math question...need help

[twankerr]

inutively i can see that all the prime numbers will be closed, right? since the first dude opens all the lockers...the only people to close the "prime" lockers will be the primes themselves. im too lazy to care about the rest of the lockers though, non-primes disinterest me.

[jason_t]

[ QUOTE ]
inutively i can see that all the prime numbers will be closed, right? since the first dude opens all the lockers...the only people to close the "prime" lockers will be the primes themselves. im too lazy to care about the rest of the lockers though, non-primes disinterest me.

[/ QUOTE ]

This is not correct.

[Scotch78]

[ QUOTE ]
inutively i can see that all the prime numbers will be closed, right? since the first dude opens all the lockers...the only people to close the "prime" lockers will be the primes themselves. im too lazy to care about the rest of the lockers though, non-primes disinterest me.


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This is not correct.

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His reasoning sounds good to me, so I'm curious why you think he's wrong?

Scott

[RJT]

Jason,

LOL. You are a ballbuster. You are the NotReady of the Math threads. Your style I mean. (I mean that as a compliment or rather I don’t mean it as a negative criticism.)

RJT

[jason_t]

[ QUOTE ]
[ QUOTE ]
inutively i can see that all the prime numbers will be closed, right? since the first dude opens all the lockers...the only people to close the "prime" lockers will be the primes themselves. im too lazy to care about the rest of the lockers though, non-primes disinterest me.


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This is not correct.

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His reasoning sounds good to me, so I'm curious why you think he's wrong?

Scott

[/ QUOTE ]

The primes will be closed that is correct; the non-primes are not "uninteresting" is the statement I had issue with. The correct answer is that door n will be open iff n is a perfect square.

The reason the primes are closed is because they have two divisors. In general, any integer with an even number of divisors will be closed. For any integer that is not a perfect square, if a is a divisor of n then there is an integer b so that a * b = n and a will not be equal to b. Thus the divisors of n can be paired off giving that n has an even number of divisors. Perfect squares n = k^2 have the extra divisor k that can't be paired off with enough divisor. So they have an odd number of divisors. Thus, only the perfect squares will be open.

[Farfenugen]

Oh I see, your counting the two guy as doing something. I was unclear on that.

[lotus776]

yeah, I was talking with my buddy today and he was trying to give me a hint b/c he solved it a little over a year ago...The perfect squares idea is something I'm going to explore more. I was originally trying to find a sequence, now I have a better idea

thanks

[eviljeff]

brute force it.

if you don't have a compiler, hopefully you have a janitor buddy

[jason_t]

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Oh I see, your counting the two guy as doing something. I was unclear on that.

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After years of experience in solving math problems, it's clear that this what was intended.

[ChromePony]

I remember doing this problem in sixth grade, it stumped us for a while but we came around eventually.

All the squares should be left open since theyre the only ones with an odd number of factors.

EDIT: oops I missed where jason actually answered it already...I thought he was just being a smart ass [img]/images/graemlins/grin.gif[/img]