What are the divisors of 2560?

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

There is a total of 20 positive divisors.
The sum of these divisors is 6138.
The arithmetic mean is 306.9.

18 even divisors

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

2 odd divisors

1, 5

How to compute the divisors of 2560?

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 2560 by each of the numbers from 1 to 2560 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)

2560 / 1 = 2560 (the remainder is 0, so 1 is a divisor of 2560)
2560 / 2 = 1280 (the remainder is 0, so 2 is a divisor of 2560)
2560 / 3 = 853.33333333333 (the remainder is 1, so 3 is not a divisor of 2560)
...
2560 / 2559 = 1.0003907776475 (the remainder is 1, so 2559 is not a divisor of 2560)
2560 / 2560 = 1 (the remainder is 0, so 2560 is a divisor of 2560)

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 2560 (i.e. 50.596442562694). 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 $\frac{N}{d} = D$ )

2560 / 1 = 2560 (the remainder is 0, so 1 and 2560 are divisors of 2560)
2560 / 2 = 1280 (the remainder is 0, so 2 and 1280 are divisors of 2560)
2560 / 3 = 853.33333333333 (the remainder is 1, so 3 is not a divisor of 2560)
...
2560 / 49 = 52.244897959184 (the remainder is 12, so 49 is not a divisor of 2560)
2560 / 50 = 51.2 (the remainder is 10, so 50 is not a divisor of 2560)