What are the divisors of 2580?

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 43, 60, 86, 129, 172, 215, 258, 430, 516, 645, 860, 1290, 2580

There is a total of 24 positive divisors.
The sum of these divisors is 7392.
The arithmetic mean is 308.

16 even divisors

2, 4, 6, 10, 12, 20, 30, 60, 86, 172, 258, 430, 516, 860, 1290, 2580

8 odd divisors

1, 3, 5, 15, 43, 129, 215, 645

How to compute the divisors of 2580?

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 2580 by each of the numbers from 1 to 2580 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)

2580 / 1 = 2580 (the remainder is 0, so 1 is a divisor of 2580)
2580 / 2 = 1290 (the remainder is 0, so 2 is a divisor of 2580)
2580 / 3 = 860 (the remainder is 0, so 3 is a divisor of 2580)
...
2580 / 2579 = 1.0003877471888 (the remainder is 1, so 2579 is not a divisor of 2580)
2580 / 2580 = 1 (the remainder is 0, so 2580 is a divisor of 2580)

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 2580 (i.e. 50.793700396801). 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 $\frac{N}{d} = D$ )

2580 / 1 = 2580 (the remainder is 0, so 1 and 2580 are divisors of 2580)
2580 / 2 = 1290 (the remainder is 0, so 2 and 1290 are divisors of 2580)
2580 / 3 = 860 (the remainder is 0, so 3 and 860 are divisors of 2580)
...
2580 / 49 = 52.65306122449 (the remainder is 32, so 49 is not a divisor of 2580)
2580 / 50 = 51.6 (the remainder is 30, so 50 is not a divisor of 2580)