What are the divisors of 1854?

1, 2, 3, 6, 9, 18, 103, 206, 309, 618, 927, 1854

There is a total of 12 positive divisors.
The sum of these divisors is 4056.
The arithmetic mean is 338.

6 even divisors

2, 6, 18, 206, 618, 1854

6 odd divisors

1, 3, 9, 103, 309, 927

How to compute the divisors of 1854?

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 1854 by each of the numbers from 1 to 1854 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)

1854 / 1 = 1854 (the remainder is 0, so 1 is a divisor of 1854)
1854 / 2 = 927 (the remainder is 0, so 2 is a divisor of 1854)
1854 / 3 = 618 (the remainder is 0, so 3 is a divisor of 1854)
...
1854 / 1853 = 1.0005396654074 (the remainder is 1, so 1853 is not a divisor of 1854)
1854 / 1854 = 1 (the remainder is 0, so 1854 is a divisor of 1854)

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 1854 (i.e. 43.058100283222). 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 $\frac{N}{d} = D$ )

1854 / 1 = 1854 (the remainder is 0, so 1 and 1854 are divisors of 1854)
1854 / 2 = 927 (the remainder is 0, so 2 and 927 are divisors of 1854)
1854 / 3 = 618 (the remainder is 0, so 3 and 618 are divisors of 1854)
...
1854 / 42 = 44.142857142857 (the remainder is 6, so 42 is not a divisor of 1854)
1854 / 43 = 43.116279069767 (the remainder is 5, so 43 is not a divisor of 1854)