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?

MathyGoo13 Sep 27, 2022

#1**+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.

Melody Sep 27, 2022

#2**+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.

Guest Sep 28, 2022

#3**+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**+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

Melody Sep 28, 2022

#5**0 **

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 \({6 \over 523776} = \color{brown}\boxed{1 \over 87296}\)

BuilderBoi Sep 29, 2022