What are the divisors of 5800?

1, 2, 4, 5, 8, 10, 20, 25, 29, 40, 50, 58, 100, 116, 145, 200, 232, 290, 580, 725, 1160, 1450, 2900, 5800

There is a total of 24 positive divisors.
The sum of these divisors is 13950.
The arithmetic mean is 581.25.

18 even divisors

2, 4, 8, 10, 20, 40, 50, 58, 100, 116, 200, 232, 290, 580, 1160, 1450, 2900, 5800

6 odd divisors

1, 5, 25, 29, 145, 725

How to compute the divisors of 5800?

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 5800 by each of the numbers from 1 to 5800 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)

5800 / 1 = 5800 (the remainder is 0, so 1 is a divisor of 5800)
5800 / 2 = 2900 (the remainder is 0, so 2 is a divisor of 5800)
5800 / 3 = 1933.3333333333 (the remainder is 1, so 3 is not a divisor of 5800)
...
5800 / 5799 = 1.0001724435247 (the remainder is 1, so 5799 is not a divisor of 5800)
5800 / 5800 = 1 (the remainder is 0, so 5800 is a divisor of 5800)

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 5800 (i.e. 76.157731058639). 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 $\frac{N}{d} = D$ )

5800 / 1 = 5800 (the remainder is 0, so 1 and 5800 are divisors of 5800)
5800 / 2 = 2900 (the remainder is 0, so 2 and 2900 are divisors of 5800)
5800 / 3 = 1933.3333333333 (the remainder is 1, so 3 is not a divisor of 5800)
...
5800 / 75 = 77.333333333333 (the remainder is 25, so 75 is not a divisor of 5800)
5800 / 76 = 76.315789473684 (the remainder is 24, so 76 is not a divisor of 5800)