You can figure out that for any given door, say door #38, you will visit it for every divisor it has. so has 1 & 38, 2 & 19. so on pass 1 i will open the door, pass 2 i will close it, pass 19 open, pass 38 close. For every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9… leaving it open at the end. only perfect square doors will be open at the end.
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