Puzzle | 100 Doors

There are 100 doors in a row, and all doors are initially closed. A person walks through all doors multiple times and toggles (if open, then close; if closed, then open)

• In the first walk, the person toggles (or opens) every door.
• In the second walk, toggles (or closes) every second door (i.e., 2nd, 4th, 6th, 8th, and so on).
• In the third walk, toggles every third door (i.e., 3rd, 6th, 9th, etc.).

This pattern continues, and in the 100th walk, the person toggles only the 100th door.

Check if you were right - full answer with solution below.

Solution: 

• A door is toggled in the i-th walk if i divides the door number, for example: Door number 45 is toggled during the 1st, 3rd, 5th, 9th, 15th, and 45th walks.
• Each door is toggled once for every divisor of its number, and divisors typically come in pairs (e.g., for 45: (1,45), (3,15), (5,9)).
• Each pair of divisors cancels out the toggle effect (open - close or close -open), therefore, doors with an even number of divisors return to their initial closed state.
• Perfect square numbers (e.g., 16) have one unpaired divisor (like 4 in 4×4), resulting in an odd number of divisors.
• An odd number of toggles leaves the door in the open position; hence, only perfect square-numbered doors (e.g., 1, 4, 9, 16, ..., 100) remain open.
• Prime numbers (e.g., 2, 3, 5, 7) have exactly two divisors (1 and itself), which is a pair - the door remains closed.
• Non-square composite numbers (e.g., 15) have divisor pairs, so they are also closed at the end.
  
  ⁛So the answer is 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100.