  with $r$ eggs in $n$ trials for sure. Obviously
$$h_0(n)=1\quad(n\geq0)\ ,\qquad h_r(0)=1\quad(r\geq0)\ .$$
The numbers $h_r(n)$ satisfy the recursion
$$h_r(n)=h_{r-1}(n-1)+h_r(n-1)\qquad(r\geq1,\ n\geq1)\ .\qquad(*)$$
Proof. Assume $C$ lies in a certain interval $I$ of length $\ell:=h_{r-1}(n-1)+h_r(n-1)$. We may as well assume that the lower end of $I$ is at level zero. Drop the first egg at height $h_{r-1}(n-1)$. If it brakes then $C$ is contained in the interval $[0,h_{r-1}(n-1)]$ and can be located with the remaining $r-1$ eggs in $n-1$ trials. If it survives then $C$ is contained in the interval $[h_{r-1}(n-1),\ell]$ of length $h_r(n-1)$ and can be located with the $r$ eggs in $n-1$ trials. This proves $h_r(n)\geq\ell$.
Conversely, assume that we know only that $C$ lies in a certain interval of length $\ell'>\ell$ and that there is an algorithm that locates $C$ with $r$ eggs in $n$ trials. This algorithm would tell us the height $k$ at which we should drop the first egg. If $k>h_{r-1}(n-1)$ and the egg brakes or if $k\leq h_{r-1}(n-1)$ and the egg survives it would be impossible to finish the task, as the remaining interval that contains $C$ is larger than allowed for the remaining resources. It follows that $h_r(n)\leq\ell$.
From $(*)$ we obtain
$$h_1(n)=n+1\ ,\quad h_2(n)={1\over2}(n^2+n+2),\quad h_3(n)={1\over6}(n^3+5n+6)\qquad(n\geq0)\ .$$
As $h_2(13)<100<h_2(14)$ we need at least $14$ trials to locate $C$ (or eliminate $C\subset[0,100]$) in the original setup. 
