How many lockers are left open?

Question

There are 1000 student and 1000 lockers.
The first student opens each of the 1000 lockers.
Second student closes every second locker.
Third student opens every third locker if it is closed or if it is open he closes it.
The fourth student goes to every fourth locker and if it is open he closes it, if it closed he opens it.
This process continues until it is completed by the 1000 students.

Answer 1

My final answer is 31.

31 is the number of perfect squares from 1 to 1000.

Now let me think of a good way to explain what perfect squares have to do with this....

The locker has to be touched an odd number of times to be left open.

Only perfect squares can be divided evenly by an odd number of different integers.

1 is only divisable by 1
4 is divisible by 4, 2, 1
9 is divisible by 9, 3, 1
16 is divisible by 16, 8, 4, 2, 1
etc...

all other numbers will have an even number of different integers that can divide it evenly, so those lockers will be closed.

2 is divisable by 2, 1
6 is divisable by 6, 3, 2, 1
72 is divisable by 72, 36, 24, 18, 12, 9, 8, 6, 4, 3, 2, 1
etc.

(Edited to add: This answer is correct. Here is a link to a similar question and answer that I just found, the reasoning is the same: http://www.braingle.com/7824.html .

Believe it or not, I solved and explained the reasoning myself before finding this other answer. I'm only adding this link to reaffirm my answer.)

Answer 2

Somebody has asked this question some weeks back, in a similar way, but missing the logic of only opens if closed and only closes if opened.

Anyway, an answer similar to q_midori's already appeared.

Just to add how 31 comes about since it was not mentioned.

31 = floor(sqrt(1000)) = floor(31.6...)

Answer 3

I'm guessing 4 open. The first one, the fourth one, and the last two. Thats my final answer.

Answer 4

All 100 will be open, You have to use the sequential derivative to solve it.

Answer 5

one open?