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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

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

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

Hi;

Yes, 2010^2 has 81 factors.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

Hi;

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Oh, thanks! Now I get the whole thing.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,204

Hi;

You are welcome!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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