Problem Statement

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.

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.

Let's start with the solution:

NDoors := proc( N :: posint )
        # Initialise, using 0 to represent "closed"
        local pass, door, doors := Array( 1 .. N, 'datatype' = 'integer'[ 1 ] );
        # Now do N passes
        for pass from 1 to N do
                for door from pass by pass while door <= N do
                        doors[ door ] := 1 - doors[ door ]
                end do
        end do;
        # Output
        for door from 1 to N do
                printf( "Door %d is %s.\n", door, `if`( doors[ door ] = 0, "closed", "open" ) )
        end do;
        # Since this is a printing routine, return nothing.
        NULL
end proc:

> NDoors( 100 );

> seq( i^2, i = 1 .. isqrt( 100 ) );
                  1, 4, 9, 16, 25, 36, 49, 64, 81, 100

> [seq]( 1 .. 10 )^~2;
                 [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]