What are the divisors of 3645?

1, 3, 5, 9, 15, 27, 45, 81, 135, 243, 405, 729, 1215, 3645

There is a total of 14 positive divisors.
The sum of these divisors is 6558.
The arithmetic mean is 468.42857142857.

14 odd divisors

1, 3, 5, 9, 15, 27, 45, 81, 135, 243, 405, 729, 1215, 3645

How to compute the divisors of 3645?

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 3645 by each of the numbers from 1 to 3645 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)

3645 / 1 = 3645 (the remainder is 0, so 1 is a divisor of 3645)
3645 / 2 = 1822.5 (the remainder is 1, so 2 is not a divisor of 3645)
3645 / 3 = 1215 (the remainder is 0, so 3 is a divisor of 3645)
...
3645 / 3644 = 1.0002744237102 (the remainder is 1, so 3644 is not a divisor of 3645)
3645 / 3645 = 1 (the remainder is 0, so 3645 is a divisor of 3645)

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 3645 (i.e. 60.373835392494). 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 $\frac{N}{d} = D$ )

3645 / 1 = 3645 (the remainder is 0, so 1 and 3645 are divisors of 3645)
3645 / 2 = 1822.5 (the remainder is 1, so 2 is not a divisor of 3645)
3645 / 3 = 1215 (the remainder is 0, so 3 and 1215 are divisors of 3645)
...
3645 / 59 = 61.779661016949 (the remainder is 46, so 59 is not a divisor of 3645)
3645 / 60 = 60.75 (the remainder is 45, so 60 is not a divisor of 3645)