Question

Prove: If k is a positive integer and 2^(k) - 1 is prime, then 2^(k-1) * (2^(k)-1) is perfect.
(Note:An integer n is perfect if the sum of all of its positive divisors (including 1 and itself) is 2n.) @-------College Al…

Answer 1

if k is a positive integer and \[2^{k-1}\] is prime, then \[2^{k-1} (2^{k}-1)\] is perfect.

Answer 2

This is the clear thing to prove. :)

Answer 3

Do you mean
\[2^k-1\]
is prime?

Answer 4

yes :)

Answer 5

\[2^{k-1} (2^{k}-1)\]

\[2^{k-1} 2^{k} - 2^{k-1}\]
\[2^{k-1+1} - 2^{k-1}\]
\[2^{k} - 2^{k-1}\]
\[k=2n\]
\[2^{2n} - 2^{2n-1}\]
\[2^{2n}(1 - 2^{-1})\]
\[2^{2n}(1/2)\]
so far :)

Answer 6

ok

Answer 7

what does, "is perfect" mean?

Answer 8

according to the defintion i have,it says"An integer n is perfect if the sum of all of its positive divisors (including 1 and itself) is 2n."

Answer 9

\[2^{2n}(1/2)\]
\[2^{2n-1}\]
i gotta see if the wolf can dbl chk this

Answer 10

ok

Answer 11

nah, i made a mistake in it somplace ....

Answer 12

ok, thats alright.

Answer 13

the first few perfect numbers are 6, 28, 496, 8128, ...

6 = 1+2+3
28 = 1 + 2 + 4 + 7 + 14

a perfect number

Answer 14

Got it! I had fun with this one. LaTeX proof forthcoming...

Answer 15

http://mathworld.wolfram.com/PerfectNumber.html

this looks to have something similar on it

Answer 16

also, i read positive integer and thought "even" integer .... gotta get some new glasses

Answer 17

I thought i saw the same thing for like ten minutes too! lol

Answer 18

good luck with it :) Jem seems capable enough ;)

Answer 19

This hinges on the following sum that I had to derive:
\[\sum_{n=0}^{k-1}2^n = 2^k-1\]
My derivation was as follows:
\[S = \sum_{n=0}^{k-1} 2^n\]
\[2S = 2\sum_{n=0}^{k-1}2^n=\sum_{n=1}^{k}2^n = \sum_{n=0}^{k-1}2^n + 2^k - 1\]
So,
\[2S-S = S = \sum_{n=0}^{k-1}2^n + 2^k - 1 - \sum_{n=0}^{k-1}2^n = 2^k-1\]

From there, we can hit this problem. Since 2^k-1 is prime, we don't have to worry about its factors. All we need to do is sum up all the factors of k^{n-1}. Note, however, that we need to consider the following: imagine that k = 3. Then the factors of 2^2 =4 are 1, 2, and 4. However, the factors of 2^2*X are 1, 2, 4, X, 2X, and 4X because when we multiply two factors together the prime bit could be a part of either one. With that in mind, the sum of the factors of 4X is (1+2+4) + (X + 2X + 4X) =(1+2+4)(1+X). In an exactly analogous way, we must find the sum of the products of 2^(k-1) and multiply that by (1+(2^k-1)). But the sum of those factors is just what I derived above, so we have
\[ (\sum_{n=0}^{k-1}2^n )(1+2^k-1) = (2^k-1)(2^k) = 2[2^{k-1}(2^k-1)] \]
but since our original number was
\[2^{k-1}(2^k-1)\]
we see that it is perfect according to the definition.

Answer 20

Thanks! I'm going to study this for a while.

Answer 21

@amistre64 thanks for the compliment :)