What are the divisors of 3618?

1, 2, 3, 6, 9, 18, 27, 54, 67, 134, 201, 402, 603, 1206, 1809, 3618

There is a total of 16 positive divisors.
The sum of these divisors is 8160.
The arithmetic mean is 510.

8 even divisors

2, 6, 18, 54, 134, 402, 1206, 3618

8 odd divisors

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

How to compute the divisors of 3618?

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 3618 by each of the numbers from 1 to 3618 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)

3618 / 1 = 3618 (the remainder is 0, so 1 is a divisor of 3618)
3618 / 2 = 1809 (the remainder is 0, so 2 is a divisor of 3618)
3618 / 3 = 1206 (the remainder is 0, so 3 is a divisor of 3618)
...
3618 / 3617 = 1.0002764722145 (the remainder is 1, so 3617 is not a divisor of 3618)
3618 / 3618 = 1 (the remainder is 0, so 3618 is a divisor of 3618)

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 3618 (i.e. 60.14981296729). 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 $\frac{N}{d} = D$ )

3618 / 1 = 3618 (the remainder is 0, so 1 and 3618 are divisors of 3618)
3618 / 2 = 1809 (the remainder is 0, so 2 and 1809 are divisors of 3618)
3618 / 3 = 1206 (the remainder is 0, so 3 and 1206 are divisors of 3618)
...
3618 / 59 = 61.322033898305 (the remainder is 19, so 59 is not a divisor of 3618)
3618 / 60 = 60.3 (the remainder is 18, so 60 is not a divisor of 3618)