Solution
Two Eggs, 100 Floors
Show the problem again
You have two identical eggs and a 100-floor building. There is some critical floor at or above which a dropped egg breaks. Broken eggs are gone; intact eggs are reusable. What is the minimum number of drops that guarantees you identify the critical floor in the worst case?
Worked solution
The answer is 14. With a first drop at floor k, a break forces up to k − 1 single-floor tests; so each successive first-egg drop should rise by one floor less to keep the worst case constant. You need k + (k−1) + ... + 1 = k(k+1)/2 ≥ 100, so k = 14 (105 ≥ 100). Drop at floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100; every scenario resolves within 14 drops.
Source: Classic egg-drop optimisation puzzle in wide circulation, long used as an interview question. Statement written for AxiomIQ.