What are the divisors of 1830?

1, 2, 3, 5, 6, 10, 15, 30, 61, 122, 183, 305, 366, 610, 915, 1830

There is a total of 16 positive divisors.
The sum of these divisors is 4464.
The arithmetic mean is 279.

8 even divisors

2, 6, 10, 30, 122, 366, 610, 1830

8 odd divisors

1, 3, 5, 15, 61, 183, 305, 915

How to compute the divisors of 1830?

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 1830 by each of the numbers from 1 to 1830 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)

1830 / 1 = 1830 (the remainder is 0, so 1 is a divisor of 1830)
1830 / 2 = 915 (the remainder is 0, so 2 is a divisor of 1830)
1830 / 3 = 610 (the remainder is 0, so 3 is a divisor of 1830)
...
1830 / 1829 = 1.0005467468562 (the remainder is 1, so 1829 is not a divisor of 1830)
1830 / 1830 = 1 (the remainder is 0, so 1830 is a divisor of 1830)

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 1830 (i.e. 42.778499272415). 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 $\frac{N}{d} = D$ )

1830 / 1 = 1830 (the remainder is 0, so 1 and 1830 are divisors of 1830)
1830 / 2 = 915 (the remainder is 0, so 2 and 915 are divisors of 1830)
1830 / 3 = 610 (the remainder is 0, so 3 and 610 are divisors of 1830)
...
1830 / 41 = 44.634146341463 (the remainder is 26, so 41 is not a divisor of 1830)
1830 / 42 = 43.571428571429 (the remainder is 24, so 42 is not a divisor of 1830)