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Two integers are selected from the first 1024 positive integral perfect squares. What is the probability that both of these numbers are fifth powers of integers?

 Sep 27, 2022
 #1
avatar+118132 
+1

Firstly, it would be polite to respond after each time someone answers a question. Especially when it is a 'real' answer.

** You already asked this question in its non-proofed form and I answered.

 

Secondly, if you copy the whole question into google you will get several readily available answers.

 Sep 27, 2022
 #2
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+1

You did it again Melody. Called someone out for something not true, and provided a non-workable solution. All solutions on google require for you to pay. 

 Sep 28, 2022
 #3
avatar+118132 
+1

This is a post I was referring to.  It is rude to not respond and then just repost.

https://web2.0calc.com/questions/probability_37639#r1

 

Yes, the second part of my comment may not be correct. I appologise for that.

Melody  Sep 28, 2022
 #4
avatar+118132 
+1

 

1024^2 = 1048576

1048576^(1/5) = 16

 

 

4/1024  * 3/1023 =  12/1047552 = 3/261888     

 

EDIT:  It is only my last disision that is incorrect    12 /1047552 =  1/87296  NOW our answers agree

Thanks Builderboi.    Oh, and your answer is a lot more elegent than mine is  wink  cool

 Sep 28, 2022
edited by Melody  Sep 29, 2022
edited by Melody  Sep 29, 2022
 #5
avatar+2448 
+1

I got a different answer, Melody.

 

If a number is both a square and a 5th power, it must also be a perfect 10th power.

 

Note that \(1024^2 = 4^{10}\), so there are 4 numbers that work: \(1^{10}\)\(2^{10}\)\(3^{10}\), and \(4^{10}\)

 

There are \({4 \choose 2} = 6\) successes and \({1024 \choose 2} = 523776\) total outcomes.

 

So, the probability is \({6 \over 523776} = \color{brown}\boxed{1 \over 87296}\)

 Sep 29, 2022

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