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?
(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
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
