What are the divisors of 1809?

1, 3, 9, 27, 67, 201, 603, 1809

There is a total of 8 positive divisors.
The sum of these divisors is 2720.
The arithmetic mean is 340.

8 odd divisors

1, 3, 9, 27, 67, 201, 603, 1809

How to compute the divisors of 1809?

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 1809 by each of the numbers from 1 to 1809 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)

1809 / 1 = 1809 (the remainder is 0, so 1 is a divisor of 1809)
1809 / 2 = 904.5 (the remainder is 1, so 2 is not a divisor of 1809)
1809 / 3 = 603 (the remainder is 0, so 3 is a divisor of 1809)
...
1809 / 1808 = 1.0005530973451 (the remainder is 1, so 1808 is not a divisor of 1809)
1809 / 1809 = 1 (the remainder is 0, so 1809 is a divisor of 1809)

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 1809 (i.e. 42.532340636273). 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 $\frac{N}{d} = D$ )

1809 / 1 = 1809 (the remainder is 0, so 1 and 1809 are divisors of 1809)
1809 / 2 = 904.5 (the remainder is 1, so 2 is not a divisor of 1809)
1809 / 3 = 603 (the remainder is 0, so 3 and 603 are divisors of 1809)
...
1809 / 41 = 44.121951219512 (the remainder is 5, so 41 is not a divisor of 1809)
1809 / 42 = 43.071428571429 (the remainder is 3, so 42 is not a divisor of 1809)