How many doors?

If a door is initially closed, that door number would need even number of factors so that the door's state would be changed even number of times to make its state finally closed.

eg. 2- 1 will come and open it, 2 will go and close it,

7- 1 will go and open it, 7 will go and close it.

50- 1 will go and open it, 2 will go and close it, 5 will go and open it, 10 will close it, 25 will open and finally 50 will close it. Similarly with other numbers.

Now think how many numbers from 1-100 have odd number of factors? The answer is all square numbers. Now, in 1-100 there are 10 square numbers, so the answer is simply 10.

A door is toggled in n-th walk if n divides the door number.

For example the door number 45 is toggled in 1st, 3rd, 5th, 9th and 15th walk.

The door is switched back to initial stage for every pair of divisors. For example 45 is toggled 6 times for 3 pairs (5, 9), (15, 3) and (1, 45).

It looks like all doors would be closed at the end. But there are door numbers which would become open, for example 16, the pair (4, 4) means only one walk. Similarly all other perfect squares like 4, 9, 25.36….etc

The final answer is 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. (There are 10 squares)

To be really simple, I'm pretty sure all the square numbers from 1 to 100 will be open. Please correct me if I'm wrong!

The squares are the key to this question... Only the square numbered door will be open.. i.e. 1,4,9,16,25,36,49,64,81,100. Thus, 10 doors will be open.