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Hi.
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In this problem, we're going to talk about something called the lockers problem, where you have 100 lockers and 100 students.
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Each student is assigned to one of the lockers from 1 to 100.
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And the lockers start off all closed.
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And then it starts with the first student opening every locker.
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And then the second student opening every other lockers, so every two.
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Third student opens the lockers with the multiples of three, and so on and so forth.
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So the question is which lockers will be left open? so what we can think of this as is, so they start off closed, and then they're opened and then closed, open closed.
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So any locker that was sort of switched, an odd number of times would be open, and any locker that was switched an even number of times would be closed.
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But think about when a locker is switched, it's when that locker number is divisible by whatever student's number is going.
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So when student four goes, lockers four, eight, 12, 16, and so on, get open or closed, they get switched.
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And so those are the ones that are divisible by four.
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So the trick here is that generally divisors come in pairs, right? for instance, you know, you take a number like 28.
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And this is, we know, one times 28, two times 14, it's got to be more.
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4 times 7.
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I think that's it.
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So they come in pairs, right? so 28 has six divisors, which means that locker 28 would be closed because it would get opened, closed.
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Or i guess if we're going in order of the numbers, it would be open.
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And then 2 would close it, 4 would open it, 7 would close it, 14 would open it, 28 would close it.
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So 28 would be closed.
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But what numbers don't satisfy this? think 25, 1 times 25, and 5 squared.
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36, 1 times 36, 2 times 18, 3 times 12, 4 times 9, but then 6 squared.
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So these have an odd number of divisors.
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So these will be left open.
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So this kind of answers a through d...