The discussion revolves around a mathematical problem involving 1000 doors, each labeled from 1 to 1000. The process involves opening and closing doors based on their factors, leading to the question of which doors remain open after all operations are completed. The key insight is that a door remains open if it has an odd number of factors, which occurs for perfect squares. Therefore, the open doors will be those labeled with perfect square numbers up to 1000, specifically doors 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961.
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mck3939 said:There are 1000 doors. Each one labeled with a number 1-1000. A person opens all doors whose number has one as a factor (which is all of them). Then she closes all doors whose number has two as a factor. Then the person continues to change the status of the doors (opening or closing them) based on the number and factors. What lockers will be open when we reach 1000? How do you show your work?
I tried first dividing 1000 by 2 to get 500
then 1000 by 3 to get approximately 333
and 1000 by 4 and so on, but I find this is taking forever and is no the best strategy to use. I cannot think of a better one.