What are the divisors of 5409?

1, 3, 9, 601, 1803, 5409

There is a total of 6 positive divisors.
The sum of these divisors is 7826.
The arithmetic mean is 1304.3333333333.

6 odd divisors

1, 3, 9, 601, 1803, 5409

How to compute the divisors of 5409?

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 5409 by each of the numbers from 1 to 5409 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)

5409 / 1 = 5409 (the remainder is 0, so 1 is a divisor of 5409)
5409 / 2 = 2704.5 (the remainder is 1, so 2 is not a divisor of 5409)
5409 / 3 = 1803 (the remainder is 0, so 3 is a divisor of 5409)
...
5409 / 5408 = 1.0001849112426 (the remainder is 1, so 5408 is not a divisor of 5409)
5409 / 5409 = 1 (the remainder is 0, so 5409 is a divisor of 5409)

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 5409 (i.e. 73.545904032788). 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 $\frac{N}{d} = D$ )

5409 / 1 = 5409 (the remainder is 0, so 1 and 5409 are divisors of 5409)
5409 / 2 = 2704.5 (the remainder is 1, so 2 is not a divisor of 5409)
5409 / 3 = 1803 (the remainder is 0, so 3 and 1803 are divisors of 5409)
...
5409 / 72 = 75.125 (the remainder is 9, so 72 is not a divisor of 5409)
5409 / 73 = 74.095890410959 (the remainder is 7, so 73 is not a divisor of 5409)