A long queue of bored mathematicians
Level
2
A queue of 2018 people decide to play a game. A move consists of moving the person at the front to the back or reversing the order of the queue. How many different orders of queues could you have after 2018 moves?
The answer is 2018.
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1 solution
First, notice that if you imagine writing the numbers on the bracelet, whatever moves you do won't change the bracelet, just rotate or turn over the bracelet. Now if a person starts in an odd position(1st, 3rd, 5th...) then after an even number of moves they will still be on an odd position, since a move swaps the parity of everyone's position. So now there are 2018 orders left, all of which turn out to be possible as follows. If the original 1st person ends up in the (2n+1)th position then either: a) move the front to the back 2n+1 times, then reverse the order 2017-2n times b) move the front to the back 2018-2n times, then reverse the order 2n times This means that there are 2018 possible orders of the queue at the end