What are the divisors of 2080?

1, 2, 4, 5, 8, 10, 13, 16, 20, 26, 32, 40, 52, 65, 80, 104, 130, 160, 208, 260, 416, 520, 1040, 2080

There is a total of 24 positive divisors.
The sum of these divisors is 5292.
The arithmetic mean is 220.5.

20 even divisors

2, 4, 8, 10, 16, 20, 26, 32, 40, 52, 80, 104, 130, 160, 208, 260, 416, 520, 1040, 2080

4 odd divisors

1, 5, 13, 65

How to compute the divisors of 2080?

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 2080 by each of the numbers from 1 to 2080 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)

2080 / 1 = 2080 (the remainder is 0, so 1 is a divisor of 2080)
2080 / 2 = 1040 (the remainder is 0, so 2 is a divisor of 2080)
2080 / 3 = 693.33333333333 (the remainder is 1, so 3 is not a divisor of 2080)
...
2080 / 2079 = 1.000481000481 (the remainder is 1, so 2079 is not a divisor of 2080)
2080 / 2080 = 1 (the remainder is 0, so 2080 is a divisor of 2080)

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 2080 (i.e. 45.607017003966). 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 $\frac{N}{d} = D$ )

2080 / 1 = 2080 (the remainder is 0, so 1 and 2080 are divisors of 2080)
2080 / 2 = 1040 (the remainder is 0, so 2 and 1040 are divisors of 2080)
2080 / 3 = 693.33333333333 (the remainder is 1, so 3 is not a divisor of 2080)
...
2080 / 44 = 47.272727272727 (the remainder is 12, so 44 is not a divisor of 2080)
2080 / 45 = 46.222222222222 (the remainder is 10, so 45 is not a divisor of 2080)