Here is the problem:
There are 100 doors and 100 people. The doors have two states (open/closed). All the doors are closed at first. Then the first person opens all the doors, the second person toggles the doors in multiples of 2 (e.g. door 2,4,6..). The third person toggles the doors in multiples of 3 (3,6,9...). This continues until person 100 toggles the door 100. So how many doors are open and how many are closed. Please give an equation for any number of doors as well. Thank you.
It is a mathematical game I heard time ago. If the door's number is a perfect square, then its state is open. Otherwise it is close. The reason is the commutative property of the product: if the j-th person toggles a door after i steps, then the i-th person toggles the same door after j steps. This way the door has even switches if its number is not a perfect square. Otherwise the door has odd switches.
It is a mathematical game I heard time ago. If the door's number is a perfect square, then its state is open. Otherwise it is close. The reason is the commutative property of the product: if the j-th person toggles a door after i steps, then the i-th person toggles the same door after j steps. This way the door has even switches if its number is not a perfect square. Otherwise the door has odd switches.
The problem as described by you will cause the following pattern (assuming I understood correctly):
TRUE 2xFALSE TRUE 4xFALSE TRUE 6xFALSE TRUE 8xFALSE ...
This puts open doors at exact square positions:
1, 4, 9, 16, 25, 36, etc....
The total number of open doors is the sum of squares below the total door number N, and includes this one if it is an exact square:
For the 100 door problem, then it is:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100 = 10 open doors
In other words, your equation is:
open doors = floor( sqrt(N) )
, where floor() is the round-down operator.
Now this can be used to check if a given number is a perfect square or not which was a question in another thread.
I think that graph theory will work. You can tray to solve the problem with complete graph.
As Tomas said, the mathematical game is solved by "number of open doors among the first N" = floor( sqrt(N) ).
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