What are the divisors of 2610?

1, 2, 3, 5, 6, 9, 10, 15, 18, 29, 30, 45, 58, 87, 90, 145, 174, 261, 290, 435, 522, 870, 1305, 2610

There is a total of 24 positive divisors.
The sum of these divisors is 7020.
The arithmetic mean is 292.5.

12 even divisors

2, 6, 10, 18, 30, 58, 90, 174, 290, 522, 870, 2610

12 odd divisors

1, 3, 5, 9, 15, 29, 45, 87, 145, 261, 435, 1305

How to compute the divisors of 2610?

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 2610 by each of the numbers from 1 to 2610 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)

2610 / 1 = 2610 (the remainder is 0, so 1 is a divisor of 2610)
2610 / 2 = 1305 (the remainder is 0, so 2 is a divisor of 2610)
2610 / 3 = 870 (the remainder is 0, so 3 is a divisor of 2610)
...
2610 / 2609 = 1.0003832886163 (the remainder is 1, so 2609 is not a divisor of 2610)
2610 / 2610 = 1 (the remainder is 0, so 2610 is a divisor of 2610)

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 2610 (i.e. 51.088159097779). 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 $\frac{N}{d} = D$ )

2610 / 1 = 2610 (the remainder is 0, so 1 and 2610 are divisors of 2610)
2610 / 2 = 1305 (the remainder is 0, so 2 and 1305 are divisors of 2610)
2610 / 3 = 870 (the remainder is 0, so 3 and 870 are divisors of 2610)
...
2610 / 50 = 52.2 (the remainder is 10, so 50 is not a divisor of 2610)
2610 / 51 = 51.176470588235 (the remainder is 9, so 51 is not a divisor of 2610)