What are the divisors of 2709?

1, 3, 7, 9, 21, 43, 63, 129, 301, 387, 903, 2709

There is a total of 12 positive divisors.
The sum of these divisors is 4576.
The arithmetic mean is 381.33333333333.

12 odd divisors

1, 3, 7, 9, 21, 43, 63, 129, 301, 387, 903, 2709

How to compute the divisors of 2709?

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 2709 by each of the numbers from 1 to 2709 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)

2709 / 1 = 2709 (the remainder is 0, so 1 is a divisor of 2709)
2709 / 2 = 1354.5 (the remainder is 1, so 2 is not a divisor of 2709)
2709 / 3 = 903 (the remainder is 0, so 3 is a divisor of 2709)
...
2709 / 2708 = 1.0003692762186 (the remainder is 1, so 2708 is not a divisor of 2709)
2709 / 2709 = 1 (the remainder is 0, so 2709 is a divisor of 2709)

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 2709 (i.e. 52.048054718692). 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 $\frac{N}{d} = D$ )

2709 / 1 = 2709 (the remainder is 0, so 1 and 2709 are divisors of 2709)
2709 / 2 = 1354.5 (the remainder is 1, so 2 is not a divisor of 2709)
2709 / 3 = 903 (the remainder is 0, so 3 and 903 are divisors of 2709)
...
2709 / 51 = 53.117647058824 (the remainder is 6, so 51 is not a divisor of 2709)
2709 / 52 = 52.096153846154 (the remainder is 5, so 52 is not a divisor of 2709)