Puzzle | 100 Doors (original) (raw)

Last Updated : 6 Aug, 2025

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.