Perfect Numbers

Date: 06/15/2002 at 15:27:29
From: Trung Ho
Subject: perfect numbers

Please show that any even perfect number ends in 6 or 8.          

Date: 06/17/2002 at 07:32:00
From: Doctor Floor
Subject: Re: perfect numbers

Hi Trung Ho,

Thanks for your question.

You should start by reading the following page from the Dr. Math 
archives:

  Perfect numbers
  http://mathforum.org/dr.math/faq/faq.perfect.html 

On this page you will find that any even perfect numbe r is of the form

   p(n) = 2^(n-1)* (2^n - 1)

where 2^n - 1 must be a prime number.

Now if we know the final digit of 2^(n-1), then it is easy to find 
the final digit for 2^n and 2^n-1 and consequently p(n). We find


          final digits table
   2 ^(n-1) |  2^n   | 2^n - 1 |    p(n)
   --------+--------+---------+----------
     2     |   4    |    3    | 2*3 -> 6
     4     |   8    |    7    | 4*7 -> 8
     6     |   2    |    1    | 6*1 -> 6
     8     |   6    |    5    | 8*5 -> 0

The table supports your theorem except when the final digit of 2^(n-1)
is an 8.  But in that case, luckily, we see that p(n) is a multiple of
10. And thus 2^n - 1 must be a multiple of 5, and since p^n - 1 must
be a prime, we see that 2^n - 1 = 5 and 2^n = 6 which is impossible.

So indeed any even perfect number ends in 6 or 8.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/