# Sum of Divisors

In number theory, many formulas exist that we never think of. The first topic I wrote about number theory is to find the number of divisors an integer has. This time let’s talk about finding the sum of those divisors without listing them individually.

Formula:

Given a positive integer $N$ with prime factor $(a^m)(b^n)(c^p)\ldots$ the sum of its divisors ( Sn) can be calculated using the formula, $S_n= (\sum_{x=0}^{m}a^m)( \sum_{n=0}^{n}a^n) ( \sum_{x=0}^{n}b^n) ( \sum_{p=0}^{p}c^p)\ldots$

Worked Problems 1:

Find the sum of divisors of 12.

Solution: $12=2^2(3)$ $S_n=(2^0+2^1+2^2)(3^0+3^1)$ $S_n=(7)(4) = 28$

It is easy to check that the answer is correct by listing down all the divisors of 12 which is {1,2,3,4,6,12}.

Worked Problem 2:

Find the sum of divisors of 5400.

Solution: $5400=2^3(3^3)(5^2)$ $S_n=(2^0+2^1+2^2+2^3)(3^0+3^1+3^2+3^3)(5^0+5^1+5^2)$ $S_n=(15)(40)(31)$ $S_n=18,600$

Worked Problem 3:

The sum of divisors of an integer N is 7623. N can be written in the form of 2x3y where x and y are all positive integers. What is N?

Solution: $S_n=(2^0+2^1+2^2+\ldots+2^x)(3^0+3^1+3^2+\ldots+3^y)$ $7623=(2^0+2^1+2^2+\ldots+2^x)(3^0+3^1+3^2+\ldots+3^y)$ $7623=(2^{x+1}-1)(\frac{3^{y+1}-1}{2})$ $15246=(2^{x+1}-1)( 3^{y+1}-1)$

We take note of the following facts; $2^{x+1}-1$ is odd and $3^{y+1}-1$ is even.

Also $15246=2(3^2)(7)(11^2)$

Since $3^{y+1}-1$ is even, 2 is one of its factors. And take note that x and y are positive integers. By inspection; $3^{y+1}-1 = 2(11^2) = 242$ $3^{y+1}= 243$ $y=4$

Consequently, $2^{x+1}-1=3^2(7)=63$ $2^{x+1}=64$ $x=5$

Since $N=2^x(3^y)$ $N=2^5(3^4)$ $N=2592$