Puzzle 16 | (100 Doors)

There are 100 doors in a row, all doors are initially closed. A person walks through all doors multiple times and toggle (if open then close, if close then open) them in the following way: 

In the first walk, the person toggles every door 

In the second walk, the person toggles every second door, i.e., 2^nd, 4^th, 6^th, 8^th, … 

In the third walk, the person toggles every third door, i.e. 3^rd, 6^th, 9^th, … 

Likewise,

In the 100^th walk, the person toggles the 100^th door. 

Which doors are open in the end? 

Solution: 
A door is toggled in an i^th walk if i divide door number. For example, door number 45 is toggled in the 1^st, 3^rd, 5^th, 9^th,15^th, and 45^th walks.
The door is switched back to an initial stage for every pair of divisors. For example, 45 is toggled 6 times for 3 pairs (5, 9), (15, 3), and (1, 45). 

It looks like all doors would become closed at the end. But there are door numbers that would open, for example, in 16, the divisors are (1,2,4,8,16) and as the pair(4,4) contributes only one divisor making the number of divisors odd, it would become open at the end. Similarly, all other perfect squares like 4, 9,…, and 100 would become open. Now, for prime numbers like 2,3,5,7… the divisors are (1, that number) and it is a pair, so they will remain closed at the end. And for all other numbers divisors are always in pairs, e.g. 15 = (1,15),(3,5), they will also remain closed.

So the answer is 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. 

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