What are the divisors of 3072?

1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768, 1024, 1536, 3072

There is a total of 22 positive divisors.
The sum of these divisors is 8188.
The arithmetic mean is 372.18181818182.

20 even divisors

2, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768, 1024, 1536, 3072

2 odd divisors

1, 3

How to compute the divisors of 3072?

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 3072 by each of the numbers from 1 to 3072 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)

3072 / 1 = 3072 (the remainder is 0, so 1 is a divisor of 3072)
3072 / 2 = 1536 (the remainder is 0, so 2 is a divisor of 3072)
3072 / 3 = 1024 (the remainder is 0, so 3 is a divisor of 3072)
...
3072 / 3071 = 1.0003256268317 (the remainder is 1, so 3071 is not a divisor of 3072)
3072 / 3072 = 1 (the remainder is 0, so 3072 is a divisor of 3072)

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 3072 (i.e. 55.425625842204). 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 $\frac{N}{d} = D$ )

3072 / 1 = 3072 (the remainder is 0, so 1 and 3072 are divisors of 3072)
3072 / 2 = 1536 (the remainder is 0, so 2 and 1536 are divisors of 3072)
3072 / 3 = 1024 (the remainder is 0, so 3 and 1024 are divisors of 3072)
...
3072 / 54 = 56.888888888889 (the remainder is 48, so 54 is not a divisor of 3072)
3072 / 55 = 55.854545454545 (the remainder is 47, so 55 is not a divisor of 3072)