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:

// 100 doors problem
// Author: Sinuhe masan (2019)
init

	// 100 elements array of boolean type
	doors:bool[100]
    
	for var i = 1 to 100
		doors[i] = false  // set all doors closed
    
    
	for var i = 1 to 100
		j:int = i
		while j <= 100 do
			doors[j] = not doors[j]
			j = j + i
    
	print("Doors open: ")
	for var i = 1 to 100
		if doors[i]
			stdout.printf ("%d ", i)