A number theory problem by Tyler I am lying 2
There are 2021 doors, each numbered 1 - 2021. 2021 students line up and:
(1) The first student opens all doors.
(2) The second student closes all doors numbered a multiple of 2.
(3) The third student closes all open doors numbered a multiple of 3, and opens all closed doors numbered a multiple of 3.
(4) The fourth student closes all open doors numbered a multiple of 4, and opens all closed doors numbered a multiple of 4.
(5) n th student ( n is natural, n ≤ 2021) closes all open doors numbered a multiple of n , and opens all closed doors numbered a multiple of n .
After all 2021 students do this, how many doors are open?
The answer is 44.
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1 solution
The status of the door should change an odd number of times. So, the number of the door should have an odd number of factors. Therefore, it should be a perfect square.
2 0 2 1 = 4 4 . 9 5 . . . − > the answer is 44.