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### Finding the Sum of the Factors of a Number

```Date: 07/25/2007 at 15:04:55
Subject: Sum of Factors

Is there a formula to find the SUM of all the factors?

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Date: 07/25/2007 at 23:00:04
From: Doctor Greenie
Subject: Re: Sum of Factors

I assume you mean a formula for finding the sum of the factors of a
given number.

Yes, there is a formula, or perhaps it should be called an algorithm.
It is difficult to put it into the form of a formula, but the process
is easy to understand.

Let's look at a relatively simple example to understand how the
"formula" works.  Consider the number 18.  Its factors are

1 2 3 6 9 18

The sum of these factors is 39.  We will find a way to find this sum
using the prime factorization of the number (without finding all the

The prime factorization of 18 is

(2^1)(3^2)

Let's write the factors of 18 again, expressing each one as a
product of powers of the prime factors 2 and 3:

1 = (2^0)(3^0)
2 = (2^1)(3^0)
3 = (2^0)(3^1)
6 = (2^1)(3^1)
9 = (2^0)(3^2)
18 = (2^1)(3^2)

Now let's use these forms of the factors to write the sum of all the
factors; then we will rearrange the terms in that sum to develop a
formula for finding the sum WITHOUT finding all the factors and

sum of factors = (2^0)(3^0) + (2^1)(3^0) + (2^0)(3^1) +
(2^1)(3^1) + (2^0)(3^2) + (2^1)(3^2)

sum of factors = (2^0)(3^0) + (2^0)(3^1) + (2^0)(3^2) +
(2^1)(3^0) + (2^1)(3^1) + (2^1)(3^2)

sum of factors = (2^0)(3^0 + 3^1 + 3^2) + (2^1)(3^0 + 3^1 + 3^2)

sum of factors = (2^0 + 2^1)(3^0 + 3^1 + 3^2)

sum of factors = (1 + 2)(1 + 3 + 9)

sum of factors = (3)(13) = 39

Look at the equation as it stands on the next-to-last line above.  The
first factor in the expression on the right is the sum of the powers
of 2 up to the first power; the second factor is the sum of the powers
of 3 up to the second power (as is easily seen in the preceding line).
When we multiply these two expressions together, we get terms which
consist of exactly every possible combination of powers of 2 from 0 to
1 and powers of 3 from 0 to 2.  But those combinations are exactly the
factors of the original number.

So the process for finding the sum of all the factors of the number
consists of the following steps:

(1) find the prime factorization
(2) form a product in which each term is the sum of all the powers
of one of the prime factors up to the exponent on that prime
factor in the prime factorization

That product is the sum of the factors of the number.

Here are a couple of more examples....

(A)  28...

(You may know that this is a perfect number; the sum of its PROPER
factors is the number itself--so the sum of ALL the factors should
be twice the number.)

28 = (2^2)(7^1)

sum of factors = (1 + 2 + 4)(1 + 7) = (7)(8) = 56

(B)  200...

200 = (2^3)(5^2)

sum of factors = (1 + 2 + 4 + 8)(1 + 5 + 25) = (15)(31) = 465

(C)  9000...

9000 = (2^3)(3^2)(5^3)

sum of factors = (1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5 + 25 + 125) =
(15)(13)(156) = 30420

I hope this helps.  Please write back if you have any further

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/

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Date: 07/25/2007 at 23:46:38
Subject: Thank you (Sum of Factors)

Thank you very much, Doctor Greenie.  More power to you.

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Associated Topics:
High School Number Theory

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