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I got really good scores on the Mathematicas, I got a medal from Byron-Germain and Pythagoras, they were really easy, I did the AMC 3 days ago, and doing my second Gauss this year.
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.
I'm better at geometry, I got 2 medals from Mathematica contests, and I'm very bad at probability, always ending being a fail at it, can't understand when to use the formulaes in differents problems, in fact, I don't really even understand matrixes!
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!
Thanks for the invitation, just the place to improve math skills.
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.
Hi guys, looks like a pretty good forum, hopefully I can learn a lot from here, thanks for creating this forum.:)
The 2 divisors do not have replacement, and the 2 divisors are distinct, thanks.
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???????