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.


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.


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.


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?


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


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


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











Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

Latest posts by Dan (see all)

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *