An Impossible Problem!

[~Dernhelm~]

Here's one which I just can't find the answer to or the mathematical reasoning for no matter how hard i may try... Can anybody do this one?

***

Midvale School has 1000 students and 1000 lockers. On the first day of class all of the lockers are closed. The slightly eccentric principal admits the students one by one and asks them to each perform a rather strange task.

The first student must walk around the school and open every locker. The second student much walk around school and close every second locker. The third student must walk around the school and at the third locker, and each third locker following, open the locker if it is closed or close it if it is open.

Similaryly, the fourth student must go and change the state of every fourth locker; the fifth student must change the state of every fifth locker; and so on until all 1000 students have been through the entire school. After the 1000th student has gone to locker #1000 and opened it or closed it; depending upon whether it was already closed or open, which lockers remain open?

***

Sounds evil, doesn't it? But don't get discouraged and give it a try, perhaps the math behind it and the answer could be found...

[Imrel]

31, all of the ones square numbered ones.

[~Dernhelm~]

That was fast, how in the world did you figure it out?

[Imrel]

I made a table of the first twelve lockers for the first ten students and looked for a patern, and I noticed that the square numbers seemed to stick being open. Then I realized that square numbers always have an odd number of factors, while others have even numbers. So only the square numbered lockers would be opened or closed an odd number of times, and since they started closed, an odd number of status changes would mean they would have to end up open.

[~Dernhelm~]

Whoa! You got me there... I've tried counting it out before, only to make tons of mistakes. Some sort of sequence with one open then two closed, one open then three closed was my original guess... Then again, I'm not too good at these things!

[Imrel]

It's just factoring. Take locker twelve, for example. Twelve can be divided by one, two, three, four, six and twelve. So the first, second, third, fourth, sixth and twelfth pupils will open/close it. So it will open then close then open then close then open then close.

The case is similar for any other number, unless it's square, because one of its factors serves as two usually do in non-square numbers. (Eg, four has the factors one, two and four, with one times four equalling four and vica versa, and two times two equalling four.) I dunno if I'm making much sense, sorry...

[~Dernhelm~]

Yup! I've got it now. ;D So locker #100 would also be open because it's square root is 10? Okay!

[Imrel]

Uh huh, you got it!

[~Dernhelm~]

Lol! *laughs*

I meant it as in "I understand the math behind it now". You were the one who figured it out and was kind enough to explain it. Please don't get mad!

[Imrel]

I'm not, I'm just glad that I actually made sense, lol.

[~Dernhelm~]

Oh. Okay. *smiles*

[~Dernhelm~]

I think I've found a mathematical equation to it!

(n+1)²

This then translates to:

(n+1)*(n+1)

Which then translates to:

n²+2n+1

Would that be it? (Just playing around... )

[Imrel]

I think that equation gives you a square number with whatever number you plug in. I don't think this is one you can actually have an equation for... Well it could be but I'm not sure how it'd work.

[~Dernhelm~]

It's a weird thing!

Should we try to find another "Impossible Problem" perhaps?

[Imrel]

It's your thread...

I'll give you one if I find one, too.