What are the divisors of 3080?

1, 2, 4, 5, 7, 8, 10, 11, 14, 20, 22, 28, 35, 40, 44, 55, 56, 70, 77, 88, 110, 140, 154, 220, 280, 308, 385, 440, 616, 770, 1540, 3080

There is a total of 32 positive divisors.
The sum of these divisors is 8640.
The arithmetic mean is 270.

24 even divisors

2, 4, 8, 10, 14, 20, 22, 28, 40, 44, 56, 70, 88, 110, 140, 154, 220, 280, 308, 440, 616, 770, 1540, 3080

8 odd divisors

1, 5, 7, 11, 35, 55, 77, 385

How to compute the divisors of 3080?

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 3080 by each of the numbers from 1 to 3080 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)

3080 / 1 = 3080 (the remainder is 0, so 1 is a divisor of 3080)
3080 / 2 = 1540 (the remainder is 0, so 2 is a divisor of 3080)
3080 / 3 = 1026.6666666667 (the remainder is 2, so 3 is not a divisor of 3080)
...
3080 / 3079 = 1.000324780773 (the remainder is 1, so 3079 is not a divisor of 3080)
3080 / 3080 = 1 (the remainder is 0, so 3080 is a divisor of 3080)

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 3080 (i.e. 55.497747702046). 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 $\frac{N}{d} = D$ )

3080 / 1 = 3080 (the remainder is 0, so 1 and 3080 are divisors of 3080)
3080 / 2 = 1540 (the remainder is 0, so 2 and 1540 are divisors of 3080)
3080 / 3 = 1026.6666666667 (the remainder is 2, so 3 is not a divisor of 3080)
...
3080 / 54 = 57.037037037037 (the remainder is 2, so 54 is not a divisor of 3080)
3080 / 55 = 56 (the remainder is 0, so 55 and 56 are divisors of 3080)