Logic Challenge

Assume you have 1 million doors, they are all closed.

You will do 1 million operations of opening/closing doors.

At operation number 1 you close (the ones opened) and open (the ones closed) that are multiple of 1

At operation number 2 you close (the ones opened) and open (the ones closed) that are multiple of 2

At operation number n you close (the ones opened) and open (the ones closed) that are multiple of n

.
.
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At operation number 1,000,000 you close (the ones opened) and open (the ones closed) that are multiple of 1,000,000

How many open doors do you have at the end?

1,000,000.

edit

1,000

1

Edit: Misread; I should've known that would be too easy.

0

Ugh, I'm not good at this type of thing. The only thing I'm sure of is that all the prime-numbered doors would be closed. But that's easy.

Okay... so basically if something has an even number of factors, it's closed. If it has an odd number of factors, it's open. Not sure how you could devise a way to do figure that out without having to systematically go through each door and figure out if it's open or closed; which is something I think the million is trying to discourage.

Uh, this seems impossible without chopping everything up with a chain of automated reiterations.

The first turn will turn everything (including the first door) open. Subsequent operations no long touch doors they already passed, thus certain (e.g the first door) or more precise every door up to 999th retains its state past a certain point.
The open/close states depend on the door's total number of factors, which so far I haven't seen the pattern for. Even the primes pose enough of a problem for me personally.

Yeah, the 1,000,000 is supposed to discourage force-solving.

If you really want discouragement, use 10^90 doors.

69.

Easy, all perfect squares of the form f(x) = x^2 are left open where x is an integer greater than 0.

Take the square root of the number, in this case it's 1*10^6, which simpliy equates to 1,000. In the case of a decimal, round down.

So what Jean/mondays said.

I see, whatever number of factors x has, the number of factors of f(x) is made even with the square. But how do you see the reverse, if X had an even number of factors, it is a perfect square?

Even prime numbers would be closed, since they can be a factor of 1 and that prime number.

So the only doors that are open are: f(x)=x^2 where x is an element of the natural number.

You still have all doors, they never went anywhere, you jsut opened and closed them.

Easy, all perfect squares of the form f(x) = x^2 are left open where x is an integer greater than 0.

Take the square root of the number, in this case it's 1*10^6, which simpliy equates to 1,000. In the case of a decimal, round down.

So what Jean/mondays said.

=3

You still have all doors, they never went anywhere, you jsut opened and closed them.

He won.

You still have all doors, they never went anywhere, you jsut opened and closed them.

How many open doors do you have at the end?

Meep.

Meep.

Damn, read that wrong then.

You still have all doors, they never went anywhere, you jsut opened and closed them.

He won.

...

1,000,000.

i was right fir--

Meep.

DAMN YOU.

5. I got lazy.