You have a 1000 students and 1000 lockers. the first student comes and
opens all the lockers. The second student comes and closes every
second locker. the third student comes and either opens or close every
third locker depending on whether it's open or closed. and the fourth
student continues to open or close the
lockers. Which lockers are open after 1000 students go through.

Hi.  I would love to answer your question.  If I understand the
question correctly, after all 1000 students have gone through, the
first locker would be the only locker open.

No. Every student up to a 1000 either opens or closes the locker. so
the 4th opens or close every 4th locker the 5th student opens or
closes every 5th locker. and so on

school has exactly 1,000 lockers and exactly 1,000 students. On the
first day of school, the students meet outside the building and agree
on the following plan: The first student will enter the school and
open all of the lockers. The second student will then enter the school
and close every locker with an even number (2,4,6,8,...). The third
student will then "reverse" every third locker. That is, if the locker
is closed, he will open it; if the locker is open, he will close it.
The fourth student will reverse every fourth locker, and so on until
all 1,000 students, in turn, have entered the building and reversed
the proper lockers. Which lockers will finally remain open?

I'm not going to post the entire list, but here's an abbreviated version:

1-c;2-o;3-o;4-c;5-c;6-c;7-o;8-c;9-o;10-o;11-o;12-c;13-o;14-c;15-o;
.....
985-o;986-c;987-o;988-c;989-o;990-c;991-o;992-c;993-o;994-c;995-o;996-c;997-o;998-c;999-o;1000-c;

Here's a little VBScript that you can run to get the full results.
Just copy and paste the text below into a text editor and save as a
.vbs file (Windows PCs only, unless you have a parser on a Mac or
Linux box)
Run it as cscript.exe <filename.vbs> to put the results into a console
window for copying/pasting.

'start here *************
dim locker(999)
dim i, j, msg
msg = ""

for i = 0 to 999
 locker(i) = "o"
next

for j = 2 to 1000
  for i = 0 to 999 step j+1
    if locker(i) = "o" then
       locker(i) = "c"
    else
       locker(i) = "o"
    end if
  next
next

for i = 0 to 999
  msg = msg & i+1 & "-" & locker(i) & ";"
next

wscript.echo msg
'end here *************

i DON'T KNOW HOW TO USE A TEXT EDITOR OR DO THE THINGS YOU SUGGEST.
hOW MANY DOORS ARE OPEN?????

All of them except for 500.

P.S. your CAPS lock is on

500 is not correct. your 1 thru 15 example is wrong. 1 is open then 4
is open then 9 is open the 16 is open. the other numbers are close.
Thanks for the warning on the cap lock.

1, 4, 9, 16...those are squares.  Is that significant?  Is it going to
be 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, etc.?  Why?

This may be of interest:

"In a certain high school there were 1000 students and 1000 lockers.
Each year for homecoming the students lined up in alphabetical order
and performed the following ritual: The first student opened every
locker. The second student went to every second locker and closed it.
The third student went to every third locker and changed it (i.e., if
the locker was open, he closed it; if it was closed, he opened it). In
a similar manner, the fourth, fifth, sixth, . . . student changed
every fourth, fifth, sixth, . . . locker. After all 1000 students had
passed by the lockers, which lockers were open?

The locker problem is really a 'problem.' For our students there is no
way to solve it by recall since the locker problem does not look like
any familiar type of 'story problem.' The only way to solve it is by
doing mathematics. One might guess and check--popular guesses are
prime number lockers, the first locker and/or the last locker. It is
easy to check that the first locker remains open, but how about the
last one? Prime numbers also don't seem to work (check 3 or 7, for
example). It is clear that we have a problem. Someone in the class
suggests that we see what will happen with 10 lockers. The class
agrees that solving a simpler and more manageable problem might lead
to some understanding of 'what's going on here?' Working in small
groups they 'open' and 'close' 10 lockers: lockers 1, 4, and 9
remained open. Then they do the same with 20 lockers--1, 4, 9, and 16
are open. Sooner or later each small group in the class has a
conjecture: Either that all the open lockers are square numbers or
that the differences between the open lockers are consecutive odd
numbers.

