Sum of factors of 1024?

Question

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.

Answer 1

1024 = 2^10.

The factors have to be a power of two, so just add up the powers of 2 from 2^0 to 2^10:

1 + 2 + 4 + ... + 256 + 512 + 1024.



You don't have to actually add the 11 numbers, since it's a geometric series, and you can sum that by formula.

1 + a + a^2 + ... + a^n = (a^(n+1) - 1) / (a - 1)

So you have

(2^11 - 1) / (2 - 1) = (2048 - 1)/(2-1) = 2047.



If you add up the powers of 2 one at a time what do you notice?

1 + 2 = 3

3 +4 = 7

7 + 8 = 15...

At each step, you get 1 less than the next highest power. So you could solve this by looking at that pattern.



Note getting the sum of all the factors of any number is easy if you have the prime factors.

If n = p^a q^b, the sum of the factors is:

(p^(a+1) - 1)/(p-1) times (q^(b+1) - 1)/(q-1).



Try that on 2^2 3^2 = 36

The factors are 1 2 3 4 6 9 12 18 36.

Those add up to 91

(8-1)/1 times (27-1)/2

= 7 times 13 = 91.



You can expand that formula to any number of primes. It works because, for 36 for example), the sum of the factors = (1 + 2 + 4)(1 + 3 + 9) = 7 x 13 as noted above. Expand those terms and you have all possible factors of 36, because you've including the product of all possible prime powers up to the highest power in 36.



I hope that's not too much information. If you go over what I wrote a few times, it should become pretty clear.

Answer 2

Since 1024 is even then divide by '1'

2(1024

2)512)

2)256

2)128

2)64

2)32

2)16

2)8

2)4

2)2

= 1



Sum = 2 x 10 = 20 is the sum of the factors.



?? I don't know where you got '20147' from ????

Answer 3

Factor into powers of prime numbers

1024 = 2¹⁰

1 is also a factor



∑ = 1 + 2¹ + 2² + ... + 2¹⁰

Answer 4

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