What are the divisors of 5445?

1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 121, 165, 363, 495, 605, 1089, 1815, 5445

There is a total of 18 positive divisors.
The sum of these divisors is 10374.
The arithmetic mean is 576.33333333333.

18 odd divisors

1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 121, 165, 363, 495, 605, 1089, 1815, 5445

How to compute the divisors of 5445?

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 5445 by each of the numbers from 1 to 5445 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)

5445 / 1 = 5445 (the remainder is 0, so 1 is a divisor of 5445)
5445 / 2 = 2722.5 (the remainder is 1, so 2 is not a divisor of 5445)
5445 / 3 = 1815 (the remainder is 0, so 3 is a divisor of 5445)
...
5445 / 5444 = 1.0001836884644 (the remainder is 1, so 5444 is not a divisor of 5445)
5445 / 5445 = 1 (the remainder is 0, so 5445 is a divisor of 5445)

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 5445 (i.e. 73.790243257493). 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 $\frac{N}{d} = D$ )

5445 / 1 = 5445 (the remainder is 0, so 1 and 5445 are divisors of 5445)
5445 / 2 = 2722.5 (the remainder is 1, so 2 is not a divisor of 5445)
5445 / 3 = 1815 (the remainder is 0, so 3 and 1815 are divisors of 5445)
...
5445 / 72 = 75.625 (the remainder is 45, so 72 is not a divisor of 5445)
5445 / 73 = 74.58904109589 (the remainder is 43, so 73 is not a divisor of 5445)