Most prospective teachers are quite happy with their surprising
solution and are willing to predict at this point what all the open
lockers are. Since inductive reasoning is used in every day life as a
mean for making predictions (e.g., Martin and Harel, 1989), most
students see this stage as the final stage of the solution of the
problem. But can we really be sure that the pattern continues? Why? To
make sure that this is the case, deductive reasoning should be used to
construct a supporting argument that is convincing. The questions
mentioned above, in addition to some others that explore the
relationship between a student's number and the locker numbers
visited, are assigned as homework. The next day a whole-group
discussion takes place. Many students discover that the relationship
between student numbers and the locker numbers visited by them can be
described as the relationship between factors and multiples.
Throughout the discussion it becomes clearer that open lockers are the
ones that have an odd number of factors. Do all square numbers have an
odd number of factors? Why? Why do nonsquare numbers have an even
number of factors? Investigating these questions by exploring factor
pairs for some specific numbers (e.g., factor pairs for 24 are 1 and
24, 2 and 12, 3 and 8, 4 and 6. Factor pairs for 25 are 1 and 25, 5
and 5) makes it clearer why square numbers (and only square numbers)
have an odd number of factors.

How about the other conjecture? Are the differences between the open
lockers consecutive odd numbers? Can we show that this is true?
Proving by mathematical induction that the sum of consecutive odd
numbers, starting from 1, is a square number, is not appropriate in
this context. But a pictorial representation (see Figure 2) can
provide a convincing argument. The number 1 is represented by one dot
at the upper left corner. By adding the number 3 which is represented
by three dots, we can form the number 4--a square of 2 x 2. Then, by
adding the number 5 (five dots), the number nine can be formed--square
of 3 x 3. One can verify that this process can continue for any given
sum of consecutive odd number starting with 1. The result is always a
square number."

http://ncrtl.msu.edu/http/craftp/html/pdf/cp893.pdf

(NOTE... AFTER DOING ALL OF THIS, I REACHED AN ANSWER THAT MAKES
INTUITIVE SENSE... SKIP TO THE BOTTOM FOR THAT)


First, it sounds like you need to determine the state (open or closed)
of EVERY SINGLE locker, so a guess-and-check system won't work.

Start by treating all lockers as closed (before student #1 opens them
all). You're going to end up keeping track of the quantity of factors
for each "Test Number", where is "Test Number" is the number of a
locker. For example, "Test Number" 6 is locker #6. The number 6 has
factors of 1, 2, 3 and 6. Thus, locker #6 has its state changed FOUR
times, because students 1,2,3,6 are the only ones who act on it. Since
it begins as a closed locker, student 1 opens it.. student 2 closes..
student 3 opens.. student 6 closes. Thus, an EVEN number of factors
for a Test Number means the locker is closed at the end. An ODD number
means the locker is opened at the end.

From this point, you can set up a table in Excel to test whether every
number 1-1000 is a factor of every number 1-1000 (the MOD function
gives you the remainder of division... where remainder is zero, the
number is a factor).

The Excel table outputs results and... lo and behold.. only perfect
squares are open. Now that I think about it, that makes sense. Every
number has its factors, and factors always come in pairs (1x20, 2x10,
4x5 for the number 20). So every locker will be opened and closed by a
pair of students. The only exceptions are perfect squares, (1x36,
2x18, 3x12, 4x9, ****6x6**** for number 36), since the 'perfect square
student' only acts on the locker once. In the case of locker #36,
student number 6 comes along and opens it without the 'corresponding'
student number 6 to come along after to close it. So perfect squares
are the only numbers where the paired factors don't cancel each other
out to re-close a locker that has been opened.

Lockers corresponding to 1 squared through 31 squared are the only open lockers.

Pinkfreud
   As immaculate as ever.. i enjoyed your post... thanks

-ddhara