You have two identical eggs and you are in a 100-storey building. For the eggs there is a "Breaking Floor", which is the lowest floor that the eggs will break if dropped from that floor. Below the Breaking Floor the eggs will not break, and above it they will break.
The eggs will survive multiple drops without degrading, so they can be used as many times as you like, until they break.
What is the optimum strategy to determine the Breaking Floor in the fewest number of drops?
With an unlimited number of eggs the solution would be to start at floor 50, and continue partitioning the building into halves, quarters, eighths etc. depending on whether the egg breaks or not.
But with two eggs you have to but a bit cleverer with your partitioning.
Start by dropping the first egg from floor 14. If it doesn't break move to floor 27 (14+13) and try again. Continue on this pattern to 39 (27+12), 50 (39+11), 60 (50+10), 69 (60+9), 77 (69+8), 84 (77+7), 90, (84+6), 95 (90+5) and 99 (95+4). If it survives all of these then try floor 100 to complete the task.
However, the likelihood is that the first egg will break on one of these floors. At which point go to the floor one above the last floor from which the egg survived, and go up floors one at a time until you find the Breaking Floor.
Because you have used smaller partitions as you have moved up the building you can find the Breaking Floor in a maximum of 14 drops.