What are the divisors of 6039?

1, 3, 9, 11, 33, 61, 99, 183, 549, 671, 2013, 6039

There is a total of 12 positive divisors.
The sum of these divisors is 9672.
The arithmetic mean is 806.

12 odd divisors

1, 3, 9, 11, 33, 61, 99, 183, 549, 671, 2013, 6039

How to compute the divisors of 6039?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 6039 by each of the numbers from 1 to 6039 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

6039 / 1 = 6039 (the remainder is 0, so 1 is a divisor of 6039)
6039 / 2 = 3019.5 (the remainder is 1, so 2 is not a divisor of 6039)
6039 / 3 = 2013 (the remainder is 0, so 3 is a divisor of 6039)
...
6039 / 6038 = 1.0001656177542 (the remainder is 1, so 6038 is not a divisor of 6039)
6039 / 6039 = 1 (the remainder is 0, so 6039 is a divisor of 6039)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 6039 (i.e. 77.711003081932). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

6039 / 1 = 6039 (the remainder is 0, so 1 and 6039 are divisors of 6039)
6039 / 2 = 3019.5 (the remainder is 1, so 2 is not a divisor of 6039)
6039 / 3 = 2013 (the remainder is 0, so 3 and 2013 are divisors of 6039)
...
6039 / 76 = 79.460526315789 (the remainder is 35, so 76 is not a divisor of 6039)
6039 / 77 = 78.428571428571 (the remainder is 33, so 77 is not a divisor of 6039)