What are the divisors of 1818?

1, 2, 3, 6, 9, 18, 101, 202, 303, 606, 909, 1818

There is a total of 12 positive divisors.
The sum of these divisors is 3978.
The arithmetic mean is 331.5.

6 even divisors

2, 6, 18, 202, 606, 1818

6 odd divisors

1, 3, 9, 101, 303, 909

How to compute the divisors of 1818?

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 1818 by each of the numbers from 1 to 1818 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)

1818 / 1 = 1818 (the remainder is 0, so 1 is a divisor of 1818)
1818 / 2 = 909 (the remainder is 0, so 2 is a divisor of 1818)
1818 / 3 = 606 (the remainder is 0, so 3 is a divisor of 1818)
...
1818 / 1817 = 1.0005503577325 (the remainder is 1, so 1817 is not a divisor of 1818)
1818 / 1818 = 1 (the remainder is 0, so 1818 is a divisor of 1818)

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 1818 (i.e. 42.638011210656). 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 $\frac{N}{d} = D$ )

1818 / 1 = 1818 (the remainder is 0, so 1 and 1818 are divisors of 1818)
1818 / 2 = 909 (the remainder is 0, so 2 and 909 are divisors of 1818)
1818 / 3 = 606 (the remainder is 0, so 3 and 606 are divisors of 1818)
...
1818 / 41 = 44.341463414634 (the remainder is 14, so 41 is not a divisor of 1818)
1818 / 42 = 43.285714285714 (the remainder is 12, so 42 is not a divisor of 1818)