What are the divisors of 5439?

1, 3, 7, 21, 37, 49, 111, 147, 259, 777, 1813, 5439

There is a total of 12 positive divisors.
The sum of these divisors is 8664.
The arithmetic mean is 722.

12 odd divisors

1, 3, 7, 21, 37, 49, 111, 147, 259, 777, 1813, 5439

How to compute the divisors of 5439?

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 5439 by each of the numbers from 1 to 5439 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)

5439 / 1 = 5439 (the remainder is 0, so 1 is a divisor of 5439)
5439 / 2 = 2719.5 (the remainder is 1, so 2 is not a divisor of 5439)
5439 / 3 = 1813 (the remainder is 0, so 3 is a divisor of 5439)
...
5439 / 5438 = 1.0001838911364 (the remainder is 1, so 5438 is not a divisor of 5439)
5439 / 5439 = 1 (the remainder is 0, so 5439 is a divisor of 5439)

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 5439 (i.e. 73.749576269969). 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 $\frac{N}{d} = D$ )

5439 / 1 = 5439 (the remainder is 0, so 1 and 5439 are divisors of 5439)
5439 / 2 = 2719.5 (the remainder is 1, so 2 is not a divisor of 5439)
5439 / 3 = 1813 (the remainder is 0, so 3 and 1813 are divisors of 5439)
...
5439 / 72 = 75.541666666667 (the remainder is 39, so 72 is not a divisor of 5439)
5439 / 73 = 74.506849315068 (the remainder is 37, so 73 is not a divisor of 5439)