# Thread: problem involving sum of divisors

1. ## problem involving sum of divisors

studying for finals and enountered this problem on a practice test.
If n is a natural number such that n = 2^k(2^(k+1) - 1) with 2^(k+1) -1 prime then 2n = sum of the divisors.

2. Originally Posted by ChrisBickle
studying for finals and enountered this problem on a practice test.
If n is a natural number such that n = 2^k(2^(k+1) - 1) with 2^(k+1) -1 prime then 2n = sum of the divisors.

What are the divisors of $\displaystyle n=2^k\left(2^{k+1}-1\right)$? Let's write them down:

** $\displaystyle 1,2,2^2,\ldots,2^k$

** $\displaystyle 2^{k+1}-1$

** $\displaystyle 2^{k+2}-2,2^{k+3}-2^2,\ldots,2^{2k+1}-2^k$

Well, let's sum up the bastards above:

$\displaystyle \sum\limits_{i=0}^k2^i+\sum\limits_{j=0}^k\left(2^ {k+1+j}-2^k\right)$ $\displaystyle =2^{k+1}\sum\limits_{j=0}^k2^j=2^{k+1}\,\frac{2^{k +1}-1}{2-1}=2^{k+1}(2^{k+1}-1)$ ...

Tonio

3. awesome..everytime you see a solution it looks so easy...im convinced math is to make smart people feel stupid

4. that is all the even perfect number of the form $\displaystyle 2^{n-1}(2^n-1) \text{provide that} 2^n-1 \text{is prime}$.

5. Originally Posted by Shanks
that is all the even perfect number of the form $\displaystyle 2^n(2^n-1) \text{provide that} 2^n-1 \text{is prime}$.

Perhaps you meant to say: an even natural number is perfect iff it is of the form $\displaystyle 2^n\left(2^{n+1}-1\right)$ AND $\displaystyle 2^{n+1}-1$ is prime .

Tonio

6. oooops,,,sorry, type error...