You and 99 friends are staying in a hotel with 100 rooms. You decide to play a g

Question

You and 99 friends are staying in a hotel with 100 rooms. You decide to play a game: you label yourselves with the numbers 1 through 100. The person labelled “1” opens all the hotel room doors. The person labelled “2” closes all the even hotel room doors. The person labelled “3” opens all the closed doors that are a multiple of 3 and closes all the open doors that are a multiple of 3 and so forth (so the nth person opens each closed door that is a multiple of n and closes each open door that is a multiple of n).

Before you even begin the process, however, you tell your friend you know exactly which doors will be open and which will be closed at the end. How? (Your answer does not involve simulating the process with a computer or in your head).

Explanation / Answer

First think who will operate each door, obviously person #2 will do all the even numbers, and say person #10 will operate all the doors that end in a zero. So who would operate for example door 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........ That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a door open there will be someone to switch it closed. This will result in the door being back at it's original state.

So why aren't all the doors closed? Think of door 36:- The factors are: 1 & 36, 2 & 18, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will openly flick the door open and so the pairs doesn't cancel. This is true of all the square numbers.

There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 doors remain open.

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