Sum of factors of 1024

How do you find the sum of all the factors of 1024?

The answer is 2047 but I'm not sure how to get there.

1024 = 2^10

The factor of 1024 are a geometric sequence :
n = 1 , A1 = 2^0
n = 2 , A2 = 2^1
n = 3 , A3 = 2^2
...
n = 9 , A9 = 2^8
n = 10 , A10 = 2^9
n = 11 , A11 = 2^10

A1 = 1
r = 2
An = A1 * r^(n-1) = 2^(n - 1)
Sn = A1(r^n - 1) / (r - 1) = 1(2^11 - 1) / (2-1) = 2^11 - 1 = 2048 - 1 = 2047

what seems to be the problem.

the sum is 2047.

1024
512
256
128
064
032
016
008
004
002
001
---------
2047



sum of unit's digits = 47 so 4 carried to tens.

tens sum is20 + 4 = 24 so 2 carried over to hundreds

hundreds sum is 8+2 = 10 so 1 carried over to thou.which is 1+1=2
--------

2^10 = 1024.
So that is ten 2's in a row. So add up ten 2's. That would be 20. Then add a 1, because that is a factor. Now we have 21. Hmmmm. Problem. I don't think 2047 is correct. What did you get?
OOOOPS, the other answers are right.

Start with 1, and list all the factor pairs:
1, 1024
2, 512
4, 256
8, 128
16, 64
32, 32
Add them up, being careful to count 32 only once.

Jen

512, 2, 4, 256, 128 ,8, 16, 64, 32, 1024, 1

512 + 2 + 4 + 256 + 128 + 8 + 16 + 64 + 32 + 1024 + 1 = 2047

1, 1024 2, 512 4,256 8, 128 16, 64, 32

Then add em'.

~Kevin.