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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 sixth powers of integers?

 Oct 17, 2022
 #1
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(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

 Oct 17, 2022
 #2
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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

 Oct 17, 2022

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