Locker Puzzle

• Student #1 closes each locker.
• Student #2 opens each second locker.
• Student #3 changes the state of each third locker (i.e. opens them if they are closed, or closes them if they are open)
• ...and so on, until all 1,000 students have had their turn.

After all 1,000 lockers are done, which lockers are open? Which are closed? Why?

      

After I read this question, the first mathematical thing is looking for patterns by induction. 

1st  try: I used 6 students, 

I found 2 locked, the position of #1,#4, the rest of them are open.

2nd try: I  used 10 students,

I found 3 locked in this try, the position is #1,#4,#9, the rest are open.

I see some pattern here, the difference the consecutive number is 3 and 5. they are odd and the difference is 2, so I hope the next difference will be 7.

I also found  I only need to draw the diagonal states since under the diagonal the states don't change any more. This reminded me of matrix diagonal, but I could not go further. 

                          #2 student has multiple of 2 positions change its state,

                         #3 student only when the position number is the multiple of this student number 3, then change its state and so on...doesn't help much

             

3rd try: I used 20 students, 

I found 4 locked, position #1,#4,#9,#16, the rest are open. If we order the position number to locked number  n=1,2,3,4... then the differences between the two consecutive position numbers are 3,5,7...which is 2n+1. I predicted that the next locked position will be #25,#49,#64 since we keep adding 2n+1 to the previous number, then we get a series of perfect square number n^2. 

 I have n^2+(2n+1)=(n+1)^2<=1000, so we have (n+1)^2<=1000, then n+1<=sqrt(1000)

so n<=31.6-1, therefore, n=30, so the last locked position will be (n+1)^2=31^2=961

therefore, it has a total of 31 locked and all the locked positions are the perfect square. The last locked is the #961.

Then I come back to the 1st try to find out the reason. I go back to the idea that #2 student has multiple of 2 positions change its state, #3 student only when the position number is the multiple of this student number 3, then change its state and so on...then the factor idea come to my mind. when having even numbers of factors which mean even numbers of changing states, so it will be open, otherwise will remain closed. Only perfect square numbers have odd numbers of factors, here you go! 

To solve this question, I  used the induction method, using mathematical knowledge about factors, perfect squares, multiples, inequality, patterns.