There are four lifts in a building. Each makes three stops, which do not have to be on consecutive floors or include the ground floor. For any two floors, there is at least one lift which stops on both of them. What is the maximum number of floors that this building can have?
I find this question really confusing.(Angry)
July 27th 2008, 05:26 AM
First off you can see that the maximum number of floors is less than 12 because then that would mean that all floors (if they have a lift at all) can only be served by 1 lift (4x3=12).
For each floor to have at least 2 lifts stopping there, there can in fact be no more than 6 floors (4x3/2). Now the job is to prove that 6 floors can be served by these 4 lifts.
So try it out, by trial and error:
yep, that works. Each of the 4 lifts (a, b, c, d) are stopping at 3 floors, and each floor has exactly 2 lifts stopping at each.
So the maximum is 6.
This question can I believe be solved rigirously using techniques from graph theory, but this question is simple enough not to have to get those big guns out.