What are the divisors of 5391?

1, 3, 9, 599, 1797, 5391

There is a total of 6 positive divisors.
The sum of these divisors is 7800.
The arithmetic mean is 1300.

6 odd divisors

1, 3, 9, 599, 1797, 5391

How to compute the divisors of 5391?

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 5391 by each of the numbers from 1 to 5391 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)

5391 / 1 = 5391 (the remainder is 0, so 1 is a divisor of 5391)
5391 / 2 = 2695.5 (the remainder is 1, so 2 is not a divisor of 5391)
5391 / 3 = 1797 (the remainder is 0, so 3 is a divisor of 5391)
...
5391 / 5390 = 1.000185528757 (the remainder is 1, so 5390 is not a divisor of 5391)
5391 / 5391 = 1 (the remainder is 0, so 5391 is a divisor of 5391)

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 5391 (i.e. 73.423429503123). 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 $\frac{N}{d} = D$ )

5391 / 1 = 5391 (the remainder is 0, so 1 and 5391 are divisors of 5391)
5391 / 2 = 2695.5 (the remainder is 1, so 2 is not a divisor of 5391)
5391 / 3 = 1797 (the remainder is 0, so 3 and 1797 are divisors of 5391)
...
5391 / 72 = 74.875 (the remainder is 63, so 72 is not a divisor of 5391)
5391 / 73 = 73.849315068493 (the remainder is 62, so 73 is not a divisor of 5391)