Perfect NumbersDate: 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/ |
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