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#1 2012-11-16 12:58:27

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Question on squares

I just want to to know how to solve this problem:

Maya lists all the positive divisors of 2012^2. She then randomly selects 2 of them divisiors. Let p be the probability that exactly one of the 2 divisors is a square. p can be expressed as m/n where m and n are relatively primes numbers. Find m+n.

I solved to a part that the p(of a square) 16/81 and p(of non-sqaure) 65/81. What to do next???????

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#2 2012-11-16 13:21:50

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

Hi;

She then randomly selects 2 of them divisiors.

With replacement or without?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#3 2012-11-16 13:36:17

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Re: Question on squares

The 2 divisors do not have replacement, and the 2 divisors are distinct, thanks.

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#4 2012-11-16 13:45:03

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

Hi;

Does your list of divisors look like this?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#5 2012-11-16 13:50:55

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Re: Question on squares

Actually, the number of divisors that 2012^2 has is (2+1)^4, the problem is to find the sum of the numerator+denomerator of the probability of the chance that only ONE of the two divisors is a perfect square, the number of divisors that 2012 has is (1+1)^4 and this is also the number of perfect square divisors 2012^2 has, I'm very bad at probability and can't figure what is the probability of only one of the divisors being a perfect square.

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#6 2012-11-16 13:55:19

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

Hi;

Can you post the divisors of 2012 ^2 please?

I am getting 16 and proper divisors only 15.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#7 2012-11-16 14:13:04

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Re: Question on squares

I can't list them all, let's see.....: 1,2,4,16,253009,1012036, 2024072,4048144, these are only the half of the perfect square factors of 2012^2, there are 8 more of them, how to solve the problem goes like this: (2+1)^4 is the number of factors that 2012^2 has and (1+1)^4 is the number of factors that 2012^2 has that are perfect squares, so the probability is [2*2^4*(3^4-2^4)]/[3^4(3^4-1)=26/81 so the answer is 26+81=107, which is m+n, this looks more like a bunch of random numbers placed together and I personally think that knowing the answer with knowing how to do it is useless, so I just need an explanation of this, you are only given 12 minutes to do this, I don't think listing out all the positive divisors of 2012^2 is a efficient idea!

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#8 2012-11-16 14:15:47

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

I am not claiming it is an efficient idea but I have to know the sample space before we can get the probability. I am getting only 15 divisors of 2012^2 not 81. That is why I am asking for your list.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#9 2012-11-16 14:31:55

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Re: Question on squares

It does look weird that 2012^2 has that many factors, but 2012 has a prime factorization of 2*3*5*67, so 2012^2 has a prime factorization of 2^2*3^2*5^2*67^2 and that leads to the fact that (2+1)^4 is the number of factors that 2012^2 has, though only 16 of them are perfect squares, so the question is simplified to what is the probability of getting only 1 perfect square out of 2 if the probability of getting a square is 16/81. I also doubt that 2012^2 has 81 factors, but that's what the formula calculates, but I think I get the question now, but anyways, sorry to bother you with all these random things I said above, the only reason why I want to have the answer to this problem is because it fascinates me and is a really good probability problem for me.

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#10 2012-11-16 14:37:51

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

Hi;

That is not correct 2012^2 only has 15 factors not 81. The list I gave in post #4 are the only positive factors.

so 2012^2 has a prime factorization of 2^2*3^2*5^2*67^2

You have factored 2012^2 incorrectly.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#11 2012-11-16 14:46:07

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Re: Question on squares

OH-NO, sorry! I got the whole problem wrong! It's suppose be 2010, not 2012 and 2010^2 has 81 factors, I can't believe that just a difference of 2 can make such a huge difference, now I hope that this problem is much more clear! I actually though that 5 was a factor of 2012!

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#12 2012-11-16 14:47:22

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

Hi;

Yes, 2010^2 has 81 factors.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#13 2012-11-16 14:52:54

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Re: Question on squares

Because this problem was in a 2010 contest so though I still thought that it was 2012, but now the problem seems to make much more sense.

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#14 2012-11-16 14:56:25

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

I am getting 26 / 81 as the probability one of the numbers is a square.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#15 2012-11-16 14:58:49

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Re: Question on squares

Yes, that is correct, I'm just trying to understand the solution by myself, so if 26/81 is the probability, then m+n is equal to 26+81 which is 107, which is the final answer, but the solution is confusing, since it's all probability.

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#16 2012-11-16 15:02:58

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

Hi;


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#17 2012-11-16 15:05:44

zehao1000
Member
Registered: 2012-11-16
Posts: 34

Re: Question on squares

Oh, thanks! Now I get the whole thing.

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#18 2012-11-16 15:14:59

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

Re: Question on squares

Hi;

You are welcome!


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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