What are the divisors of 320?

1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160, 320

There is a total of 14 positive divisors.
The sum of these divisors is 762.
The arithmetic mean is 54.428571428571.

12 even divisors

2, 4, 8, 10, 16, 20, 32, 40, 64, 80, 160, 320

2 odd divisors

1, 5

How to compute the divisors of 320?

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 320 by each of the numbers from 1 to 320 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)

320 / 1 = 320 (the remainder is 0, so 1 is a divisor of 320)
320 / 2 = 160 (the remainder is 0, so 2 is a divisor of 320)
320 / 3 = 106.66666666667 (the remainder is 2, so 3 is not a divisor of 320)
...
320 / 319 = 1.0031347962382 (the remainder is 1, so 319 is not a divisor of 320)
320 / 320 = 1 (the remainder is 0, so 320 is a divisor of 320)

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 320 (i.e. 17.888543819998). 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 $\frac{N}{d} = D$ )

320 / 1 = 320 (the remainder is 0, so 1 and 320 are divisors of 320)
320 / 2 = 160 (the remainder is 0, so 2 and 160 are divisors of 320)
320 / 3 = 106.66666666667 (the remainder is 2, so 3 is not a divisor of 320)
...
320 / 16 = 20 (the remainder is 0, so 16 and 20 are divisors of 320)
320 / 17 = 18.823529411765 (the remainder is 14, so 17 is not a divisor of 320)