What are the divisors of 5120?

1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2560, 5120

There is a total of 22 positive divisors.
The sum of these divisors is 12282.
The arithmetic mean is 558.27272727273.

20 even divisors

2, 4, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2560, 5120

2 odd divisors

1, 5

How to compute the divisors of 5120?

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 5120 by each of the numbers from 1 to 5120 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)

5120 / 1 = 5120 (the remainder is 0, so 1 is a divisor of 5120)
5120 / 2 = 2560 (the remainder is 0, so 2 is a divisor of 5120)
5120 / 3 = 1706.6666666667 (the remainder is 2, so 3 is not a divisor of 5120)
...
5120 / 5119 = 1.0001953506544 (the remainder is 1, so 5119 is not a divisor of 5120)
5120 / 5120 = 1 (the remainder is 0, so 5120 is a divisor of 5120)

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 5120 (i.e. 71.554175279993). 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 $\frac{N}{d} = D$ )

5120 / 1 = 5120 (the remainder is 0, so 1 and 5120 are divisors of 5120)
5120 / 2 = 2560 (the remainder is 0, so 2 and 2560 are divisors of 5120)
5120 / 3 = 1706.6666666667 (the remainder is 2, so 3 is not a divisor of 5120)
...
5120 / 70 = 73.142857142857 (the remainder is 10, so 70 is not a divisor of 5120)
5120 / 71 = 72.112676056338 (the remainder is 8, so 71 is not a divisor of 5120)