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

Guest Oct 17, 2022

#1**0 **

(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32)

(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024) =perfect squares of above 32 numbers

There are only 2 perfect squares that can be expressed as 6th powers of some integers. They are:

64 =2^6 and 729 =3^6

Therefore, the probability is: 2 / 32 =**1 / 16**

Guest Oct 17, 2022

#2**0 **

Here is my attempt:

1024^2 ==1,048,576

There are: 1^6, 2^6, 3^6, 4^6, 5^6, 6^6, 7^6, 8^6, 9^6, 10^6 =10 perfect squares that can be expressed as 6th powers of some integers.

Since there are 1024 perfect squares, then the probability is: [10 / 1024] * [9 / 1023] =90 / 1,047,552=** 15 / 174592**

Guest Oct 17, 2022