What are the divisors of 1806?

1, 2, 3, 6, 7, 14, 21, 42, 43, 86, 129, 258, 301, 602, 903, 1806

There is a total of 16 positive divisors.
The sum of these divisors is 4224.
The arithmetic mean is 264.

8 even divisors

2, 6, 14, 42, 86, 258, 602, 1806

8 odd divisors

1, 3, 7, 21, 43, 129, 301, 903

How to compute the divisors of 1806?

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 1806 by each of the numbers from 1 to 1806 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)

1806 / 1 = 1806 (the remainder is 0, so 1 is a divisor of 1806)
1806 / 2 = 903 (the remainder is 0, so 2 is a divisor of 1806)
1806 / 3 = 602 (the remainder is 0, so 3 is a divisor of 1806)
...
1806 / 1805 = 1.0005540166205 (the remainder is 1, so 1805 is not a divisor of 1806)
1806 / 1806 = 1 (the remainder is 0, so 1806 is a divisor of 1806)

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 1806 (i.e. 42.497058721752). 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 $\frac{N}{d} = D$ )

1806 / 1 = 1806 (the remainder is 0, so 1 and 1806 are divisors of 1806)
1806 / 2 = 903 (the remainder is 0, so 2 and 903 are divisors of 1806)
1806 / 3 = 602 (the remainder is 0, so 3 and 602 are divisors of 1806)
...
1806 / 41 = 44.048780487805 (the remainder is 2, so 41 is not a divisor of 1806)
1806 / 42 = 43 (the remainder is 0, so 42 and 43 are divisors of 1806)