Locker Problem Rough Thoughts~

 It is obviously to see from the first will stay close until the last, as nobody can touch it after the first person closed it. And we can also easily to see the second locker will remain open after the second person opened it again and no one else could touch it. And the third locker would be open as well because the first and the third person are the only people are allowed to touch it.

If I label each locker by number and connect the locker to the number, then my thought turning to: "what matters the open and close to the each specific locker?" And my answer is: how many divisor of each specific number from 1-1000. My first quick thought leads me that: obviously, every prime number locker should remain open as only the first and the corresponding number student are the only two person who turned the locker close and open. So how many people can touch the locker is the key point to figure out each locker's state at the end.

Then, my brain suddenly related to number theory, or relative field, that there is a specific formula to calculate the number of divisor for each number, as long as we know the idea of primes, prime factorization, etc. And I also realized that, if I can figure out how many divisors for each number, then even number of divisors would lead to open and odd number of divisors would lead to a close, as a locker has only open and close state, if one person open it, the next person will close it, no matter what happens as long as under this game's rule.

So my goal turns out to: figure out the number of divisor of each number for 1-1000, and even number of divisors means an open state at the end of that number related locker, where odd number of divisors means a close state at the end of that number related locker. 

Now I am trying to figure out, what number's factorization would lead to a even number, and what would lead to odd.

By looking at number of divisor's formula, and I realized that only perfect squares would leads to odd number of divisors, and the rest will be even number of divisors, as all perfect squares have all even exponents in their prime factorization, and then by the formula, which is products of their exponents+1, they will be all odds multiplying odds, which leads odds products. And non-perfect squares will have odd exponents which gives even multiplier while calculating the number of divisors. Therefore, only perfect square between 1-1000, which are 1, 4, 9, 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 will be closed, and the rest will be open


3 comments:

  1. Very nice, You! One thing I'm not quite clear about: do perfect squares have all even exponents -- or just one even exponent? (One would be enough I think, but just wondering!)

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    1. For prime factorization of perfect squares, they all must all even number of exponents. Let's assume a perfect square contains one of odd exponent in its prime factorization, then it won't be able to be split by two integer exponents prime factorization for its square root, a contradiction! So it should be all even exponents and by then, the number of divisors formula will be products of all odds, and therefore will be odd number of divisors, which leads to close state for the corresponding locker.

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