What are the divisors of 539?

1, 7, 11, 49, 77, 539

There is a total of 6 positive divisors.
The sum of these divisors is 684.
The arithmetic mean is 114.

6 odd divisors

1, 7, 11, 49, 77, 539

How to compute the divisors of 539?

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 539 by each of the numbers from 1 to 539 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)

539 / 1 = 539 (the remainder is 0, so 1 is a divisor of 539)
539 / 2 = 269.5 (the remainder is 1, so 2 is not a divisor of 539)
539 / 3 = 179.66666666667 (the remainder is 2, so 3 is not a divisor of 539)
...
539 / 538 = 1.0018587360595 (the remainder is 1, so 538 is not a divisor of 539)
539 / 539 = 1 (the remainder is 0, so 539 is a divisor of 539)

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 539 (i.e. 23.216373532488). 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 $\frac{N}{d} = D$ )

539 / 1 = 539 (the remainder is 0, so 1 and 539 are divisors of 539)
539 / 2 = 269.5 (the remainder is 1, so 2 is not a divisor of 539)
539 / 3 = 179.66666666667 (the remainder is 2, so 3 is not a divisor of 539)
...
539 / 22 = 24.5 (the remainder is 11, so 22 is not a divisor of 539)
539 / 23 = 23.434782608696 (the remainder is 10, so 23 is not a divisor of 539)