There are 100 doors in a row that are all initially closed.

You make 100 passes by the doors.

The first time through, visit every door and  toggle  the door  (if the door is closed,  open it;   if it is open,  close it).

The second time, only visit every 2nd door   (door #2, #4, #6, ...),   and toggle it.

The third time, visit every 3rd door   (door #3, #6, #9, ...), etc,   until you only visit the 100th door.



Task

Answer the question:   what state are the doors in after the last pass?   Which are open, which are closed?


Alternate:
As noted in this page's   discussion page,   the only doors that remain open are those whose numbers are perfect squares.

Opening only those doors is an   optimization   that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.




proc nonrec main() void:
    byte DOORS = 100;
    [DOORS+1] bool door_open;
    unsigned DOORS i, j;
 
    /* make sure all doors are closed */
    for i from 1 upto DOORS do door_open[i] := false od;
 
    /* pass through the doors */
    for i from 1 upto DOORS do
        for j from i by i upto DOORS do
            door_open[j] := not door_open[j]
        od
    od;
 
    /* show the open doors */
    for i from 1 upto DOORS do
        if door_open[i] then
            writeln("Door ", i, " is open.")
        fi
    od
corp