What are the divisors of 3090?

1, 2, 3, 5, 6, 10, 15, 30, 103, 206, 309, 515, 618, 1030, 1545, 3090

There is a total of 16 positive divisors.
The sum of these divisors is 7488.
The arithmetic mean is 468.

8 even divisors

2, 6, 10, 30, 206, 618, 1030, 3090

8 odd divisors

1, 3, 5, 15, 103, 309, 515, 1545

How to compute the divisors of 3090?

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 3090 by each of the numbers from 1 to 3090 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)

3090 / 1 = 3090 (the remainder is 0, so 1 is a divisor of 3090)
3090 / 2 = 1545 (the remainder is 0, so 2 is a divisor of 3090)
3090 / 3 = 1030 (the remainder is 0, so 3 is a divisor of 3090)
...
3090 / 3089 = 1.0003237293623 (the remainder is 1, so 3089 is not a divisor of 3090)
3090 / 3090 = 1 (the remainder is 0, so 3090 is a divisor of 3090)

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 3090 (i.e. 55.587768438749). 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 $\frac{N}{d} = D$ )

3090 / 1 = 3090 (the remainder is 0, so 1 and 3090 are divisors of 3090)
3090 / 2 = 1545 (the remainder is 0, so 2 and 1545 are divisors of 3090)
3090 / 3 = 1030 (the remainder is 0, so 3 and 1030 are divisors of 3090)
...
3090 / 54 = 57.222222222222 (the remainder is 12, so 54 is not a divisor of 3090)
3090 / 55 = 56.181818181818 (the remainder is 10, so 55 is not a divisor of 3090)