What are the divisors of 960?

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 64, 80, 96, 120, 160, 192, 240, 320, 480, 960

There is a total of 28 positive divisors.
The sum of these divisors is 3048.
The arithmetic mean is 108.85714285714.

24 even divisors

2, 4, 6, 8, 10, 12, 16, 20, 24, 30, 32, 40, 48, 60, 64, 80, 96, 120, 160, 192, 240, 320, 480, 960

4 odd divisors

1, 3, 5, 15

How to compute the divisors of 960?

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 960 by each of the numbers from 1 to 960 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)

960 / 1 = 960 (the remainder is 0, so 1 is a divisor of 960)
960 / 2 = 480 (the remainder is 0, so 2 is a divisor of 960)
960 / 3 = 320 (the remainder is 0, so 3 is a divisor of 960)
...
960 / 959 = 1.0010427528676 (the remainder is 1, so 959 is not a divisor of 960)
960 / 960 = 1 (the remainder is 0, so 960 is a divisor of 960)

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 960 (i.e. 30.983866769659). 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 $\frac{N}{d} = D$ )

960 / 1 = 960 (the remainder is 0, so 1 and 960 are divisors of 960)
960 / 2 = 480 (the remainder is 0, so 2 and 480 are divisors of 960)
960 / 3 = 320 (the remainder is 0, so 3 and 320 are divisors of 960)
...
960 / 29 = 33.103448275862 (the remainder is 3, so 29 is not a divisor of 960)
960 / 30 = 32 (the remainder is 0, so 30 and 32 are divisors of 960)