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










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