What are the divisors of 3048?

1, 2, 3, 4, 6, 8, 12, 24, 127, 254, 381, 508, 762, 1016, 1524, 3048

There is a total of 16 positive divisors.
The sum of these divisors is 7680.
The arithmetic mean is 480.

12 even divisors

2, 4, 6, 8, 12, 24, 254, 508, 762, 1016, 1524, 3048

4 odd divisors

1, 3, 127, 381

How to compute the divisors of 3048?

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 3048 by each of the numbers from 1 to 3048 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)

3048 / 1 = 3048 (the remainder is 0, so 1 is a divisor of 3048)
3048 / 2 = 1524 (the remainder is 0, so 2 is a divisor of 3048)
3048 / 3 = 1016 (the remainder is 0, so 3 is a divisor of 3048)
...
3048 / 3047 = 1.0003281916639 (the remainder is 1, so 3047 is not a divisor of 3048)
3048 / 3048 = 1 (the remainder is 0, so 3048 is a divisor of 3048)

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 3048 (i.e. 55.208694967369). 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 $\frac{N}{d} = D$ )

3048 / 1 = 3048 (the remainder is 0, so 1 and 3048 are divisors of 3048)
3048 / 2 = 1524 (the remainder is 0, so 2 and 1524 are divisors of 3048)
3048 / 3 = 1016 (the remainder is 0, so 3 and 1016 are divisors of 3048)
...
3048 / 54 = 56.444444444444 (the remainder is 24, so 54 is not a divisor of 3048)
3048 / 55 = 55.418181818182 (the remainder is 23, so 55 is not a divisor of 3048)