Two eggs, one hundred floors

Idea. Ask the reverse question: with $d$ drops, how many floors can two eggs cover? The first egg should jump by smaller and smaller steps, so that every possible break leaves the second egg exactly enough drops.

Step 1: one egg. With a single egg you cannot risk skipping a floor: you must go up one floor at a time. So with $m$ drops left and one egg, you can handle $m$ floors.

Step 2: two eggs. With $d$ drops, drop the first egg from floor $d$. If it breaks, the second egg checks floors $1$ to $d-1$ one by one with the $d-1$ drops left. If it survives, you have $d-1$ drops and two eggs for the floors above, so you can go up by $d-1$ floors for the next drop, then $d-2$, and so on. The number of floors covered is $$d+(d-1)+\cdots +1=\frac{d(d+1)}{2}.$$ No strategy covers more: if the first drop were higher than floor $d$ and the egg broke, the second egg would not have enough drops.

Step 3: 100 floors. We need $\frac{d(d+1)}{2}\ge 100$. Since $\frac{13\cdot 14}{2}=91$ and $\frac{14\cdot 15}{2}=105$, the answer is 14 drops.

The strategy. Drop the first egg from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100, stopping when it breaks, then check the floors above the last safe drop one at a time with the second egg. Each step up is one floor shorter than the previous one, so the total is never more than 14.

With three eggs, the same reasoning shows that 9 drops are enough for 100 floors.