What are the divisors of 5985?

1, 3, 5, 7, 9, 15, 19, 21, 35, 45, 57, 63, 95, 105, 133, 171, 285, 315, 399, 665, 855, 1197, 1995, 5985

There is a total of 24 positive divisors.
The sum of these divisors is 12480.
The arithmetic mean is 520.

24 odd divisors

1, 3, 5, 7, 9, 15, 19, 21, 35, 45, 57, 63, 95, 105, 133, 171, 285, 315, 399, 665, 855, 1197, 1995, 5985

How to compute the divisors of 5985?

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 5985 by each of the numbers from 1 to 5985 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)

5985 / 1 = 5985 (the remainder is 0, so 1 is a divisor of 5985)
5985 / 2 = 2992.5 (the remainder is 1, so 2 is not a divisor of 5985)
5985 / 3 = 1995 (the remainder is 0, so 3 is a divisor of 5985)
...
5985 / 5984 = 1.0001671122995 (the remainder is 1, so 5984 is not a divisor of 5985)
5985 / 5985 = 1 (the remainder is 0, so 5985 is a divisor of 5985)

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 5985 (i.e. 77.362781749366). 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 $\frac{N}{d} = D$ )

5985 / 1 = 5985 (the remainder is 0, so 1 and 5985 are divisors of 5985)
5985 / 2 = 2992.5 (the remainder is 1, so 2 is not a divisor of 5985)
5985 / 3 = 1995 (the remainder is 0, so 3 and 1995 are divisors of 5985)
...
5985 / 76 = 78.75 (the remainder is 57, so 76 is not a divisor of 5985)
5985 / 77 = 77.727272727273 (the remainder is 56, so 77 is not a divisor of 5985)