iplraf wrote:The divisors of a natural number, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect.
Now, let P is a number of the form [2^(n-1)]*[(2^n)-1]. Is P an even perfect number?
(1) [(2^n)-1] is an odd number.
(2) [(2^n)-1] is a prime number.
We'll need to use the following:
2 + 2 + 2^2 + 2^3 + ... + 2^x = 2^(x+1)
You can see that this is true by adding from the left: 2+2 = 2^2, and 2^2 + 2^2 = 2^3, and so on. So we have, subtracting 1 from both sides above:
1+ 2 + 2^2 + 2^3 + ... + 2^x = 2^(x+1) - 1
Onto the question:
Notice first that 1) really doesn't tell us much- it only tells us that n is greater than 1.
If we assume 2) is true, then 2^n - 1 is prime. Let's call it p for now. We want to know if
p*2^(n-1)
is prime. Notice the above is a prime factorization. What are its proper divisors? Well, it has divisors that are not divisible by p:
1, 2, 2^2, 2^3, ..., 2^(n-1)
and these add to (2^n) - 1, by the result at the start of my post.
It also has divisors that are divisible by p:
p , 2*p, (2^2)*p, (2^3)*p, ..., (2^(n-2))*p
and these add to (again using the result from the start of this post)
p(1 + 2 + 2^2 + ... + 2^(n-2) ) = (2^(n-1) - 1)*p
So the sum of all the divisors is
[(2^n) - 1] + [(2^(n-1) - 1]*p
Plug back (2^n)-1 for p:
(2^n) - 1 + [(2^(n-1) - 1]*[(2^n) - 1]
= 2^n - 1 + [2^(n-1)]*[(2^n) - 1] - 2^n + 1
= [2^(n-1)]*[(2^n)-1]
So the number is perfect, and Statement 2 is sufficient.
Notice that if 2^n - 1 isn't prime, we'll have extra divisors, and the sum of the divisors will be larger, so statement 1 is not sufficient.
And, I think it's a pretty difficult question to complete in two minutes, at least if you don't know the result in advance. Where is it from?
