#1**+8 **

If, for some *k* > 1, 2^{k} - 1 is prime then 2^{k-1}(2^{k} - 1) is a perfect number

For instance, the first perfect number, 6 = 2^(2 - 1) (2^2 - 1) = 2^1 ( 3) = 6

So far, only even perfect numbers have been found......no one knows if there are any odd ones...if there are, the above formula would not applly......[it produces only even numbers ]

Here's a fascinating look at perfect numbers........http://www-history.mcs.st-and.ac.uk/HistTopics/Perfect_numbers.html

CPhill
Jun 15, 2015

#1**+8 **

Best Answer

If, for some *k* > 1, 2^{k} - 1 is prime then 2^{k-1}(2^{k} - 1) is a perfect number

For instance, the first perfect number, 6 = 2^(2 - 1) (2^2 - 1) = 2^1 ( 3) = 6

So far, only even perfect numbers have been found......no one knows if there are any odd ones...if there are, the above formula would not applly......[it produces only even numbers ]

Here's a fascinating look at perfect numbers........http://www-history.mcs.st-and.ac.uk/HistTopics/Perfect_numbers.html

CPhill
Jun 15, 2015