What are the divisors of 520?

1, 2, 4, 5, 8, 10, 13, 20, 26, 40, 52, 65, 104, 130, 260, 520

There is a total of 16 positive divisors.
The sum of these divisors is 1260.
The arithmetic mean is 78.75.

12 even divisors

2, 4, 8, 10, 20, 26, 40, 52, 104, 130, 260, 520

4 odd divisors

1, 5, 13, 65

How to compute the divisors of 520?

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 520 by each of the numbers from 1 to 520 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)

520 / 1 = 520 (the remainder is 0, so 1 is a divisor of 520)
520 / 2 = 260 (the remainder is 0, so 2 is a divisor of 520)
520 / 3 = 173.33333333333 (the remainder is 1, so 3 is not a divisor of 520)
...
520 / 519 = 1.0019267822736 (the remainder is 1, so 519 is not a divisor of 520)
520 / 520 = 1 (the remainder is 0, so 520 is a divisor of 520)

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 520 (i.e. 22.803508501983). 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 $\frac{N}{d} = D$ )

520 / 1 = 520 (the remainder is 0, so 1 and 520 are divisors of 520)
520 / 2 = 260 (the remainder is 0, so 2 and 260 are divisors of 520)
520 / 3 = 173.33333333333 (the remainder is 1, so 3 is not a divisor of 520)
...
520 / 21 = 24.761904761905 (the remainder is 16, so 21 is not a divisor of 520)
520 / 22 = 23.636363636364 (the remainder is 14, so 22 is not a divisor of 520)